Chapter 25 Trigonometrical Ratios, Functions & Identities PDF

60 2 Trigonometrical Ratios, Functions and Identities 1.1 Definitions. (1) Angle : The motion of any revolving line in a plane from its initial position (initial side) to the final position (terminal side) is called angle. The end point O about which the line rotates is called the vertex of the angl...

60 2 Trigonometrical Ratios, Functions and Identities 1.1 Definitions. (1) Angle : The motion of any revolving line in a plane from its initial position (initial side) to the final position (terminal side) is called angle. The end point O about which the line rotates is called the vertex of the angle. E3 B Terminal side (2) Measure of an angle : The measure of an angle is the amount of O rotation from the initial side to the terminal side. Initial side A ID (3) Sense of an angle : The sense of an angle is determined by the direction of rotation of the initial side into the terminal side. The sense of an angle is said to be U positive or negative according as the initial side rotates in anticlockwise or clockwise direction to get the terminal side. B  A D YG O O Positive angle  A B Negative angle (4) Right angle : If the revolving ray starting from its initial position to final position describes one quarter of a circle. Then we say that the measure of the angle formed is a right angle. ST U (5) Quadrants : Let X ' OX and YOY ' be two lines at right angles in the plane of the paper. These lines divide the plane of paper into four equal parts. Which are Y known as quadrants. The lines X ' OX and YOY ' are known as x-axis and y-axis. These two lines taken together are known as the coII I quadrant quadrant ordinate axes. (6) Angle in standard position : An angle is said to be in standard position if its vertex concides with the origin O and the initial side concides with OX i.e., the positive direction of x-axis. X X O III quadrant IV quadrant Y (7) Angle in a quadrant : An angle is said to be in a particular quadrant if the terminal side of the angle in standard position lies in that quadrant. (8) Quadrant angle : An angle is said to be a quadrant angle if the terminal side concides with one of the axes. 1.2 System of Measurement of Angles There are three system for measuring angles Trigonometrical Ratios, Functions and Identities 3 (1) Sexagesimal or English system : Here a right angle is divided into 90 equal parts known as degrees. Each degree is divided into 60 equal parts called minutes and each minute is further 1 right angle = 90 degree ( 90 o ) divided into 60 equal parts called seconds. Therefore, 1o  60 minutes ( 60 ' ) 1'  60 second ( 60 ' ' ) 60 (2) Centesimal or French system : It is also known as French system, here a right angle is divided into 100 equal parts called grades and each grade is divided into 100 equal parts, called minutes and each minute is further divided into 100 seconds. Therefore, E3 1 right angle = 100 grades ( 100 g ) 1 grade = 100 minutes ( 100 ' ) 1 minute = 100 seconds ( 100 ' ' ) ID (3) Circular system : In this system the unit of measurement is radian. One radian, written as c 1 , is the measure of an angle subtended at the centre of a circle by an arc of length equal to the radius of the circle. U P A D YG O Consider a circle of radius r having centre at O. Let A be a point on the circle. Now cut off an arc AP whose length is equal to the radius r of the circle. Then by the definition the measure of AOP is 1 radian ( 1c ). 1.3 Relation between Three Systems of Measurement of an Angle. U Let D be the number of degrees, R be the number of radians and G be the number of grades in an angle . 90 o = 1 right angle  Do  ST Now, D right angles 90  1o  1 right angle 90   D right angles 90 ……..(i) Again,  ……..(ii)  radians = 2 right angles R radians  2R  right angles  1 radian    2R  2  right angles right angles 4 Trigonometrical Ratios, Functions and Identities and 100 grades = 1 right angle  G grades   1 grade  G right angles 100   1 right angle 100 G right angles 100 ……..(iii) 60 D G 2R   90 100 π From (i), (ii) and (iii) we get, This is the required relation between the three systems of measurement of an angle. :  One radian  180 o    radians  180 o  1 radian = 57o 1 7 44.8   57 o1 7 4 5 . E3 Note ID 1.4 Relation between an Arc and an Angle. If s is the length of an arc of a circle of radius r, then the angle  (in radians) subtended by this arc at the centre of the circle is given by   U × angle in radians s or s  rθ i.e., arc = radius r B  O C r s A Sectorial area : Let OAB be a sector having central angle  C and radius r. 1 2 r θ. 2 D YG Then area of the sector OAB is given by Important Tips  The angle between two consecutive digits in a clock is 30o (= /6 radians). The hour hand rotates through an angle of 30o in one hour. The minute hand rotate through an angle of 6o in one minute. U  Example: 1 The circular wire of radius 7 cm is cut and bend again into an arc of a circle of radius 12 cm. The angle subtended by an arc at the centre of the circle is [Kerala (Engg.) 2002] ST (a) 50 o Solution: (b) Given the diameter of circular wire = 14 cm. Therefore length of wire = 14  cm Hence, required angle  Example: 2 (b) 22o (c) 23o c (d) 24o We have,  radians  180 o o  Example: 3 7 180 o Arc 4 7     210 o. radian  Radius 12 6  6  2  The degree measure corresponding to the given radian    15  (a) 21o Solution: (d) (d) 60 o (c) 100 o (b) 210 o  180  1c    ;    c o  2   2 180  o        24.  15 15     The angles of a quadrilateral are in A.P. and the greatest angle is 120o, the angles in radians are Trigonometrical Ratios, Functions and Identities 5 (a) 3 , 9 , 9 , (b) 3   2 3 3 , 2 , 3 , (c) 3 5 8 11 12 (d) None of these , , , 18 18 18 18 Let the angles in degrees be   3 ,    ,    ,   3 Sum of the angles  4  360 o  Also greatest angle    3  120 o , Hence, 3  120 o    120 o  90 o  30 o    10 o   90 o Hence the angles are 90 o  30 o ,90 o  10 o ,90 o  10 o and 90 o  30 o That is, the angles in degrees are 60 o , 80 o , 100 o and 120 o  In terms of radians the angles are 60   180 , 80  10  3 (b) 20  3 (c) that is  4  5 3 , 9 , 9 and 30  3 40  3 (d) 20  20   20 cm  cm. 60 3 U The angle subtended at the centre of radius 3 metres by the arc of length 1 metre is equal to [UPSEAT 1973] (b) 60o (c) 1/3 radian (d) 3 radian D YG (a) 20o Solution: (c)  180 We know that the tip of the minute hand makes one complete round in one hour i.e. 60 minutes since the length of the hand is 10 cm. the distance moved by its tip in 60 minutes  2  10 cm  20 cm Hence the distance in 20 minutes  Example: 5 and 120  The minute hand of a clock is 10 cm long. How far does the tip of the hand move in 20 minutes (a) Solution: (b)  180 ID Example: 4 , 100  E3 2. 