Trigonometric Functions PDF

11. TRIGONOMETRIC FUNCTIONS TRIGONOMETRIC FUNCTIONS In this chapter, we will learn about trigonometric functions and their graphs, inverse trigonometric functions, trigonometric identities and solving trigonometric equations. Some physical changes such as tides, annual temperatures and phases of th...

11. TRIGONOMETRIC FUNCTIONS TRIGONOMETRIC FUNCTIONS In this chapter, we will learn about trigonometric functions and their graphs, inverse trigonometric functions, trigonometric identities and solving trigonometric equations. Some physical changes such as tides, annual temperatures and phases of the Moon are described as cyclic or periodic because they repeat regularly. Trigonometric functions are also periodic and we can use them to model real-life situations. CHAPTER OUTLINE 11.01 Angles of any magnitude 11.02 Trigonometric identities 11.03 EXT1 Further trigonometric identities 11.04 Radians 11.05 Trigonometric functions 11.06 Trigonometric equations 11.07 Applications of trigonometric functions 11.08 EXT1 Inverse trigonometric functions 11.09  EXT1 Properties of inverse trigonometric functions IN THIS CHAPTER YOU WILL: evaluate trigonometric ratios for angles of any magnitude in degrees and radians use reciprocal trigonometric ratios and trigonometric identities EXT1 manipulate sums, differences and products of trigonometric ratios θ EXT1 derive and use expressions of trigonometric ratios in terms of t where t = tan 2 solve trigonometric equations understand trigonometric functions and sketch their graphs examine practical applications of trigonometric functions EXT1 identify the domain and range over which inverse trigonometric functions exist and draw their graphs EXT1 understand and apply properties of inverse trigonometric functions Shutterstock.com/Jon Beard  TERMINOLOGY amplitude: The height from the centre of period: The length of one cycle of a periodic a periodic function to the maximum or function on the x-axis, before the function minimum values (peaks and troughs of its graph 2π repeats itself. For y = k sin ax the period is respectively) For y = k sin ax the amplitude is k a centre: The mean value of a periodic function that periodic function: A function that repeats itself is equidistant from the maximum and minimum regularly values. For y = k sin ax + c the centre is c phase: A horizontal shift (translation). identity: An equation that shows the equivalence For y = k sin [a(x + b)], the phase is b, that is, the of 2 algebraic expressions for all values of the graph of y = k sin ax shifted b units to the left variables reciprocal trigonometric ratios: The cosecant, inverse trigonometric functions: The sin-1x, secant and cotangent ratios, which are the cos-1x and tan-1x functions, which are the reciprocals of sine, cosine and tangent respectively inverse functions of the sine, cosine and tangent t-formulas: Formulas for sin A, cos A and tan A A functions respectively in terms of t = tan 2 WS 11.01 Angles of any magnitude In Chapter 5, Trigonometry, we examined acute and obtuse obtuse angles by looking at angles Homework Angles of any magnitude turning around a unit circle. We can find angles of any size by continuing around the circle. 1st quadrant: acute angles (between 0° and 90°) y You can see from the triangle in the unit circle with 90° 1st quadrant angle θ that: (x, y) sin θ = y 1 unit y cos θ = x 180° θ 0° x y x 360° tan θ = x In the 1st quadrant, x and y are both positive so all ratios are positive in the 1st quadrant. 270° 2nd quadrant: obtuse angles y (between 90° and 180°) 2nd quadrant 90° sin θ = y (positive) (−x, y) cos θ = −x (negative) y 1 unit 180° − θ y 180° θ 0° x tan θ = (negative) x 360° −x The angle that gives θ in the triangle is 180° − θ. 270° 592 MATHS IN FOCUS 11. Mathematics Extension 1 ISBN 9780170413299  2nd quadrant sin (180° − θ) = sin θ cos (180° − θ) = −cos θ tan (180° − θ) = −tan θ 3rd quadrant: angles between 180° and 270° sin θ = −y (negative) y cos θ = −x (negative) 90° −y y tan θ = = (positive) −x x The angle that gives θ in the triangle is 180° + θ. 180° + θ 180° 0° x x θ 360° 3rd quadrant y 1 unit sin (180° + θ) = −sin θ (−x, −y) cos (180° + θ) = −cos θ 3rd quadrant tan (180° + θ) = tan θ 270° 4th quadrant: angles between 270° and 360° sin θ = −y (negative) y cos θ = x (positive) 90° −y tan θ = (negative) x The angle that gives θ in the triangle is 360° − θ. 180° 0° x θ x 360° − θ 360° y 4th quadrant 1 unit sin (360° − θ) = −sin θ (x, −y) cos (360° − θ) = cos θ 4th quadrant 270° tan (360° − θ) = −tan θ ISBN 9780170413299 11. Trigonometric functions 593  Putting all of these results together gives a rule for all 4 quadrants that we usually call the ASTC rule. ASTC rule A: All ratios are positive in the y 1st quadrant. 