3  180 60 Solution: (a)  4 5 2 Required angle = Arc 1  radian. radius 3 1.5 Trigonometrical Ratios or Functions. U In the right angled triangle OMP, we have base = OM = x, perpendicular =PM = y and hypotenues = OP =r. We define the following trigonometric ratio which are also known as trigonometric function. Perpendicu lar y Base x cos   sin     Y Hypotenues Hypotenues r r A Perpendicu lar y  Base x cot   r Hypotenues  x Base cosec   ST tan   sec   Base x  , Perpendicu lar y P(x, y) r Hypotenues r  Perpendicu lar y (1) Relation between trigonometric ratio (function) (i) sin .cosec  1 (ii) tan . cot   1 cos  sin  (iii) cos . sec  1 (iv) tan   (v) cot   sin  cos  (2) Fundamental trigonometric identities (i) sin 2   cos 2   1 (ii) 1  tan 2   sec 2  Important Tips y  O x (iii) 1  cot 2   cosec 2 M X 6 Trigonometrical Ratios, Functions and Identities  If x  sec  + tan  , then 1  sec   tan . x  If x  coesc   cot  , then 1  cosec   cot . x (iii) In third quadrant : x  0, y  0  sin   y x y r  0, cos    0, tan    0, cosec    0 , r r x y x r  0 and cot    0. Thus, in the third quadrant all trigonometric functions are y x U sec   ID E3 60 (3) Sign of trigonometrical ratios or functions : Their signs depends on the quadrant in which the terminal side of the angle lies. r y x y (i) In first quadrant : x  0, y  0  sin    0, cos    0, tan    0, cosec    0 , y r r x x r sec    0 and cot    0. Thus, in the first quadrant all trigonometric functions are y x positive. (ii) In second quadrant : y x y r x r x  0, y  0  sin    0, cos    0, tan    0, cosec    0, sec    0 and cot    0. r r y x y x Thus, in the second quadrant sin and cosec function are positive and all others are negative. (iv) D YG negative except tangent and cotangent. In fourth quadrant : x  0, y  0  sin   y  0, r r r y x cos   0, tan    0, cosec    0 , sec    0 and y x x r x cot    0 Thus, in the fourth quadrant all trigonometric y Y II quadrant S X’ functions are negative except cos and sec. U In brief : A crude aid to memorise the signs of x < 0, y > 0 sin and cosec are positive O III quadrant T x < 0, y < 0 tan and cot are positive Y’ ST trigonometrical ratio in different quadrant. "Add Sugar To Coffee". I quadrant A x > 0, y > 0 All are positive IV quadrant C x > 0, y < 0 cos and sec are positive X Important Tips  First determine the sign of the trigonometric function.  If  is measured from X OX i.e., {(  , 2 – )} then retain the original name of the function.  3      , then change sine to cosine, If  is measured from Y OY i.e.,    , 2 2  Y cosine to sine, tangent to cotangent, cot to tan, sec to cosec and cosec to sec. (4) Variations in values of trigonometric functions in different quadrants : Let X ' OX and YOY ' be the coordinate axes. Draw a circle with centre at origin O and radius unity. B (0,1) M (x, y) X (–1, 0) A  O x Xy N B(0, – 1) Y A (1, 0) X Trigonometrical Ratios, Functions and Identities 7 Let M (x , y ) be a point on the circle such that AOM   then x  cos  and y  sin  ; I-Quadrant (A) sin  decreases from 1 to 0 sin  increases from 0 to 1 cos  decreases from 0 to – 1 cos  decreases from 1 to 0 tan  increases from –  to 0 tan  increases from 0 to  cot  decreases from 0 to –  cot  decreases from  to 0 sec  increases from –  to – 1 sec  increases from 1 to  cosec increases from 1 to  cosec decreases from  to 1 ID E3 II-Quadrant (S) IV-Quadrant (C) sin  decreases from 0 to – 1 sin  increases from – 1 to 0 cos  increases from – 1 to 0 cos  increases from 0 to 1 U III-Quadrant (T) tan  increases from 0 to  D YG tan  increases from –  to 0 cot  decreases from  to 0 cot  decreases from 0 to –  sec  decreases from – 1 to – sec  decreases from  to 1 cosec increases from –  to –1 cosec decreases from – 1 to –   and –  are two symbols. These are not real number. When we say that tan  U Note :  60  1  cos  1 and  1  sin   1 for all values of .  2 it means that tan  increases in the interval ST increases from 0 to  for as  varies from 0 to   . Similarly for other trigonometric  0,  and it attains large positive values as  tends to 2  2 functions. Example: 6 If sin   cosec   2 , then sin2   cosec 2  (a) 1 (b) 4 [UPSEAT 2002; MP PET 1992; MNR 1990] (c) 2 Solution: (c) (sin   cosec  )  (sin   cosec  )  2 sin .cosec   2  2  2. Example: 7 If sin   cos   m and sec   cosec   n , then n(m  1)(m  1) equal to 2 (a) m Solution: (c) 2 2 (b) n n(m  1)  (sec   cosec  ).2 sin . cos  2 (d) None of these 2 (c) 2m [m  1  2 sin . cos  ] 2 [MP PET 1986] (d) 2n 8 Trigonometrical Ratios, Functions and Identities sin   cos .2 sin . cos   2m. sin . cos   (a) Solution: (b) x sin  y sin  and tan   , then x / y equal to 1  x cos  1  y cos  sin  sin  (b) (c) tan  sin   sin   cos  tan  sin  cos   cos  sin  y Similarly, sin  ; sin  cos   cos  sin  The equation sec 2    x sin  . y sin   cos 2   1  sec 2   4 xy  1  4 xy  (x  y )2  (x  y)2  0 (x  y)2 (1  sin 2  )  (b) 191 U If sin x  cos x  ST 25 17 sin x  cos x  sin 2 x   Example: 13 (c) 80 (d) 194 24 25  tan 4 A  cot 4 A  194. 1 , then tan 2 x is 5 (b) 7 25 [UPSEAT 2003] (c) 25 7 (d) 24 7 1 1  sin 2 x  cos 2 x  2 sin x cos x  25 5 7 24 24  cos 2 x    tan 2 x . 25 7 25 If sin x   (a) 24 , then the value of tan x is 25 (b) 24 7 2 Solution: (b) [Kerala (Engg.) 2002] tan A  cot A  4  tan 2 A  cot 2 A  2 tan A cot A  16 (a) Solution: (d) (d) sec . tan  If tan A  cot A  4 , then tan 4 A  cot 4 A is equal to  tan 2 A  cot 2 A  14  tan 4 A  cot 4 A  2  196 Example: 12 (d) None of these 1  sin   sec   tan . cos  (a) 110 Solution: (d) (c) sec   tan  (b) 1 (1  sin  ) 2 sin  1  cos  ( x , y  R) U 1  sin  equals 1  sin  D YG Example: 11 ID (b) x  y (a) 0 Solution: (c) (c) x  y (a) x  y Which is possible only when x  y Example: 10 (d) 4 xy is only possible when [MP PET 1986; IIT 1996; Karnataka CET 1997; AMU (x  y)2 1987, 1991] Solution: (a) sin  1  cos  x sin   tan   x cos . tan  x Example: 9 sin  sin  [MP PET 1991] 60 If tan   E3 Example: 8 [UPSEAT 2003] (c) 25 24 7 sin x 24   24  cos x  1  sin 2 x  1     tan x .   25 25 cos x 7   (d) None of these Trigonometrical Ratios, Functions and Identities 9 If tan   sec   e x , then cos  equals (a) Solution: (b) (e x  e  x ) 2 (b) [AMU 2002] 2 (e  e  x )........(ii) 2  , if x   n 0 (a) xyz  xz  y From s    cos 2n , y   sin 2n , z  1 1   cos 2   sin 2   1  x  y  xy ; From (i), xyz = x + y + z. x y If P  cos  2 sin  and Q  , then 1  sin   cos  1  sin ......(i) D YG U Also, PQ (d) Both (b) and (c) 1 1 1 1 1 1 xy   , y , z   xy  1 1  cos 2  sin 2  1  sin 2  cos 2  1  cos 2  sin 2  1  1 xy (a) PQ  1 Solution: (d) (c) xyz  x  y  z (b) xyz  xy  z  xyz  z  xy  xyz = xy + z Example: 16  sin 2n  ,then 2n n 0 n 0 a 1r We get, x =  cos E3  2. e x  e x ID For 0    (e x  e  x ) (e x  e  x )........(i) x From (i) and (ii),  2 sec   e x  e  x  cos   Solution: (d) (d) tan   sec   e x  sec   tan   e Example: 15 (e x  e  x ) 2 (c) x 60 Example: 14 (b) Q 1 P (c) Q  P  1 [MP PET 2001] (d) Q  P  1 2 sin  cos   1  sin   cos  1  sin  After solving, P  Q  1. Example: 17 The value of 6(sin 6   cos 6  )  9(sin 4   cos 4  )  4 equals to (a) – 3 (c) 1 (d) 3 = 6(sin6   cos 6  )  9(sin4   cos 4  )  4 U Solution: (c) (b) 0 [MP PET 2001, 1997] = 6[(sin2   cos 2  )3  3 sin 2  cos 2  (sin 2   cos 2  )]  9[(sin2   cos 2  )2  2 sin 2 . cos 2  ]  4 ST = 6[1  3 sin 2  cos 2  ]  9[1  2 sin 2  cos 2  ]  4 = 6  9  4  1. Example: 18 sin  cos   equals to 1  cot  1  tan  (a) 0 Solution: (d) (b) 1 [Karnataka CET 1998] (c) cos   sin  (d) cos   sin  cos 2   sin 2  sin 2  cos 2  sin . sin  cos . cos    cos   sin . = =  (sin   cos  ) (cos   sin  ) (cos   sin  ) sin  (1  cot  ) (1  tan  ) cos  1.6 Trigonometrical Ratios of Allied Angles. Two angles are said to be allied when their sum or difference is either zero or a multiple of 90. o 10 Trigonometrical Ratios, Functions and Identities (1) Trigonometric ratios of (–): Let a revolving ray starting from its initial position OX , trace out an angle XOA  . Let P(x, y) be a point on OA such that OP = r. Draw PM  from P on x-axis. Angle XOA '   in the clockwise sense. Let P' be a point on OA ' such that OP '  OP. Clearly M and M  coincide and OMP is congruent to OMP ' then P' are (x, – y). y y x y sin(  )     tan    sin  ; cos( )   cos  ; tan(  )  r x r r Taking the reciprocal of these trigonometric ratios; cosec (  )   cosec  , sec(  )  sec  and cot(  )   cot  Y A P(x, y) r  X – M O r Note : E3 60 P (x, – y) A  A function f (x ) is said to be an even function if f ( x )  f (x ) for all x in its domain.  A function f (x ) is said to be an odd function if f ( x )   f (x ) for all x in its domain.  sin  , tan  , cot  , cosec are odd functions and cos  , sec  are even functions. ID (2) Trigonometric function of (90 –  ) : Let the revolving line, starting from OA, trace out any acute angle AOP, equal to . From any point P, draw PM  to OA. Three angles of a triangle are together equal to two right angles, and since OMP is a right angle, the sum of the two angles P 90o–  90 U D YG U o A MOP and OPM is right angle. OPM  90 o   O M [When the angle OPM is consider, the line PM is the ‘base’ and MO is the ‘perpendicular’] MO PM cos(90 o   )  cos MPO  sin( 90 o   )  sin MPO   cos AOP  cos  ,  sin AOP  sin  PO PO MO PM tan( 90 o   )  tan MPO   cot AOP  cot  , cot(90 o   )  cot MPO   tan AOP  tan  PM MO PO PO cosec(90 o   )  cosec MPO   sec AOP  sec  , sec(90 o   )  sec MPO   cosec AOP  cosec MO PM (3) Trigonometric function of (90+ ) : Let a revolving ray OA starting from its initial poisiton OX, trace out an angle XOA   and let another revolving Y ray O A starting from the same initial position OX, first A A trace out an angle . So as to coincide with OA and then it ST revolves through an angle of 90 o in anticlockwise direction to form an angle XOA '  90 o  . X Let P and P' be points on OA and OA ' respectively such that OP  OP '  r. Draw perpendicular PM and PM ' from P and P' respectively on OX. Let the coordinates of P be (x, y). Then OM  x and PM  y clearly, OM '  PM  y and P ' M '  OM  x. (–y, x)   M O  P(x, y) M X Y So the coordinates of P ' are –y, x M ' P' x OM '  y sin( 90   )    cos  , cos(90   )     sin  OP ' r OP ' r M ' P' x x tan( 90   )      cot  , cot(90   )   tan  , sec(90   )  cosec , cosec(90   )  sec OM '  y y Trigonometrical Ratios, Functions and Identities 11 or Trigo. Ratio (180  θ) or ( 90  θ) or π   θ 2  (180  θ) (270  θ) or or ( π  θ)  3π  (π  θ) π   θ 2   θ   2  (270  θ) or  3π  θ   2  ( 360  θ) or (2 π  θ) 60 (90 – ) (θ) Allied angles cos cos  sin – sin – cos – cos  – sin cos cos sin – sin – cos – cos – sin sin cos tan  – tan cot – cot – tan tan cot – cot – tan E3 – sin sin Important Tips  sin n   0, cos n   (1)  sin(n    )  (1)n sin  , cos( n    )  (1)n cos    n  sin     (1) 2   n cos  , if n is odd ID n 1 2 = (1)n / 2 sin  , if n is even n 1 2 sin  , if n is odd U   n      (1) cos   2   (–1)n / 2 cos  , if n is even  sin cos tan D YG 1.7 Trigonometrical Ratios for Various Angles. 0 0 /6 1/2 /4 1/ 2 1 3 /2 0 1/ 3 1/ 2 1 /3 /2  3 /2 1/2 1 0 3/2 –1 2 0 0 –1 0 1 3  0  0 1.8 Trigonometrical Ratios in terms of Each other U sin sin ST sin cos tan sec cos sin  1  cos 2  cos  cos  2 1  sin 2  sin  1 1  sin  1 sin  2 cosec 1  cos 2  1  sin 2  1  sin  cot  cos 1  cos  2 tan tan  1  cot  1 cot  1  tan  2 tan 1 tan  1  tan 2  1 1  tan 2  tan  1  cos  sec 1  tan  2 1 cos  2 cot 1 2 1  cot  1 cot  2 cot 1  cot 2  cot  1  cot 2  sec 2   1 sec  1 sec  sec 2   1 1 cosec 1 co sec cosec 2  1 cosec 1 cosec 2  1 cosec 2  1 sec   1 2 sec sec  sec   1 2 cosec cosec 2  1 cosec 12 Trigonometrical Ratios, Functions and Identities Important Tips Values for some standard angles ; cos 15 o  sin 75 o  5 1 ; 4 cos 36 o  sin 54 o  sin 18 o  cos 72 o  2 2 , 2 sin 22 1o 1o  cos 67  2 2 tan 22 1o 1o  cot 67  2 1 2 2 Example: 19 cos 22 2 3 2 5 1 ; 4 1o 1o  sin 67  2 2 3 1 (b) tan 15 o  cot 75 o  2  3 ; ; tan 75 o  cot 15 o  2  3 1o 1o 2 2  tan 67  ; cot 22 2 2 2 sin 75 o = (a) Solution: (b) 3 1 2 2 2 2 1 3 1 2 2 3 1 1 3 1   . 2 2 2 2 2 5 is 4 D YG 1 (b) 2 2 Example: 21 tan A  cot(180 o  A)  cot(90 o  A)  cot(360 o  A) equal to (a) 0 (b) 2 tan A tan A  cot A  ( tan A)  ( cot A)  0. Example: 22 The value of cos 15 o  sin 15 o equal to U Solution: (a) 1 (a) ST 2 2  3 1 2 2 = 1 2 (b) 2 3 1 Solution: (a) [MP PET 1990] (c) 0 (d) 1 5 5   1 1  sin   cos  sin    0. 4 4 4 4 2 2 cos 1 2 [MP PET 1992] (c) 2 cot A (d) 2(tan A  cot A) [UPSEAT 1975; MP PET 1994; MP PET (c) 1 2 (d) Zero.      3        sin 4 3     2 sin6      sin6 5    3 sin 4   2  2      (a) 0 Solution: (b)  [MNR 1979] U The value of cos A  sin A , when A  Solution: (c) Example: 23 (d) 2 2 sin 75 o  sin(45 o  30 o )  sin 45 o cos 30 o  sin 30 o cos 45 o  (a) 2002] 3 1 (c)  2 Example: 20 2 1 E3 2 2 60 3 1 sin 15 o  cos 75 o  ID  (b) 1 [IIT 1986] (c) 3 (d) sin 4  sin 6 = 3[( cos  )  ( sin  ) ]  2[ cos   sin  ] = 3[(cos 2   sin 2  )2  2 sin 2  cos 2  ]  2[(cos 2   sin 2  )3  3 cos 2  sin 2  (cos 2   sin 2  )] = 3  6 sin 2  cos 2   2  6 sin 2  cos 2  = 1. 