90° 2nd quadrant 1st quadrant S: Sin is positive in the 2nd quadrant θ (cos and tan are negative). 180° − θ S A T: Tan is positive in the 3rd quadrant (sin and cos are negative). 180° 0° x 360° C: Cos is positive in the 4th quadrant (sin and tan are negative). T C 180° + θ 360° − θ ASTC can be remembered using the phrase ‘All Stations To Central’. 3rd quadrant 4th quadrant 270° EXAMPLE 1 a Find all quadrants where: i sin θ > 0 ii cos θ < 0 iii tan θ < 0 and cos θ > 0 b Find the exact value of: i tan 330° ii sin 225° c Simplify cos (180° + x). 3 d If sin x = − and cos x > 0, find the value of tan x. 5 Solution a   i Using the ASTC rule, sin θ > 0 in the 1st and 2nd quadrants. ii cos θ > 0 in the 1st and 4th quadrants, so cos θ < 0 in the 2nd and 3rd quadrants. iii tan θ > 0 in the 1st and 3rd quadrants so tan θ < 0 in the 2nd and 4th quadrants. Also cos θ > 0 in the 1st and 4th quadrants. So tan θ < 0 and cos θ > 0 in the 4th quadrant. 594 MATHS IN FOCUS 11. Mathematics Extension 1 ISBN 9780170413299  b i 330° lies in the 4th quadrant. y The angle inside the triangle in the 4th quadrant is 360° − 330° = 30° and tan is negative in the 4th quadrant. tan 330° = −tan 30° x 330° 30° 1 =− 3 ii 225° is in the 3rd quadrant. y The angle in the triangle in the 3rd quadrant is 225° − 180° = 45° and sin is negative in the 3rd quadrant. sin 225° = −sin 45° 225° x 1 45° =− 2 c 180° + x is in the 3rd quadrant where cos x is negative. So cos (180° + x) = −cos x. d sin x < 0 and cos x > 0 so x is in the y 4th quadrant. 3 sin x = − 5 So the opposite side is 3 and the hypotenuse is 5. x x By Pythagoras’ theorem, the adjacent side is 4. (3, 4, 5 triangle). 3 5 tan x < 0 in the 4th quadrant. 3 So tan x = −. 4 ISBN 9780170413299 11. Trigonometric functions 595  We can find trigonometric ratios of angles greater than 360° by turning around the circle more than once. EXAMPLE 2 Find the exact value of cos 510°. y Solution To find cos 510°, we move around the circle more than once. 150° 30° x 510° cos (510° − 360°) = cos (150°) The angle is in the 2nd quadrant where cos is negative. The angle inside the triangle is 180° − 150° = 30°. So cos 510° = cos 150° = −cos 30° 3 =−. 2 Negative angles The ASTC rule also works for negative angles. These are measured in the opposite direction (clockwise) from positive angles as shown. Negative angles y −270° 2nd quadrant 1st quadrant −(360° − θ) −(180° + θ) S A −180° −360° x 0° T C −(180° − θ) −θ 3rd quadrant 4th quadrant −90° 596 MATHS IN FOCUS 11. Mathematics Extension 1 ISBN 9780170413299  In the 4th quadrant: In the 3rd quadrant: sin (-θ) = −sin θ sin (−(180° − θ)) = −sin θ cos (-θ) = cos θ cos (−(180° − θ)) = −cos θ tan (-θ) = −tan θ tan (−(180° − θ)) = tan θ In the 2nd quadrant: In the 1st quadrant: sin (−(180° + θ)) = sin θ sin (−(360° − θ)) = sin θ cos (−(180° + θ)) = −cos θ cos (−(360° − θ)) = cos θ tan (−(180° + θ)) = −tan θ tan (−(360° − θ)) = tan θ EXAMPLE 3 Find the exact value of tan (−120°). Solution Moving clockwise around the circle, the y angle is in the 3rd quadrant, with 180° − 120° = 60° in the triangle. tan is positive in the 3rd quadrant. tan (−120°) = tan 60° = 3 x 60° 120° ISBN 9780170413299 11. Trigonometric functions 597  Exercise 11.01 Angles of any magnitude 1 Find all quadrants where: a cos θ > 0 b tan θ > 0 c sin θ > 0 d tan θ < 0 e sin θ < 0 f cos θ < 0 g sin θ < 0 and tan θ > 0 h cos θ < 0 and tan θ < 0 i cos θ > 0 and tan θ < 0 j sin θ < 0 and tan θ < 0 2 a Which quadrant is the angle 240° in? b Find the exact value of cos 240°. 3 a Which quadrant is the angle 315° in? b Find the exact value of sin 315°. 4 a Which quadrant is the angle 120° in? b Find the exact value of tan 120°. 5 a Which quadrant is the angle −225° in? b Find the exact value of sin (−225°). 6 a Which quadrant is the angle −330° in? b Find the exact value of cos (−330°). 