4 4 6 6 Trigonometrical Ratios, Functions and Identities 13 Trick : Put   0, Example: 24  2 ; then the value of expression remains constant i.e., it is independent of . Which of the following number is rational (a) sin 15 o Solution: (c) [IIT 1998] (c) sin 15 o. cos 15 o (b) cos 15 o sin 15 o = sin(45 o  30 o ) = 3 1 = irrational  cos 15 o  cos( 45 o  30 o )  sin 15 o. cos 15 o  2 2 = irrational 60 2 2 3 1 (d) sin 15 o. cos 75 o 1 1 1 (2 sin 15 o cos 15 o )  sin 30 o  = rational 2 2 4 2 If sin x  sin 2 x  1 , then the value of cos 12 x  3 cos 10 x  3 cos 8 x  cos 6 x  2 is (a) 0 Solution: (c) (c)  1 (b) 1 Since sin x  sin 2 x  1  sin x  1  sin 2 x  cos 2 x [MP PET 2001] (d) 2........(i) ID Example: 25 E3  3 1   4  2 3  irrational. sin 15 o. cos 75 o  sin 15 o. sin 15 o  sin 2 15 o =   2 2  8   From given expression, cos 6 x (cos 6 x  3 cos 4 x  3 cos 2 x  1) – 2 = cos 6 x (cos 2 x  1)3  2 From (i) sin x  cos 2 x If 4 sin   3 cos  then (a) Solution: (b) sec 2  4[1  tan 2  ] equals to D YG Example: 26 U  sin 3 x (sin x  1)3  2 = (sin 2 x  sin x )3  2  1  2  1. 25 16 (b) Given 4 sin   3 cos   tan   25 28 (c) 1 4 (d) 1 3 4 U 9 1 sec 2  1  tan 2  25 16  The given expression is =.  2 2 9 4 [1  tan  ] 4 (1  tan  )   28 4 1   16   1.9 Formulae for the Trigonometric Ratios of Sum and Differences of Two Angles. (2) sin( A  B)  sin A cos B  cos A sin B (3) cos( A  B)  cos A cos B  sin A sin B (4) cos( A  B)  cos A cos B  sin A sin B ST (1) sin( A  B)  sin A cos B  cos A sin B (5) tan( A  B)  tan A  tan B 1  tan A tan B (6) tan( A  B)  tan A  tan B 1  tan A tan B (7) cot( A  B)  cot A cot B  1 cot A  cot B (8) cot( A  B)  cot A cot B  1 cot B  cot A (9) sin( A  B). sin( A  B)  sin 2 A  sin 2 B  cos 2 B  cos 2 A (10) cos( A  B). cos( A  B)  cos 2 A  sin 2 B  cos 2 B  sin 2 A (11) tan A  tan B  sin A sin B sin A cos B  cos A sin B sin( A  B)    cos A cos B cos A cos B cos A. cos B     A  n  , B  m   2   14 Trigonometrical Ratios, Functions and Identities    A  n , B  m    2  sin( B  A) sin A. sin B (12) cot A  cot B  1.10 Formulae for the Trigonometric Ratios of Sum and Differences of Three Angles. (1) sin( A  B  C)  sin A cos B cos C  cos A sin B cos C  cos A cos B sin C  sin A sin B sin C 60 or sin ( A  B  C)  cos A cos B cos C(tan A  tan B  tan C  tan A. tan B. tan C) (2) cos( A  B  C)  cos A cos B cos C  sin A sin B cos C  sin A cos B sin C  cos A sin B sin C (3) tan( A  B  C)  tan A  tan B  tan C  tan A tan B tan C 1  tan A tan B  tan B tan C  tan C tan A cot A cot B cot C  cot A  cot B  cot C cot A cot B  cot B cot C  cot C. cot A  1 ID (4) cot( A  B  C)  E3 cos( A  B  C)  cos A cos B cos C(1  tan A tan B  tan B tan C  tan C tan A) In general; (5) sin( A1  A 2 ......  A n ) = cos A1 cos A 2..... cos A n (S 1  S 3  S 5  S 7 ...) Where; angles. S 1  S 3  S 5  S 7 .... 1  S 2  S 4  S 6 .... D YG (7) tan( A1  A 2 .....  A n )  U (6) cos( A1  A 2 ....  A n )  cos A1 cos A 2... cos A n (1  S 2  S 4  S 6....) S 1  tan A1  tan A 2 ....  tan A n = The sum of the tangents of the separate S 2  tan A1 tan A 2  tan A1 tan A 3 .... = The sum of the tangents taken two at a time. so on. U S 3  tan A1 tan A2 tan A3  tan A2 tan A3 tan A4 ... = Sum of tangents three at a time, and If A1  A 2 ....  A n  A, then S 1  n tan A , S 2  n C 2 tan 2 A , S 3  n C 3 tan 3 A,.... ST (8) sin nA  cos n A(n C1 tan A  n C 3 tan 3 A  n C 5 tan 5 A ....) (9) cos nA  cos n A(1  n C 2 tan 2 A  n C 4 tan 4 A ...) (10) tan nA  C1 tan A  n C 3 tan 3 A  n C 5 tan 5 A .... 1  n C 2 tan 2 A  n C 4 tan 4 A  n C 6 tan 6 A ... n (11) sin nA  cos nA  cos n A(1 nC1 tan A nC2 tan 2 A nC3 tan 3 A nC4 tan 4 A nC5 tan 5 A nC6 tan 6 A .....) (12) sin nA  cos nA  cos n A(1  n C1 tan A  n C 2 tan 2 A  n C 3 tan 3 A  n C 4 tan 4 A  n C 5 tan 5 A  n C 6 tan 6 A...) (13) sin( )  sin(   )  sin(  2  ) .....  sin(  (n  1) ) = sin{  (n  1) ( / 2)}. sin( n / 2) sin(  / 2) Trigonometrical Ratios, Functions and Identities 15 1.11 Formulae to Transform the Product into Sum or Difference. 60         cos   (n  1) . sin n   2     2  (14) cos( )  cos(   )  cos(  2  ) ....  cos(  (n  1) ) =  sin   2 (1) 2 sin A cos B  sin( A  B)  sin( A  B) (2) 2 cos A sin B  sin( A  B)  sin( A  B) (3) 2 cos A cos B  cos( A  B)  cos( A  B) (4) 2 sin A sin B  cos( A  B)  cos( A  B) Then, A  E3 Let A  B  C and A  B  D CD CD and B  2 2 Therefore, we find out the formulae to transform the sum or difference into product. CD CD cos 2 2 (6) sin C  sin D  2 cos CD CD cos 2 2 (8) cos C  cos D  2 sin U (7) cos C  cos D  2 cos CD CD sin 2 2 ID (5) sin C  sin D  2 sin CD DC CD CD sin  2 sin sin 2 2 2 2 D YG Important Tips 1  sin(60 o   ). sin  sin(60 o   )   o o tan( 60  ). tan  tan( 60  ) tan 3  sin 2 n A cos A. cos 2 A. cos 2 2 A. cos 2 3 A....... cos 2 n1 A  n , if A  n 2 sin A 4  cos(60   ). cos  cos(60 o   )  sin 3 1 4 cos 3 U = 1, if A  2n cos 12 o  sin 12 o ST Example: 27 cos 12  sin 12 o (a) 2 tan 33 o 1  tan 12 o o  sin 147 o cos 147 o = 1, if A  (2n  1)  [MP PET 1991] (b) 1 = Example: 28 If sin  1  sin  2  sin  3  3 , then cos  1  cos  2  cos  3  (a) 3 Solution: (d) (b) 2 (c) 1 (d) 0 We know | sin  |  1 ; So, each 1 ,  2 and  3 must be equal to  / 2  Example: 29 (d) 0  tan 147 o = tan( 45 o  12)  tan(180 o  33 o )  tan 33 o  ( tan 33 o )  0. Solution: (d) 1  tan 12 o (c) – 1 cos  1  cos  2  cos  3  0. cos A  cos(240 o  A)  cos(240 o  A)  [MP PET 1991] 16 Trigonometrical Ratios, Functions and Identities (a) cos A Solution: (b) (b) 0 (c) (d) 3 sin A 3 cos A cos A  [2 cos 240 o cos A] = cos A  2( cos 60 o ) cos A   1  = cos A 1  2   0.  2   sin 2 A  sin 2 B  sin A cos A  sin B cos B Example: 31 (c) cot( A  B) (b) tan( A  B) (a) tan( A  B) Solution: (b) 60 [MP PET 1993] 2(sin 2 A  sin 2 B) 2 sin( A  B). sin( A  B) 2 sin( A  B) sin( A  B)  tan( A  B). =  sin 2 A  sin 2 B 2 sin A cos A  2 sin B cos B 2 sin( A  B) cos( A  B) The expression cos 2 ( A  B)  cos 2 B  2 cos( A  B) cos A cos B is (a) Dependent on B (b) Dependent on A and B (c) Dependent on A (d) Independent of A and B cos 2 ( A  B)  cos 2 B  cos( A  B)[cos( A  B)  cos( A  B)]  ID Solution: (c) (d) cot( A  B) E3 Example: 30 cos 2 B  cos( A  B) cos( A  B)  cos 2 B  (cos 2 A  sin 2 B)  1  cos 2 A Trick : Put A  90 o and 0o the value is sin 2 B  cos 2 B  1 and 0 again put B  0 o , 90o and the value is If tan   (a) Solution: (b) m 1 and tan   then     2m  1 m 1 D YG Example: 32 U sin 2 A and sin 2 A means expression depends on A.  3 (b) We have tan    4 [IIT 1978] (c) U 2m 2  2m  1 2m  2m  1 ST Hence     2 (d) None of these m 1 and tan   2m  1 m 1 m 1  2m 2  m  m  1 m  1 2 m 1  tan(   )  m 1 2m 2  m  2m  1  m 1. (m  1) (2m  1)   6  1  tan(   )  tan  tan   tan    tan(   )   1  tan  tan     4  4 Trick : As    is independent of m, therefore put m  1, then tan   1 1   Therefore, tan(   )  2 3  1 , Hence     1 4 1 6 Example: 33 Solution: (d) 1 1 and tan  . 