7 Find the exact value of: a tan 225° b cos 315° c tan 300° d sin 150° e cos 120° f sin 210° g cos 330° h tan 150° i sin 300° j cos 135° 8 Find the exact value of: a cos (−225°) b cos (−210°) c tan (−300°) d cos (−150°) e sin (−60°) f tan (−240°) g cos (−300°) h tan (−30°) i cos (−45°) j sin (−135°) 9 Find the exact value of: a cos 570° b tan 420° c sin 480° d cos 660° e sin 690° f tan 600° g sin 495° h cos 405° i tan 675° j sin 390° 3 10 If tan θ = and cos θ < 0, find sin θ and cos θ as fractions. 4 4 11 Given sin θ = and tan θ < 0, find the exact value of cos θ and tan θ. 7 5 12 If sin x < 0 and tan x = − , find the exact value of cos x. 8 598 MATHS IN FOCUS 11. Mathematics Extension 1 ISBN 9780170413299  2 13 Given cos x = and tan x < 0, find the exact value of sin x and tan x. 5 5 14 If cos x < 0 and sin x > 0, find cos x and sin x in surd form if tan x =. 7 4 15 If sin θ = − and 270° < θ < 360°, find the exact value of tan θ and cos θ. 9 3 16 If cos x = − and 180° < x < 270°, find the exact value of tan x and sin x. 8 17 Given sin x = 0.3 and tan x < 0: a express sin x as a fraction b find the exact value of cos x and tan x. 18 If tan α = −1.2 and 270° < α < 360°, find the exact values of cos α and sin α. 19 Given that cos θ = −0.7 and 90° < θ < 180°, find the exact value of sin θ and tan θ. 20 Simplify: a sin (180° − θ) b cos (360° − x) c tan (180° + β) d sin (180° + α) e tan (360° − θ) f sin (−θ) g cos (−α) h tan (−x) 11.02 Trigonometric identities WS The reciprocal trigonometric ratios Homework Trigonometric identities The reciprocal trigonometric ratios are the reciprocals of the sine, cosine and tangent ratios. Simplifying The reciprocal trigonometric ratios trigonometric functions 1 hypotenuse cosec θ can also Cosecant    cosec θ = = sin θ opposite be written as csc θ. 1 hypotenuse Secant           sec θ = = cos θ adjacent 1 adjacent Cotangent      cot θ = = tan θ opposite The reciprocal ratios have the same signs as their related ratios in the different quadrants. For example, in the 3rd and 4th quadrants, sin θ < 0, so cosec θ < 0. ISBN 9780170413299 11. Trigonometric functions 599  EXAMPLE 4 a Find cosec α, sec α and cot α for this triangle. B 2 b If sin θ = − and tan > 0, find the exact ratios of cot θ, 5 7 3 sec θ and cosec θ. c State the quadrants where cosec θ is negative. C α A 4 Solution 1 1 1 a cosec α = sec α = cot α = sin α cos α tan α hypotenuse hypotenuse adjacent = = = opposite adjacent opposite 5 5 4 = = = 3 4 3 b sin θ < 0 and tan θ > 0 in the 3rd quadrant. So cos θ < 0. By Pythagoras’ theorem: tenu se hy p o opposite             72 = a2 + 22 7 2 a2 + 4 = 49 θ              a2 = 45 a adjacent        a = 45            = 3 5 1 1 1 cot θ = sec θ = cosec θ = tan θ cos θ sin θ adjacent hypotenuse hypotenuse = = = opposite adjacent opposite 3 5 7 7 = =− =− 2 3 5 2 7 5 =- 15 c sin θ < 0 in the 3rd and 4th quadrants. So cosec θ < 0 in the 3rd and 4th quadrants. 600 MATHS IN FOCUS 11. Mathematics Extension 1 ISBN 9780170413299 Complementary angles In n ABC if ∠B = θ then ∠A = 90° − θ (by the angle A 90° sum of a triangle). ∠B and ∠A are complementary −θ angles because they add up to 90°. c b b a sin θ = sin (90° − θ) = c c θ a b C a B cos θ = cos (90° − θ) = c c b a tan θ = tan (90° − θ) = a b c c sec θ = sec (90° − θ) = a b c c cosec θ = cosec (90° − θ) = b a a b cot θ = cot (90° − θ) = b a Notice the pairs of trigonometric ratios that are equal: Complementary angle results sin θ = cos (90° − θ) tan θ = cot (90° − θ) sec θ = cosec (90° − θ) cos θ = sin (90° − θ) cot θ = tan (90° − θ) cosec θ = sec (90° − θ) EXAMPLE 5 a Simplify tan 50° − cot 40°. b Find the value of m if sec 55° = cosec (2m − 15)°. Solution a tan 50° − cot 40° = tan 50° − cot (90° − 50°) b sec 55° = cosec (90° − 55°) = tan 50° − tan 50° = cosec 35° =0 So 2m − 15 = 35 2m = 50 m = 25 ISBN 9780170413299 11. Trigonometric functions 601  The tangent identity y In the work on angles of any magnitude, we saw that y sin θ = y, cos θ = x and tan θ =. x From this we get the following trigonometric identities: P(x, y) 1 y The tangent identity θ x x For any value of θ: sin θ cos θ tan θ =   cot θ = cos θ sin θ An identity is an equation that shows the equivalence of 2 algebraic expressions for all values of the variables, for example, a2 - b2 = (a + b)(a - b) is an identity. EXAMPLE 6 Simplify