2 3 (Also check for other values of m) If tan   cot   a and sin   cos   b , then (b 2  1) 2 (a 2  4 )  (a) 2 (b)  4 Given that tan   cot   a …….(i) [WB JEE 1979] (c)  4 and sin   cos   b …….(ii) (d) 4 Trigonometrical Ratios, Functions and Identities 17 Now, (b 2  1) 2 (a 2  4 )  {(sin   cos  ) 2  1} 2 {tan   cot  ) 2  4} 1   1 4  [1  sin 2  1]2 [tan 2   cot 2   2  4 ]  sin 2 2 (cosec 2  sec 2  )  4 sin 2  cos 2   2  2   sin  cos   Trick : Obviously the value of expression (b 2  1) 2 (a 2  4 ) is independent of  , therefore put any If sin B = (a) Solution: (c) tan( A  B) 1 sin(2 A  B) , then  5 tan A 5 3 (b) 2 3 (c) (d) 1 (b) 3 (c) [Karnataka CET 1986; MP PET sin 70 o  cos 40 o sin 60 o 3 2 sin 70 o  sin 50 o 2 sin 60 o cos 10 o .  3. = =  o o o o o o 2 1 sin 30 o cos 70  sin 40 sin 20  sin 40 2 sin 30 cos(10 ) Example: 36 sin 47 o  sin 61o  sin 11o  sin 25 o  D YG Solution: (c) 2001] (a) sin 36 o (b) sin 7 o (c) cos 36 o MP PET (d) cos 7 o 5 1 5 1  cos 7 o.. 4 4 U cos 10 o  sin 10 o  cos 10 o  sin 10 o (b) cot 55 o (c)  tan 35 o (d)  cot 35 o cos 10 o  sin 10 o 1  tan 10 o   tan 35 o  tan( 90 o  35 o ) = cot 55 o. cos 10 o  sin 10 o 1  tan 10 o ST Example: 38 2003; sin 47 o  sin 61o  (sin 11o  sin 25 o ) = 2 sin 54 o. cos 7 o  2 sin 18 o cos 7 o (a) tan 55 o Solution: (b) 1 2 [EAMCET = 2 cos 7 o (sin 54 o  sin 18 o ) = 2 cos 7o.2 cos 36 o. sin 18 o = 4. cos 7 o. Example: 37 (d) 3 U (a) 1 ID sin 70 o  cos 40 o  cos 70 o  sin 40 o 1999] Solution: (d) 3 5 sin(2 A  B) 5 sin(2 A  B)  sin B 5  1  by componendo and Dividendo.  sin B 1 sin(2 A  B)  sin B 5  1 tan( A  B) 3 2 sin( A  B). cos A 6 .   tan A 2 2 cos( A  B). sin A 4 Example: 35 3 2 E3 Example: 34 60 suitable value of . Let   45 o , we get a  0 , b  2 so that [( 2 )2  1]2 (0 2  4 )  4. If tan( A  B)  p and tan( A  B)  q then the value of tan 2 A  (a) pq pq (b) pq 1  pq Solution: (d) 2 A  {( A  B)  ( A  B)}  tan 2 A  Example: 39 sin 163 o cos 347 o  sin 73 o sin 167 o  (a) 0 (b) (c) 1  pq 1 p (d) pq 1  pq tan( A  B)  tan( A  B) pq  tan 2 A  1  tan( A  B). tan( A  B) 1  pq [MP PET 2000] 1 2 (c) 1 (d) None of these Solution: (b) sin(90 o  73 o ). cos(360 o  13 o )  sin 73 o. sin(180 o  13 o ) = cos 73 o. cos 13 o  sin 73 o. sin 13 o  cos(73 o  13 o )  cos 60 o  1. Example: 40 The value of cot 70 o  4 cos 70 o is 2 18 Trigonometrical Ratios, Functions and Identities 1 (a) cot 70 o  4 cos 70 o   6 (b)  4 (d)  2 The value of ID 2 x  2.2 x  x  2 x  1    tan(   )  1  tan . 4 4 1  2 x  2.2 x  2.2 x  x  2 x tan 70 o  tan 20 o  tan 50 o U tan(   )  (a) 1 (b) 2 [Karnataka CET 2003] (c) 3 (d) 0 sin 70 o sin 20 o sin 70 o cos 20 o  cos 70 o sin 20 o  o o 2 sin(70 o  20 o ) cos 50 o 2 sin 50 o. cos 50 o cos 70 cos 20  cos 70 o. cos 20 o   = 2 cos 70 o. cos 20 o. sin 50 o sin 50 o sin 50 o 2 cos 70 o cos 20 o. sin 50 o o o cos 50 cos 50 = D YG Solution: (b)  3 1  1 1  2 x 1 1 x tan   tan  2 tan(   )   tan(   )  1 1 1  tan  tan  1. 1 1  2 x 1 1 x 2  Example: 42 [AMU 2002] (c) 1 60 If tan   (1  2  x )1 , tan   (1  2 x 1 )1 , then    equals (a) Solution: (b) 1 2 cos 70 o  2 sin 140 o cos 70 o  4 sin 70 o. cos 70 o  o sin 70 sin 70 o sin 80 o  sin 40 o 2 sin 60 o cos 20 o 2 sin 30 o cos 10 o  sin 40 o    3. sin 70 o sin 70 o sin 70 o  (d) cos 70 o  2 sin(180  40 o ) sin 20 o  sin 40 o  sin 40 o  sin 70 o sin 70 o  Example: 41 (c) 2 3 3 E3 Solution: (b) (b) 3 2 cos 50 o 2 cos 50 o  2. cos 90 o  cos 50 o 0  cos 50 o 1.12 Trigonometric Ratio of Multiple of an Angle. 2 tan A 1  tan 2 A U (1) sin 2 A  2 sin A cos A  ST (2) cos 2 A  2 cos 2 A  1  1  2 sin 2 A  cos 2 A  sin 2 A  (3) tan 2 A  2 tan A 1  tan 2 A 1  tan 2 A  ; where A  (2n  1). 2 1  tan A 4 (4) sin 3 A  3 sin A  4 sin 3 A  4 sin( 60 o  A). sin A. sin( 60 o  A) (5) cos 3 A  4 cos 3 A  3 cos A  4 cos(60 o  A). cos A. cos(60 o  A) (6) tan 3 A  3 tan A  tan 3 A  tan( 60 o  A). tan A. tan( 60 o  A) , where A  n   / 6 1  3 tan 2 A (7) sin 4  4 sin . cos 3   4 cos  sin 3  (8) cos 4  8 cos 4   8 cos 2   1 (9) tan 4  4 tan   4 tan 3  1  6 tan 2   tan 4  (10) sin 5 A  16 sin 5 A  20 sin 3 A  5 sin A (11) cos 5 A  16 cos 5 A  20 cos 3 A  5 cos A Trigonometrical Ratios, Functions and Identities 19 1.13 Trigonometric Ratio of Sub-multiple of an Angle. 3  A A A A , If 2n    / 4  A / 2  2n    cos  1  sin A or sin  cos   1  sin A i.e.,  4 2 2 2 2   , otherwise (2) sin 5  A A A A , If 2n    / 4  A / 2  2n    cos  1  sin A or (sin  cos )   1  sin A i.e.,  4 2 2 2 2   , otherwise 60 (1) sin (ii) cot A 1  cos A 1  cos A , where A  2n   2 1  cos A sin A E3 A  tan 2 A  1  1 1  cos A 1  cos A    , where A  (2n  1) 2 tan A 1  cos A sin A (3) (i) tan A + 2 cos A sin – 2 cos 3 4  sin A is +ve 2 A is+ve 2 sin A A is –ve + cos 2 2 A – cos A is 2 2 +ve sin 5 4 A A + cos is +ve 2 2 sin sin A – cos A is –ve 2 2 A A + cos is –ve 2 2 A – cos 2 U 3 7 4 A is – 2 ve 2 A 1  cos A ; where A  (2n  1)  2 1  cos A ST (4) tan 2  4 D YG sin sin A lies or you can 2 U  2 follow the following figure, ID The ambiguities of signs are removed by locating the quadrants in which (5) cot 2 A 1  cos A ; where A  2n  2 1  cos A Important Tips n   (1)n A. 2  Any formula that gives the value of sin A 2  Any formula that gives the value of cos A 2n   A in terms of cos A shall also give the value of cos of. 2 2  Any formula that gives the value of tan A n  A in terms of tan A shall also give the value of tan of. 2 2 Example: 43 If sin   in terms of sin A shall also give the value of sine of 3  3 where     , then cos equal to 5 2 2 [MP PET 1998] 20 Trigonometrical Ratios, Functions and 1 (a) (b)  1 3 (c) 10 10 Solution: (d)  3   3  4  cos         cos  ve   cos   2 2 2 2 4 5 2 Example: 44 2 sin 2   4 cos(   ) sin  sin   cos 2(   ) equal to Solution: (c) (c) cos 2 (b) cos 2  2 4 5  9 3 . 