sin θ cot θ. Solution cos θ sin θ cot θ = sin θ × sin θ = cos θ The Pythagorean identities The unit circle above has equation x2 + y2 = 1, because of Pythagoras’ theorem. But sin θ = y and cos θ = x, so We used this identity in Chapter 7 when (cos θ)2 + (sin θ)2 = 1 examining the parametric equations of a circle. A shorter way of writing this is: cos2 θ + sin2 θ = 1 This formula is called a Pythagorean identity because it is based on Pythagoras’ theorem in the unit circle. There are 2 other identities that can be derived from this identity. Dividing each term by cos2 θ: Dividing each term by sin2 θ: cos2 θ sin 2 θ 1 cos2 θ sin 2 θ 1 + = + = cos θ cos θ cos2 θ 2 2 sin θ sin θ sin 2 θ 2 2 1 + tan2 θ = sec2 θ cot2 θ + 1 = cosec2 θ 602 MATHS IN FOCUS 11. Mathematics Extension 1 ISBN 9780170413299  Pythagorean identities For any value of θ: cos2 θ + sin2 θ = 1 1 + tan2 θ = sec2 θ 1 + cot2 θ = cosec2 θ cos2 θ + sin2 θ = 1 can also be rearranged to give: cos2 θ = 1 − sin2 θ or sin2 θ = 1 − cos2 θ EXAMPLE 7 Prove that: 1 − cos x 1 a b cot x + tan x = cosec x sec x        2 = sin x 1 + cos x Solution 1 − cos x a LHS = cot x + tan x b LHS = sin 2 x cos x sin x = + 1 − cos x sin x cos x = 1 − cos2 x cos2 x + sin 2 x 1 − cos x = = sin x cos x (1 + cos x )(1 − cos x ) 1 = 1 sin x cos x = 1 + cos x 1 1 = × = RHS sin x cos x = cosec x sec x = RHS 1 − cos x 1 ∴ 2 = ∴ cot x + tan x = cosec x sec x sin x 1 + cos x ISBN 9780170413299 11. Trigonometric functions 603  Exercise 11.02 Trigonometric identities 1 For this triangle, find the exact ratios of sec x, cot x and cosec x. 5 9 2 If sin θ = , find cosec θ, sec θ and cot θ. 13 x 4 5 3 If cos θ = , find exact values of cosec θ, sec θ and cot θ. 7 6 4 If sec θ = −  and sin θ > 0, find exact values of tan θ, cosec θ and cot θ. 5 5 If cot θ = 0.6 and cosec θ < 0, find the exact values of sin θ, cosec θ, tan θ and sec θ. 6 Show sin 67° = cos 23°. 7 Show sec 82° = cosec 8°. 8 Show tan 48° = cot 42°. 9 Simplify: a cos 61° + sin 29° b sec θ − cosec (90° − θ) sin55° c tan 70° + cot 20° − 2 tan 70° d cos35° cot 25° + tan 65° e cot 25° 10 Find the value of x if sin 80° = cos (90 − x)° 11 Find the value of y if tan 22° = cot (90 − y)° 12 Find the value of p if cos 49° = sin (p + 10)° 13 Find the value of b if sin 35° = cos (b + 30)° 14 Find the value of t if cot (2t + 5)° = tan (3t − 15)° 15 Find the value of k if tan (15 − k)° = cot (2k + 60)° 16 Simplify: a tan θ cos θ b tan θ cosec θ c sec x cot x d 1 − sin2 x e 1 − cos2 α f cot2 x + 1 g 1 + tan2 x h sec2 θ − 1 i 5 cot2 θ + 5 1 j   k sin2 α cosec2 α l cot θ − cot θ cos2 θ cosec 2 x 604 MATHS IN FOCUS 11. Mathematics Extension 1 ISBN 9780170413299 17 Prove that: 1 + sin θ a cos2 x − 1 = −sin2 x b sec θ + tan θ = cos θ 3 c 3 + 3 tan2 α = d sec2 x − tan2 x = cosec2 x − cot2 x 1 − sin 2 α e (sin x − cos x)3 = sin x − cos x − 2 sin2 x cos x + 2 sin x cos2 x 1 − sin 2 θ + 2sin θ f cot θ + 2 sec θ = g cos2 (90° − θ) cot θ = sin θ cos θ sin θ cos θ 1 − sin 2 θ cos2 θ h (cosec x + cot x)(cosec x − cot x) = 1 i 2 = tan2 θ + cos2 θ cos θ EXT1 11.03 Further trigonometric identities WS Sums and differences of angles Homework Further trigonometric There are special formulas or identities for the trigonometric ratios of sums and differences identities of angles. cos (A − B) = cos A cos B + sin A sin B Trigonometric identities Proof Since cos θ = x and sin θ = y in the unit y circle, we can write the coordinates of points (x, y) on the unit circle as (cos θ, sin θ). 1 Let point P have coordinates (cos B, sin B) P(cos B, sin B) and Q have coordinates (cos A, sin A), where Q(cos A, sin A) 1 B and A are the angles of inclination of OP 1 A−B B x and OQ respectively. −1 O 1 Let’s now find the length of PQ2. By the distance formula: −1 2 2 2 d   = (x2 − x1) + (y2 − y1) PQ2 = (cos A − cos B)2 + (sin A − sin B)2 = cos2 A − 2 cos A cos B + cos2 B + sin2 A − 2 sin A sin B + sin2 B = (cos2 A + sin2 A) + (cos2 B + sin2 B) − 2 cos A cos B − 2 sin A sin B = 1 + 1 − 2(cos A cos B + sin A sin B) = 2 − 2(cos A cos B + sin A sin B) ISBN 9780170413299 11. Trigonometric functions 605  By the cosine rule: c2 = a2 + b2 − 2ab cos C PQ2 = 12 + 12 − 2(1)(1) cos (A − B) = 2 − 2 cos (A − B) From and : 2 − 2 cos (A − B) = 2 − 2(cos A cos B + sin A sin B) So cos (A − B) = cos A cos B + sin A sin B cos (A + B) = cos A cos B − sin A sin B Proof Substitute −B for B in the formula cos (A − B) = cos A cos B + sin A sin B: cos (A − (− B)) = cos A cos (−B) + sin A sin (− B) cos (A + B) = cos A cos B + sin A (−sin B)   since cos (−B) = cos B and sin (−B) = −sin B = cos A cos B − sin A sin B sin (A + B) = sin A cos B + cos A sin B Proof Substitute 90° − A for A in the formula cos (A − B) = cos A cos B + sin A sin B: cos (90° − A − B) = cos (90° − A) cos B + sin (90° − A) sin B cos (90° − (A + B)) = sin A cos B + cos A sin B using complementary angle results sin (A + B) = sin A cos B + cos A sin B sin (A − B) = sin A cos B − cos A sin B Proof Substitute −B for B in the formula sin (A + B) = sin A cos B + cos A sin B: sin (A + (−B)) = sin A cos (− B) + cos A sin (− B) = sin A cos B + cos A (− sin B)   since cos (−B) = cos B and sin (−B) = −sin B = sin A cos B − cos A sin B 606 MATHS IN FOCUS 11. Mathematics Extension 1 ISBN 9780170413299  tan A + tan B tan (A + B) = 1 − tan A tan B Proof sin( A + B ) tan (A + B) = cos( A + B ) sin A cos B + cos B sin A = cos A cos B − sin A sin B sin A cos B + cos A sin B = cos A cos B cos A cos B − sin A sin B cos A cos B sin A sin B + = cos A cos B sin A sin B 1− cos A cos B tan A + tan B = 1 − tan A tan B tan A − tan B tan (A − B) = 1 + tan A tan B Proof tan A + tan B Substitute −B for B in the formula : 1 - tan A tan B tan A + tan ( − B ) tan (A − B) = 1 − tan A tan ( − B ) tan A − tan B = 1 + tan A tan B Summarising all these results gives: The sum and difference identities sin (A ± B) = sin A cos B ± cos A sin B cos (A ± B) = cos A cos B ∓ sin A sin B tan A ± tan B tan (A ± B) = 1 ∓ tan A tan B ISBN 9780170413299 11. Trigonometric functions 607  EXAMPLE 8 a Find the exact value of cos 75° by expanding cos (30° + 45°). b Simplify sin 2θ cos θ − cos 2θ sin θ. Solution a cos 75° = cos (30° + 45°) = cos 30° cos 45° − sin 30° sin 45° 3 1 1 1 = × − × 2 2 2 2 3 1 = − 2 2 2 2 3 −1 = 2 2 b sin 2θ cos θ − cos 2θ sin θ = sin (2θ − θ) = sin θ Shutterstock.com/Andrii Chagovets 608 MATHS IN FOCUS 11. Mathematics Extension 1 ISBN 9780170413299 Double angles By using the sum of angles, we can find the trigonometric ratios for double angles. sin 2A = 2 sin A cos A Proof sin 2A = sin (A + A) = sin A cos A + cos A sin A = 2 sin A cos A cos 2A = cos2 A − sin2 A Proof cos 2A = cos (A + A) = cos A cos A − sin A sin A = cos2 A − sin2 A 2tan A tan 2A = 1 − tan 2 A Proof tan 2A = tan (A + A) tan A + tan A = 1 − tan A tan A 2tan A = 1 − tan 2 A Summarising all these results gives: The double angle identities sin 2A = 2 sin A cos A cos 2A = cos2 A − sin2 A 2tan A tan 2A = 1 − tan 2 A ISBN 9780170413299 11. Trigonometric functions 609  EXAMPLE 9 a Simplify cos2 2x − sin2 2x. 4 b If sin x = , find the exact value of sin 2x. 7 c Show that cos 2x = 1 − 2 sin2 x. Solution a cos2 2x − sin2 2x = cos 2(2x) = cos 4x b sin 2x = 2 sin x cos x B Construct a triangle to find cos x: 7 AC2 = 72 − 42 4 = 33 x A C AC = 33 sin 2x = 2 sin x cos x 4 33 = 2× × 7 7 8 33 = 49 c cos 2x = cos2 x − sin2 x = (1 − sin2 x) − sin2 x = 1 − 2 sin2 x PROBLEM 1 Ulug Beg (1393−1449) used the relation sin3 θ = (3 sin θ − sin 3θ) to draw up a table of 4 sine ratios. Can you prove this relation? 610 MATHS IN FOCUS 11. Mathematics Extension 1 ISBN 9780170413299 Products to sums and differences WS We can rearrange the sum and difference identities to find the product of trigonometric ratios. Homework Products to sums and differences 1 cos A cos B = [cos (A + B) + cos (A − B)] 2 Proof cos (A + B) + cos (A − B) = cos A cos B − sin A sin B + cos A cos B + sin A sin B = 2 cos A cos B 1 [cos(A + B) + cos (A − B)] = cos A cos B 2 1 sin A sin B = [cos (A − B) − cos (A + B)] 2 Proof cos (A − B) − cos (A + B) = cos A cos B + sin A sin B − (cos A cos B − sin A sin B) = 2 sin A sin B 1 [cos(A − B) − cos (A + B)] = sin A sin B 2 1 sin A cos B = [sin (A + B) + sin (A − B)] 2 Proof sin (A + B) + sin (A − B) = sin A cos B + cos A sin B + sin A cos B − cos A sin B = 2 sin A cos B 1 [sin(A + B) + sin (A − B)] = sin A cos B 2 ISBN 9780170413299 11. Trigonometric functions 611  1 cos A sin B = [sin (A + B) − sin (A − B)] 2 Proof sin (A + B) − sin (A − B) = sin A cos B + cos A sin B − (sin A cos B − cos A sin B) = 2 cos A sin B 1 [sin (A + B) − sin (A − B)] = cos A sin B 2 We can summarise these rules: Products to sums and