10 10 1993; IIT (d) sin 2  Since 2 cos(   )  2 cos (   )  1, 2 sin   1  cos 2    cos 2   2 cos(   )[2 sin  sin   cos(   )] 2 2 (b) 3 2 2 1   3   1  1  3  = 2  1  1  3   1  3  3 3 4 (d) 3 2 3 1  3  1  [ 3  1]2  [ 3  1]2 4 3 3    2 2 2 2 8 [ 3  1 ]  [ 3  1 ]  3 1  3  1  U  1  1 1   1  tan 2 15 1  [tan( 45 o  30 o )]2    2 o o 2 1  tan 15 1  [tan( 45  30 )]  1  1 1    (c) [MP PET 1998] ID 1 2 E3 cot 2 15 o  1 = cot 2 15 o  1 (a) Solution: (b) 1 [UPSEAT   cos 2   2 cos(   ). cos(   )   cos 2   cos 2  cos 2   cos 2. Example: 45 10 1  cos   2 1977] (a) sin 2 3 (d) 10 60 Identities Example: 46 1  tan 2  1  tan 2 15 o 3   cos 30 o . 2 2 o 2 1  tan  1  tan 15 D YG Trick : cos 2  If sin 6  32 cos 5 . sin  32 cos 3  sin  3 x , then x = (b) cos 2 (a) cos  Solution: (d) [EAMCET 2003] (c) sin  (d) sin 2 sin 6  2 sin 3. cos 3 = 2[3 sin   4 sin 3  ][4 cos 3   3 cos  ] = 24 sin . cos  (sin 2   cos 2  )  18 sin  cos   32 sin 3  cos 3  = 32 cos 5 . sin  32 cos 3 . sin  3 sin 2 U On comparing, x  sin 2 Trick : Put   0 o , then x  0. So, option (c) and (d) are correct. ST Now put   30 o , then x  Example: 47 If x 1 3. Therefore, Only option (d) is correct. 2  2 cos  , then x 6  x 6  [Karnataka CET 2003] x (a) 2 cos 6 Solution: (b) Given, x (b) 2 cos 12 1 (c) 2 cos 3  2 cos  (d) 2 sin 3........(i) x On squaring both sides we get, x  x 1  2(2 cos 2   1)  2 cos 2 x Again squaring both sides, 1 1  2  4 cos 2   x   4 cos 2   2 x x........(ii) Trigonometrical Ratios, Functions and Identities 21 x2  1 1 1  2  4 cos 2 2  x 2  2  4 cos 2 2  2  2(2 cos 2 2  1)  x 2  2  2 cos 4 2 x x x......(iii) 3 1  1 1  1   Now taking cube of both sides;  x 2  2   (2 cos 4 )3  x 6  6  3 x 2. 2  x 2  2   8 cos 3 4 x x  x  x   1  3(2 cos 4 ) = 8 cos 3 4 x6  x6  1  2(4 cos 3 4  3 cos 4 )  2 cos 3(4 )  2 cos 12. x6 1 x6  8 cos 3 4  6 cos 4 A is equal to 2 (a)  1  sin A  1  sin A (c) 1  sin A  1  sin A (d) 1  sin A  1  sin A 1  sin A  sin sin A A  cos  0 2 2 A A  cos......(i) and 2 2 Subtract (ii) from (i) we get, 2 cos Example: 49 If 2 tan A  3 tan B, then U ST sin 2 B   5  cos 2 B Example: 50 3 3 tan B  t 2 2     1 2 sin A  2t  If 90 o  A  180 o and sin A  (a) Solution: (d)  2t    1  t2  1  t2 5   2 1  t 4  6t 2  (Let tan B  t ) t 2  3t 2 [AMU 2001] (d) tan( A  2 B) (c) tan( A  B)  sin 2 B  2t 1t 2 , cos 2 B  1  t2 1  t2  tan( A  B). A 4 , then tan is equal to 5 2 (b) 4 4  tan A   , 5 3 A 2 , tan A  2 A 1  tan 2......(ii) A  1  sin A  1  sin A. 2 (b) tan( A  B) 2 tan A  3 tan B  tan A  A A  cos 2 2 sin 2 B is equal to 5  cos 2 B (a) tanA – tanB Solution: (b) 1  sin A  sin D YG Hence, A  66.5 o  2 U For A  133 o , ID (b)  1  sin A  1  sin A Solution: (c) 60 For A = 133 o , 2 cos  x6  E3 Example: 48  x6  2 tan Let tan 3 5 (90 o  A  180 o ) A P 2 [AMU 2001] (c) 3 2 (d) 2 22 Trigonometrical Ratios, Functions and  1 1 4 2P  4 P 2  6 P  4  0  P   , 2  P   (impossible)  2 2 3 1  P2 So, P  2 i.e., tan If tan   1 1 and sin   , then tan(  2  ) is equal to 7 10 (a) 1 Solution: (a) (b) 0 (c) 2 1 1 1 3 3 tan   , sin    tan    tan 2  =  1 3 7 4 10 1 9  2  t, then 1  t2 is equal to 1  t2 (b) sin  (a) cos  Example: 53 1  t2 = 1  tan The value of 1998; IIT 1992, 97] (a)  2 2  ( tan  2 (c) sec  (d) cos 2   t) = cos( 2. )  cos . 2 2 tan x when ever defined never lie between tan 3 x 1 and 3 3 Let, y  3 4 (d) (b) 1 and 4 4 (c) [Haryana CEE 1 and 5 5 (d) 5 and 6 tan x tan x  tan 3 x 3 tan x  tan 3 x 1  3 tan 2 x U Solution: (a) 1  tan 2 D YG Solution: (a) 1  t2 U If tan ID 1 3  7 4  4  21  1  tan(  2  )  3 25 1 28 Example: 52 1 2 E3 Example: 51 A  2. 2 60 Identities ST 1  tan 2 x 1  3 tan 2 x 3 y  1 3  tan 2 x 1  tan 2 x 3 Hence, y should never lie between Example: 54 If tan   t , then tan 2  sec 2 equal to (a) Solution: (a) 1 and 3 whenever defined. 3 1t 1t tan 2  sec 2  (b) 1t 1t [MP PET 1999] (c) 2t 1t 2 tan  1  tan 2   1  tan 2  1  tan 2  Given tan   t   tan 2  sec 2  2t 1  t2 2 t  1  t 2 (t  1)2 1t    =. 2 2 2 2 1t 1t 1t 1t 1t (d) 2t 1t Trigonometrical Ratios, Functions and Identities 23 If sin 2  sin 2  (a) Solution: (b) 1 3 and cos 2  cos 2  , then cos 2 (   ) equal to 2 2 3 8 (b) Given, sin 2  sin 2  1 2 5 8 3 4 (c).......(i) and cos 2  cos 2  3 2 [MP PET 2000].......(ii) 1 9  4 4 Squaring and adding, (sin 2 2  cos 2 2 )  (sin 2 2  cos 2 2 )  2[sin 2. sin 2  cos 2. cos 2 ]  If tan x  (a) Solution: (b) b , then a ab  ab 2 sin x (b) sin 2 x Given tan x  ab equal to ab b  a Now, multiplying by 2 cos x [MP PET 1990, 2002] 2 cos x (c) cos 2 x ab ab 1  b /a 1 b /a =    ab ab 1 b /a 1  b /a 1  tan 2 x in N'r and D'r = (d) sin 2 x 1  tan x 1  tan x   1  tan x 1  tan x ID Example: 56 5 1 1  cos( 2  2 )   cos 2 (   ) . 8 4 4 E3  cos 2. cos 2  sin 2. sin 2  5 4 (d) 60 Example: 55 2 1  tan x 2. 1  tan x 2 1  tan 2 x = 2 cos 2 x. sec x 2 cos x cos 2 x. U 1  tan 2 x cos 2 x 2  2 2 sin x 1.14 Maximum and Minimum Value of a cos + b sin. a  r cos  and b  r sin  D YG Let ……..(i) ……..(ii) Squaring and adding (i) and (ii), then a 2  b 2  r 2 or, r  a 2  b 2  a sin   b cos  = r(sin  cos   cos  sin  ) = r sin(   ) But  1  sin   1 So,  1  sin(   )  1 ; Then  r  r sin(   )  r Hence,  a 2  b 2  a sin   b cos   a 2  b 2 Note :  a 2  b 2 and  a 2  b 2. U Then the greatest and least values of a sin   b cos  are respectively sin 2 x  cosec 2 x  2, for every real x. ST  cos 2 x  sec 2 x  2, for every real x.  tan 2 x  cot 2 x  2 , for every real x. Important Tips Use of  (Sigma) and  (Pie) notation  sin( A  B  C)   sin A cos B cos C   sin A , tan( A  B  C )    tan A   tan A. 1   tan A tan B sin   sin(   )  sin(  2  ) ......... n terms cos( A  B  C )   cos A   cos A sin B sin C , (  denotes summation) (  denotes product) 24 Trigonometrical Ratios, Functions and Identities sin[  (n  1) / 2] sin[n  / 2]  sin( / 2) n 1  nB  sin A  B  sin 2 2  . B sin 2 n or  sin( A  r  1B)  r 1  sin A / 2  cos A / 2  2 sin / 4  A   2 cos A   / 4 .  cos   cos   cos   cos(     )  4 cos  sin   sin   sin   sin(     )  4 sin  tan   2 tan 2  4 tan 4  8 cot 8  cot . 2 sin     2 2. (d) 2 (b) tan 8 A tan 2 A [MP PET 1995] (c) cot 8 A cot 2 A (d) None of these Solution: (b) 2 sin 4. A cos 4 A. sin 4 A sin 8 A. 2 sin 2 A. cos 2 A tan 8 A 1  cos 8 A cos 4 A 2 sin 2 4 A cos 4 A =.  . . tan 2 A cos 8 A 1  cos 4 A cos 8 A 2 sin 2 2 A cos 8 A. 2 sin 2 2 A cos 8 A. 2 sin 2 2 A Example: 59 If tan   sin  cos  a  , then equal to 8 b cos  sin 8  [WB JEE 1986] (a 2  b 2 )4  a b   8  2 2 b8 a  a b (b)  (a 2  b 2 )4  a b   8  2 2 b8 a  a b (c)  (a 2  b 2 )4  a b   8  2 2 b8 a  a b (d)  (a 2  b 2 )4  a b   8  2 2 b8 a  a b U (a)  ST Solution: (a) Given , tan   a / b  cos 2  sin    a a b 2 2 [EAMCET 1994]  xy  yz  zx  22  4 2  22  0 x   , y  2, z  2  ; tan 2 A tan 8 A. (c) 1 x y z     (say) 1 2 2 sec 8 A  1 equal to sec 4 A  1 (a) 2 cos D YG Example: 58   (b) 0 We have  2 2 4  z cos , then xy  yz  zx  3 3 (a) 1 Solution: (b) sin   E3  cos ID If x  y cos 2 U Example: 57  n 1  nB  cos  A  B  sin 2 2  . B sin 2 60  n cos[  (n  1) / 2] sin[n  / 2] cos   cos(   )  cos(  2  ) ......... n terms  or  cos( A  r  1B)  sin[  / 2] r 1 1  tan 2  b 2  a2  2 2 1  tan  b  a2 ; cos    b a  b2 2     a b     2 2 4  2  2 2  2  a(a 2  b 2 )4 b(a 2  b 2 )4  sin   cos    a  b    a  b  = 8 2 =  (a  b )  8 2 2 1 / 2 2 1/2 8 8 8 8 2 b ( a  b ) a ( a  b ) cos  sin   a  b2    b a      2  2 2  2   a b   a b  b .  a  8  8 b a   Trigonometrical Ratios, Functions and Identities 25 Example: 60 The minimum value of 3 cos x  4 sin x  5 = (a) 5 Solution: (d) [UPSEAT 1991] (b) 9 (c) 7 (d) 0 Minimum value of 3 cos x  4 sin x   3 2  4 2   5 Minimum value of 3 cos x  4 sin x  5  5  5  0. The greatest and least value of sin x cos x are Solution: (b) Example: 62 1 1 , 2 2 (b) (c) The value of sin   cos  will be greatest when (a)   30 o  2 sin(  )1   2  4 )  1  sin  2    )  4  4 ) 2    4   2   D YG sin(  4 (d)   90 o ID 4  (c)   60 o U  2 sin(  If f (x ) is maximum then,  4. The maximum value of sin 2 x  3 cos 2 x is (a) 3 Solution: (b) (d) 2, 2 [UPSEAT 1977, 83; RPET (b)   45 o Let f (x )  sin   cos    1  sin(  Example: 63 1 1 , 4 4 1 sin 2 x 1 1 1  . [2 sin x cos x ]  sin 2 x ; 1  sin 2 x  1 ; 2 2 2 2 2 1995] Solution: (b) 60 (a) 1,  1 [UPSEAT 1975] E3 Example: 61 (b) 4 [Karnataka CET 2003] (c) 5 (d) 7 f (x )  4 sin 2 x  3 cos 2 x  sin 2 x  3 and 0 | sin x |  1  Maximum value of sin 2 x  3 cos 2 x is 4. Example: 64 U EAMCET 1994] If A  cos 2   sin 4  , then for all values of  ST (a) 1  A  2 Solution: (d) (b) 13  A 1 16 (c) A  cos 2 x  sin 4 x  cos 2   sin 2  sin 2   A  cos 2   sin 2  Again [  sin 2   1 ]  A 1 A  cos 2   sin 4   (1  sin 2  )  sin 4  3 1 3 3  A   sin 2      2 4 4  Hence, Example: 65 3  A 1. 4 The value of 5 cos   3 cos(   3 [UPSEAT 2001; IIT 1980; Roorkee 1992; )  3 lies between 3 13 A 4 16 (d) 3  A 1 4 26 Trigonometrical Ratios, Functions and Solution: (d) (b)  4 and 6 (a)  4 and 4 5 cos   3 cos(  = [5 cos    3 )  3 = 5 cos   3[cos  cos (c)  4 and 8    sin . sin ]  3 3 3  13  3 3 3 3 3 sin    3 cos   sin  ]  3 =  cos     2 2 2 2   2 2 2          13    3 3    13 cos   3 3 sin    13    3 3       2  2   2   2   2   2   2 cos    13   3 3 3 3 cos   sin    3  7  3   4   sin    3  10    2 2  2   So, the value lies between – 4 and 10. Solution: (d) sin 14 3 5 7 9 11 13. sin. sin. sin. sin. sin is equal to 14 14 14 14 14 14 1 8  14 = sin Example: 67. sin (b). sin  14 (c) 1 32. sin 1 64 2 3 5 5  3     3 5 7  1    . sin  1  sin  . sin. sin. sin . sin   . sin    = sin   64. 14 14 14 14 14 14 14 14 14         a2  b 2 4  a2  b 2 U (b)   2 equal to 4  a2  b 2 a2  b 2 Given that, sin   sin   a........(i) and (c) cos   cos   b a2  b 2 4  a2  b 2......(ii) ST Squaring, sin 2   sin 2   2 sin  sin   a 2 and cos 2   cos 2   2 cos  cos   b 2 Adding, 2  2(sin  sin   cos . cos  )  a 2  b 2 a2  b 2  2  2 cos(   )  a  b  2  cos(   )  2 2 2  (a 2  b 2 )  (a 2  b 2 ) tan 2  (d) 3 sin 5 7 9 11 13.. sin. sin. sin. sin 14 14 14 14 14 14 If sin   sin   a and cos   cos   b then tan (a) Soluton: (b) 1 16 U (a)  D YG sin ID  13   7  3   2 E3  13  3 3 7   cos   sin    7  2  2   Example: 66 (d)  4 and 10 60 Identities   2  2  2 tan 2 4  a2  b 2 (   ) (   )  tan 2  tan  2 2 2 2 a b   2 (   ) a2  b 2  2 2   (   ) 2 1  tan 2 2 1  tan 2  2  2 tan 2 4  a2  b 2 a2  b 2   2 (d) 4  a2  b 2 a2  b 2 Trigonometrical Ratios, Functions and Identities 27  tan  2 ,   0 o , then a  1  b   Again putting      , we get tan   2  0 , which is given by (b). The maximum value of 3 cos   4 sin  equal to (a) 3 Solution: (c) 4 [MP PET 2002; UPSEAT 1990] (b) 4 Maximum value of 3 cos   4 sin  is (c) 5 32  4 2  5. 1.15 Conditional Trigonometrical Identitites. (d) None of these E3 Example: 68  1, which is given by (a) and (b). 2 60 Trick : Put   ID We have certain trigonometric identities. Like, sin 2   cos 2   1 and 1  tan 2   sec 2  etc. Such identities are identities in the sense that they hold for all value of the angles which satisfy the given condition among them and they are called conditional identities. U If A, B, C denote the anlges of a triangle ABC, then the relation A + B + C =  enables us to establish many important identities involving trigonometric ratios of these angles. D YG (1) If A + B + C = , then A + B =  – C, B + C =  – A and C + A =  – B. (2) If A + B + C = , then sin( A  B)  sin(  C)  sin C Similarly, sin( B  C)  sin(  A)  sin A and sin( C  A)  sin(  B)  sin B (3) If A  B  C   , then cos( A  B)  cos(  C)   cos C Similarly, cos(B  C)  cos(  A)   cos A and cos(C  A)  cos(  B)   cos B U (4) If A + B + C = , then tan( A  B)  tan(  C)   tan C Similarly, tan( B  C)  tan(  A)   tan A and tan( C  A)  tan(  B)   tan B ST (5) If A  B  C   , then AB  C BC  A CA  B and and       2 2 2 2 2 2 2 2 2  A B  C  C  A B  C  C sin    sin     cos   , cos    cos     sin   ,  2  2 2 2  2  2 2 2  A B  C  C tan    tan     cot    2  2 2 2 All problems on conditional identities are broadly divided into the following three types 1. Identities involving sine and cosine of the multiple or sub-multiple of the angles involved Working Method 28 Trigonometrical Ratios, Functions and Identities Step (i) : Use C  D formulae. Step (ii) : Use the given relation (A + B + C = ) in the expression obtained in step-(i) such that a factor can be taken common after using multiple angles formulae in the remaining term. Step (iii) : Take the common factor outside. so that we can apply C  D formulae. Step (v) : Find the result according to the given options. 60 Step (iv) : Again use the given relation (A + B + C = ) within the bracket in such a manner E3 2. Identities involving squares of sine and cosine of multiple or sub-multiples of the angles involved Working Method ID Step (i) : Arrange the terms of the identity such that either sin 2 A  sin 2 B  sin( A  B). sin( A  B) or cos 2 A  sin 2 B  cos( A  B). cos( A  B) can be used. U Step (ii) : Take the common factor outside. Step (iii) : Use the given relation (A  B  C   ) within the bracket in such a manner so that we can apply C  D formulae. D YG Step (iv) : Find the result according to the given options. 