differences 1 cos A cos B = [cos (A + B) + cos (A − B)] 2 1 sin A sin B = [cos (A − B) − cos (A + B)] 2 1 sin A cos B = [sin (A + B) + sin (A − B)] 2 1 cos A sin B = [sin (A + B) − sin (A − B)] 2 EXAMPLE 10 a Write cos 2x sin 5x as a sum of trigonometric ratios. b Find the exact value of sin 75° sin 15°. Solution 1 a cos A sin B = [sin (A + B) − sin (A − B)] 2 1 cos 2x sin 5x = [sin (2x + 5x) − sin (2x − 5x)] 2 1 = [sin 7x − sin (−3x)] 2 1 = [sin 7x − (−sin 3x)] since sin (−θ) = − sin θ 2 1 = [sin 7x + sin 3x] 2 612 MATHS IN FOCUS 11. Mathematics Extension 1 ISBN 9780170413299  1 b   sin A sin B = [cos (A − B) − cos (A + B)] 2 1 sin 75° sin 15° = [cos (75° − 15°) − cos (75° + 15°)] 2 1 = [cos 60° − cos 90°] 2 11  =  − 0 22  1 = 4 The t -formulas A The t-formulas express sin A, cos A and tan A in terms of t = tan. 2 The t -formulas A If t = tan , then: 2 2t sin A = 1+ t 2 1− t 2 cos A = 1+ t 2 2t tan A = 1− t 2 Proof for tan A 2tan A tan 2A = 1 - tan 2 A A 2tan tan A = 2 2 A 1 - tan 2 2t = 1- t 2 ISBN 9780170413299 11. Trigonometric functions 613  Proof for sin A A Drawing a right-angled triangle for tan A 2 √1 + t 2 tan = t t 2 A t 2 = 1 1 By Pythagoras’ theorem, the hypotenuse is 1 + t 2. sin 2A = 2 sin A cos A A A sin A = 2 sin cos 2 2  t  1  = 2    1+ t 2   1+ t 2  2t = 1+ t 2 Proof for cos A cos 2A = cos2 A – sin2 A A A cos A = cos2 – sin2 2 2 2 2  1   t  =  -  (from the triangle)  1+ t 2   1+ t 2  1 t2 = 2 - 1+ t 1+ t 2 1- t 2 = 1+ t 2 614 MATHS IN FOCUS 11. Mathematics Extension 1 ISBN 9780170413299  EXAMPLE 11 2tan15o a Find the exact value of = 1 + tan 2 15o A A b Prove that cot – 2 cot A = tan. 2 2 Solution 2t A a sin A = 2 where t = tan 1+ t 2 2tan15o So = sin 30° A = 30° 1 + tan 2 15o 1 = 2 A b LHS = cot – 2 cot A 2 1 2 = − A tan A tan 2 1 2 = − t 2t 1− t 2 1 2(1 − t 2 ) = − t 2t 1 1− t 2 = − t t t2 = t =t A = tan 2 = RHS A A So cot – 2 cot A = tan 2 2 ISBN 9780170413299 11. Trigonometric functions 615  EXT1 Exercise 11.03 Further trigonometric identities 1 Expand: a sin (a − b) b cos (p + q) c tan (α + β) d sin (x + 20°) e tan (48° + x) f cos (2θ − α) g cos (x + 75°) h tan (5x − 7y) i sin (4α − β) 2 Simplify: tan36° + tan 29° a sin a cos b + cos a sin b b 1 − tan36° tan 29° c cos 28° cos 27° − sin 28° sin 27° d sin 2x cos 3y + cos 2x sin 3y tan3θ − tan θ e f sin 74° cos 42° − cos 74° sin 42° 1 + tan3θ tan θ g sin (45° + 30°) + sin (45° − 30°) h sin (x + y) − sin (x − y) i cos (x − y) − cos (x + y) j cos (m + n) + cos (m − n) A 3 Simplify each expression, given t = tan. 2 2t 1- t2 2tan10° a b c 1- t2 1+ t 2 1 - tan 2 10° A 1 - tan 2 1 - tan 2 25° 2tan A 2 d e f 1 + tan 2 25° 1 + tan 2 A 2 A 1 + tan 2 4 Find the exact value of: a sin 75° b cos 15° c tan 75° d tan 105° e cos 105° f sin 15° g sin 105° h tan 285° i sin (x + 30°) + cos (x + 30°) j cos (45° − y) + cos (45° + y) tan ( x + y ) + tan ( x − y ) 5 Simplify. 1 − tan ( x + y ) tan ( x − y ) 2 3 6 If sin x = and cos y = , find the exact value of: 3 4 a sin (x + y) b cos (x − y) c tan (x + y) 7 By taking 2θ = θ + θ, find an expression for: a sin 2θ b cos 2θ c tan 2θ 8 By writing 3θ as 2θ + θ, find an expression in terms of θ for: a sin 3θ b cos 3θ c tan 3θ 616 MATHS IN FOCUS 11. Mathematics Extension 1 ISBN 9780170413299  tan 7θ − tan 3θ 9 a Simplify. 1 + tan 7θ tan 3θ b Find an expression for sin 4θ in terms of 7θ and 3θ. 10 Find an expression for cos 9x in terms of 2x and 7x. 11 Find the exact value of each expression. 2tan30° 1 - tan 2 22.5° a b 1 + tan 2 30° 1 + tan 2 22.5° 1 - tan 2 30° 2tan 60° c d 1 + tan 2 30° 1 - tan 2 60° 12 Write each expression as a sum or difference of trigonometric ratios. a sin 3a sin 2b b cos 5y sin 3z c cos 2p cos 3q d sin 4x cos 9y e cos 7x cos 2x f sin 4y sin y g sin 6a cos 5a h cos 2x sin 5x 13 Find the exact value of: tan 85° − tan 25° a cos 23° cos 22° − sin 23° sin 22° b 1 + tan 85° tan 25° c sin 180° cos 60° + cos 180° sin 60° d cos 290° cos 80° + sin 290° sin 80° tan 11° + tan 19° e f cos 165° cos 15° 1 − tan 11° tan 19° g sin 105° cos 75° 3 5 14 If sin x = and cos y = , find the value of: 5 13 a cos x b sin y c sin (x − y) d tan y e tan (x + y) 15 a Write an expression for cos (x + y) + cos (x − y). b Hence write an expression for cos 50° cos 65°. A 16 Write each expression in terms of t, where t = tan. 2 a cosec A b sec A c cot A d sin A + cos A A e 1 + tan A f 1 + tan A tan 2 1 + sin A + cos A g 3 cos A + 4 sin A h 1 + sin A - cos A i tan A + sec A j sin 2A ISBN 9780170413299 11. Trigonometric functions 617  17 Find an expression for: a sin (x + y) + sin (x − y) b cos (x + y) − cos (x − y) c sin (x − y) − sin (x + y) d tan (x + y) + tan (x − y) 18 Simplify: a 2 cos 3x sin 3x b cos2 7y − sin2 7y 2tan5θ c d 1 − 2 sin2 y 1 − tan 2 5θ e sin 6θ cos 6θ f (sin x + cos x)2 g 2 cos2 3α − 1 h 1 − 2 sin2 40° 2tan β i j (sin 3x − cos 3x)2 1 − tan 2 β sin 2x 19 a Simplify. 