3. Identities for tangent and cotangent of the angles Working Method Step (i) : Express the sum of the two angles in terms of third angle by using the given relation ( A  B  C   ). Step (ii) : Taking tangent or cotangent of the angles of both the sides. U Step (iii) : Use sum and difference formulae in the left hand side. Step (iv) : Use cross multiplication in the expression obtained in the step (iii). ST Step (v) : Arrange the terms as per the result required. Example: 69 Solution: (a) If A  B  C   , then cos 2 A  cos 2 B  cos 2 C equal to (a) 1  2 sin A sin B cos C (b) 1  2 cos A cos B sin C (c) 1  2 sin A sin B cos C (d) 1  2 cos A cos B sin C cos 2 A  cos 2 B  cos 2 C  cos 2 A  (1  sin 2 B)  cos 2 C  1  [cos 2 A  sin 2 B]  cos 2 C  1  cos( A  B) cos( A  B)  cos 2 C  1  cos   C) cos( A  B)  cos 2 C  1  cos C[cos( A  B)  cos C]  1  cos C[cos( A  B)  cos{  ( A  B)}]  1  cos C[cos( A  B)  cos( A  B)]  1  cos C[2 sin A sin B]  1  2 sin A sin B cos C. Trigonometrical Ratios, Functions and Identities 29 sin 2 A  sin 2 B  sin 2C equal to sin A  sin B  sin C (a) sin A sin B cos C A B C cos cos sin 2 2 2 (b) (c)  2 sin( A  B) cos( A  B)  sin 2C (sin 2 A  sin 2 B)  sin 2C = (sin A  sin B)  sin C  AB  AB 2 sin  cos    sin C  2   2  = = = cos A cos B sin C A B C sin sin cos 2 2 2 2 sin C cos( A  B)  2 sin C cos C C C   C   AB 2 sin  cos    2 sin cos 2 2  2   2  2 sin C[2 cos A cos B] 2 sin C[cos( A  B)  cos( A  B)] cos A cos B sin C = =. A B C C A B   C   A B  A B  sin sin cos 2 cos cos     cos    2 cos  2 sin sin  2 2 2 2  2 2 2  2 2  2 2  ID and sin A  sin B  sin C  4 sin A B C sin cos  sin 2 A  sin 2 B  sin 2C  cos A cos B sin C. 2 2 2 A B C sin A  sin B  sin C sin 2  tan  2  tan  2  tan  2  tan  2  tan  2 tan  2  tan  2  tan  2  2  tan  tan tan 2  2   2  2  2 tan   2 tan  2 sin 2 cos 2 [IIT 1979] (b) tan       tan  tan tan  tan tan  1 2 2 2 2 2 2 (d) None of these 2        tan      tan   0 2  2 2 2 . tan  2. tan  2 0  tan  2  tan  2  tan  2  tan  2. tan  2. tan  2 U If A  B  C   , then cos 2 A  cos 2B  cos 2C equal to (a) 1  4 cos A cos B sin C (b) 1  4 sin A sin B cos C [EAMCET 1982] (c) 1  4 cos A cos B cos C (d) None of these cos 2 A  cos 2B  cos 2C = 2 cos( A  B). cos( A  B)  (2 cos 2 C  1) =  1  2 cos C. cos( A  B)  2 cos 2 C ST Solution: (c) 2   tan We have       2    D YG (c) tan Example: 72  U If       2 , then (a) tan Solution: (a) sin A sin B cos C A B C cos cos sin 2 2 2 C C    sin C  2 sin cos   2 2    A  B ( A  B )   sin C / 2  sin     cos  2  2  2 2 sin C[cos( A  B)  cos C] C   A B  A B  2 cos cos     cos    2  2 2  2 2  Trick :  sin 2 A  sin 2B  sin 2C  4 cos A cos B sin C Example: 71 (d)  60 Solution: (a) cos A cos B sin C A B C sin sin cos 2 2 2 E3 Example: 70 = 1  2 cos c[cos( A  B)  cos( A  B)] = 1  4 cos A. cos B. cos C Example: 73 If A  B  C  180 o , then (a) 8 sin Solution: (b) A B C sin sin 2 2 2 sin 2 A  sin 2 B  sin 2C equal to cos A  cos B  cos C  1 (b) 8 cos A B C cos cos 2 2 2 (c) 8 sin A B C cos cos 2 2 2 (d) 8 cos sin 2 A  sin 2 B  sin 2C 2 sin( A  B). cos( A  B)  2 sin C cos C 2 sin C cos( A  B)  2 cos C sin C =  C AB C cos A  cos B  cos C  1 2 cos A  B cos A  B  1  2 sin 2 C  1 2 sin cos  2 sin 2 2 2 2 2 2 2 = 2 sin C[cos( A  B)  cos( A  B)] 4 sin A sin B sin C = A B C C ( A  B) ( A  B)  4 sin sin sin 2 sin cos  cos  2 2 2 2  2 2  A B C sin sin 2 2 2 30 Trigonometrical Ratios, Functions and Identities  A A B B C C  cos  2 sin cos  2 sin cos 2 2 2 2 2 2  8 cos A cos B cos C. A B C 2 2 2 4 sin sin sin 2 2 2 If A  B  C  180 o , then the value of (cot B  cot C)(cot C  cot A)(cot A  cot B) will be Solutio: (b) cot B  cot C  (b) cosec A cosec B cosec C sin B sin C and cot A  cot B  sin C. sin A sin A sin B Therefore, (cot B  cot C)(cot C  cot A)(cot A  cot B) sin A sin B sin C..  cosec A.cosec B.cosec C. sin B. sin C sin C. sin A sin A sin B (a) 2 cot A  B  C  180 o   (b) 4 cot A B C cot cot 2 2 2 (c) cot A B C cot cot 2 2 2 (d) 8 cot A B C cot cot 2 2 2 A B C   90 o  2 2 2 A B. cot  1 C 1 2 2  tan  B A C 2 cot  cot cot 2 2 2 cot D YG Solution: (c) A B C cot cot 2 2 2 A B C  cot  cot will be 2 2 2 ID If A  B  C  180 o , then the value of cot U Example: 75 (d) 1 sin C cos B  sin B cos C sin(B  C) sin(180 o  A) sin A =   sin B. sin C sin B. sin C sin B. sin C sin B. sin C Similarly, cot C  cot A  = (c) tan A tan B tan C 60 (a) sec A sec B sec C E3 Example: 74 4  2 sin C  A B  cot     cot  90 o   or 2 2 2    A B C B A A B C C B A   or  cot. cot  1  cot  cot  cot ; cot. cot. cot  cot  cot  cot 2 2 2 2 2 2 2 2 2 2 2   Example: 76 If A, B, C are angles of a triangle, then sin 2 A  sin 2B  sin 2C is equal to (a) 4 sin A cos B cos C (c) 4 sin A cos A (d) 4 cos A cos B sin C sin 2 A  sin 2 B  sin 2C  2 sin A cos A  2 cos( B  C) sin(B  C) U Solution: (d) (b) 4 cos A [MP PET 2003] [ A  B  C   , B  C    A, cos(B  C)  cos(  A), cos(B  C)   cos A, sin(B  C)  sin A] ST = 2 cos A[sin A  sin(B  C)] = 2 cos A[sin(B  C )  sin(B  C)] = 2 cos A.2 cos B. sin C  4 cos A. cos B. sin C Trick: First put A  B  C  60 o , for these values. Options (a) and (b) satisfies the condition. Now put A  B  45 o and C  90 o , then only (d) satisfies. Hence (d) is the answer. Example: 77 Solution: (c) In any triangle ABC sin 2 A B C  sin 2  sin 2 is equal to 2 2 2 [MP PET 2003] (a) 1  2 cos A B C cos cos 2 2 2 (b) 1  2 sin A B C cos cos 2 2 2 (c) 1  2 sin A B C sin sin 2 2 2 (d) 1  2 cos A B C cos sin 2 2 2 Trick: For A  B  C  60 o only option (c) satisfies the condition. Trigonometrical Ratios, Functions and Identities 31 Important Tips Method of componendo and dividendo p a  , then by componendo and dividendo q b p q ab  p q ab We can write Example: 78 q p ba  q  p b a If tan   cos . tan  then tan 2 (a) Solution: (a) or sin(   ) sin(   )  2 (b) The given relation is p q ab  p q ab or sin(   ) sin(   ) (c) (d) cos(   ) cos(   )   (d) nm nm sin(   )   tan 2. sin(   ) 2 U 2 sin 2 sin(   ) 2   sin(   ) 2 cos 2  2 If m cos(   )  n cos(   ) , then cot  cot  equal to m n m n (b) m n m n D YG (a) Solution: (c) cos(   ) cos(   ) ID Applying componendo and dividendo rule, then Example: 79 q  p b a.  q p ba equal to tan  1  tan  cos  tan   tan  1  cos    tan   tan  1  cos  or 60 If E3  (c) m n nm m cos(   )  n cos(   ) By componendo and dividendo rule, m n. nm U cot  cos   m n 2 cos  cos  m  n cos(   )  cos(   )    m  n  2 sin  sin  m  n cos(   )  cos(   ) If cosec   ST Example: 80 (a) Solution: (b) pq    , then cot     p q 4 2 p q Given, co sec  (b) q p [EAMCET 2001] (c) (d) pq pq pq 1 pq   , p q sin  p q 2     cos 2  sin 2  1  sin  p q  p q p   Apply componendo and dividendo,     1  sin  p q  p q q  cos  sin  2 2   2  p  p  q  1  tan  / 2  2  2   1  tan  / 2   q  tan  4  2   q  cot  4  2   p       32 Trigonometrical Ratios, Functions and Identities Note :    cot     4 2 q    only if cot     0. p 4 2 ST U D YG U ID E3 60 ***