1 + cos 2x b Hence, find the exact value of tan 15°. 20 Find the exact value of tan 22 ° by using the expression for tan 2x. 1 2 21 Prove that: 1 θ 1 − cos θ a sin2 θ = sin 2θ tan θ b tan = 2 2 sin θ 1 + sin A - cos A A c = tan 1 + sin A + cos A 2 22 Show that sin2 7θ − sin2 4θ = sin 11θ sin 3θ. 23 Prove that cos 3θ = 4 cos3 θ − 3 cos θ. A 24 Find an expression for sin 2A − cos 2A in terms of t = tan. 2 618 MATHS IN FOCUS 11. Mathematics Extension 1 ISBN 9780170413299 11.04 Radians WS The rules and formulas learned in this chapter can also be expressed in radians, which we Homework Radians of any learned about in Chapter 5, Trigonometry. magnitude ASTC rule Positive angles Negative angles y y π 3π − 2nd quadrant 2 1st quadrant 2nd quadrant 2 1st quadrant π−θ θ − (π + θ) − (2π − θ) S A S A π 0 −π −2π x x 2π 0 T C − (π − θ) T C −θ π+θ 2π − θ 3rd quadrant 4th quadrant 3rd quadrant π 4th quadrant 3π − 2 2    In the 2nd quadrant: In the 4th quadrant: sin (π − θ) = sin θ sin (−θ) = −sin θ cos (π − θ) = −cos θ cos (−θ) = cos θ tan (π − θ) = −tan θ tan (−θ) = −tan θ In the 3rd quadrant: In the 3rd quadrant: sin (π + θ) = −sin θ sin (−(π − θ)) = − sin θ cos (π + θ) = −cos θ cos (−(π − θ)) = −cos θ tan (π + θ) = tan θ tan (−(π − θ)) = tan θ In the 4th quadrant: In the 2nd quadrant: sin (2π − θ) = −sin θ sin (−(π + θ)) = sin θ cos (2π − θ) = cos θ cos (−(π + θ)) = −cos θ tan (2π − θ) = −tan θ tan (−(π + θ)) = −tan θ In the 1st quadrant: sin (−(2π − θ)) = sin θ cos (−(2π − θ)) = cos θ tan (−(2π − θ)) = tan θ ISBN 9780170413299 11. Trigonometric functions 619  EXAMPLE 12 Find the exact value of: 5π 11π 5π a sin b cos EXT1 c   sin 4 6 12 Solution 5π 4π π  5π   π a = + sin   = sin  π +  4 4 4  4  4 π π = π+ = − sin 4 4 in the 3rd quadrant, so sin θ < 0. 1 =− 2 11π 12π π  11π   π b = − cos  = cos  2π −  6 6 6  6   6 π π = 2π − = cos 6 6 in the 4th quadrant, so cos θ > 0. 3 = 2 5π  2π 3π  5π c sin = sin  + 12  12 12  If it’s hard to see how to break 12 into 2 other angles, try converting it to 75°  π π and work in degrees. = sin  +   6 4 π π π π = sin cos + cos sin 6 4 6 4 1 1 3 1 = × + × 2 2 2 2 1+ 3 = 2 2 620 MATHS IN FOCUS 11. Mathematics Extension 1 ISBN 9780170413299 Exercise 11.04 Radians 1 Find the exact value of each expression. π π π a cosec b sec c cot 4 6 3 π sin 3 π π π d e 1 – cos2 f tan cos π 4 3 3 cos 3 π 3π cot + tan π2 π 2 5 10 g 1 + tan h cosec –1 i 4 6 π cot 5 3π π 2 a Show that =π−. 4 4 3π b In which quadrant is the angle ? 4 3π c Find the exact value of cos. 4 5π π 3 a Show that =π−. 6 6 5π b In which quadrant is the angle ? 6 5π c Find the exact value of sin. 6 7π π 4 a Show that = 2π −. 4 4 7π b In which quadrant is the angle ? 4 7π c Find the exact value of tan. 4 4π π 5 a Show that =π+. 3 3 4π b In which quadrant is the angle ? 3 4π c Find the exact value of cos. 3 5π π 6 a Show that = 2π −. 3 3 5π b In which quadrant is the angle ? 3 5π c Find the exact value of sin. 3 ISBN 9780170413299 11. Trigonometric functions 621  13π π 7 a i Show that = 2π +. 6 6 13π ii In which quadrant is the angle ? 6 13π iii Find the exact value of cos. 6 b Find the exact value of: 9π 7π 11π i sin ii tan iii cos 4 3 4 19π 10π iv tan v sin 6 3 8 Copy and complete each table with exact values. a π 2π 4π 5π 7π 8π 10π 11π 3 3 3 3 3 3 3 3 sin cos tan b π 3π 5π 7π 9π 11π 13π 15π 4 4 4 4 4 4 4 4 sin cos tan c π 5π 7π 11π 13π 17π 19π 23π 6 6 6 6 6 6 6 6 sin cos tan 9 Copy and complete the table where possible. 0 π π 3π 2π 5π 3π 7π 4π 2 2 2 2 sin cos tan 622 MATHS IN FOCUS 11. Mathematics Extension 1 ISBN 9780170413299 10 EXT1 Find the exact value of: π π π π a sin  +  b cos  -  4 3 6 3 2π π 2π π π π c sin cos – cos sin d 2 sin cos 3 4 3 4 8 8 π 2tan e 6 π 1 - tan 2 6 11 EXT1 Simplify: π π π π 5π π 5π π a cos cos – sin sin b sin cos – cos sin 9 5 9 5 7 8 7 8 π tan π - tan 5 π π c d sin cos π 11 9 1 + tan π tan 5 π 2tan e 7 2 π 1 + tan 7 π π 5π 12 EXT1 a i Show that + =. 6 4 12 5π ii Find the exact value of tan. 12 2π π 11π b i Show that + =. 3 4 12 11π ii Find the exact value of cos. 12 9 π 5π 7π c   i Show that - =. 4 3 12 7π   ii Find the exact value of sin. 12 d Find the exact value of: π i sin 12 13π ii cos 12 ISBN 9780170413299 11. Trigonometric functions 623  WS 11.05 Trigonometric functions Homework INVESTIGATION Sine and cosine curves WS TRIGONOMETRIC RATIOS OF 0°, 90°, 180°, 270° AND 360° y Remember the results from the unit circle: Homework Trigonometric graphs sin θ = y 2nd quadrant 90 ° 1st quadrant cos θ = x y tan θ = 180 ° − θ θ Trigonometric graphs x S A match-up 180 ° 0° x 360 ° WS T C Homework Sketching 180 ° + θ 360 ° − θ periodic functions: amplitude and 3rd quadrant period 270 ° 4th quadrant WS 1 Angle 0° is at the point (1, 0) on the unit circle. Use the circle results to find sin 0°, Homework Sketching periodic cos 0° and tan 0°. functions: 2 Angle 90° is at the point (0, 1). Use the circle results to find sin 90°, cos 90° and tan 90°. phase and vertical shift Discuss the result for tan 90° and why this happens. 3 Angle 180° is at the point (−1, 0). Find sin 180°, cos 180° and tan 180°. Amplitude and period 4 Angle 270° is at the point (0, −1). Find sin 270°, cos 270° and tan 270°. Discuss the result for tan 270° and why this happens. 5 What are the results for sin 360°, cos 360° and tan 360°? Why? 6 Check these results on your calculator. The sine function Using all the results from the investigation, we can draw up a table of values for y = sin x. x 0° 90° 180° 270° 360° y 0 1 0 −1 0 We could add in all the exact value results we know for a more accurate graph. Remember that sin x is positive in the 1st and 2nd quadrants and negative in the 3rd and 4th quadrants. x 0° 30° 45° 60° 90° 120° 135° 150° 180° 210° 225° 240° 270° 300° 315° 330° 360° 1 1 3 3 1 1 1 1 3 3 − 1 −1 y 0 1 0 − − − −1 − 0 2 2 2 2 2 2 2 2 2 2 2 2 624 MATHS IN FOCUS 11. Mathematics Extension 1 ISBN 9780170413299 Drawing the graph gives a smooth ‘wave’ curve. y 1 y = sin x 0 x 90° 180° 270° 360° −1 As we go around the unit circle and graph the y values of the points on the circle, the graph should repeat itself every 360°. y = sin x has domain (−∞, ∞) and range [−1, 1]. It is an odd function. y 1 y = sin x 0° x −360° −270° −180° −90° 90° 180° 270° 360° 450° 540° 630° 720° −1 The cosine function Similarly for y = cos x, which is positive in the 1st and 4th quadrants and negative in the 2nd and 3rd quadrants. Its graph has the same shape as the graph of the sine function. x 0° 30° 45° 60° 90° 120° 135° 150° 180° 210° 225° 240° 270° 300° 315° 330° 360° 3 1 1 1 1 3 3 − 1 1 1 1 3 y 0 1 − − − 0 − − −1 0 2 2 2 2 2 2 2 2 2 2 2 2 y y = cos x 1 x 90° 180° 270° 360° −1 As we go around the unit circle and graph the x values of the points on the circle, the graph should repeat itself every 360°. y = cos x has domain (−∞, ∞) and range [−1, 1]. It is an even function. ISBN 9780170413299 11. Trigonometric functions 625  y y = cos x 1 0° x −450° −360° −270° −180° −90° 90° 180° 270° 360° 450° 540° 630° 720° −1 The tangent function y = tan x is positive in the 1st and 3rd quadrants and negative in the 2nd and 4th quadrants. It is also undefined for 90° and 270° so there are vertical asymptotes at those x values, where the function is discontinuous. x 0° 30° 45° 60° 90° 120° 135° 150° 180° 210° 225° 240° 270° 300° 315° 330° 360° 1 1 1 1 y 0 1 3 – − 3 −1 − 0 1 3 – − 3 −1 − 0 3 3 3 3 y 1 x −1 90° 180° 270° 360° y = tan x y As we go around the unit circle and graph the values of of the points on the circle, the x graph repeats itself every 180°. y = tan x has domain (−∞, ∞) except for 90°, 270°, 540°, … (odd multiples of 90°) and range (−∞, ∞). It is an odd function. y = tan x y 1 −360° −270° −180° −90° 0° 90° 450° 540° 630° 720° x 180° 270° 360° −1 626 MATHS IN FOCUS 11. Mathematics Extension 1 ISBN 9780170413299 The cosecant function 1 cosec x = sin x We learned about reciprocal functions in Chapter 7, Further functions. Each y value of y = cosec x will be the reciprocal of y = sin x. Because sin x = 0 at x = 0°, 180°, 360°, …, y = cosec x will have vertical asymptotes at those values. We can use a table of values and explore the limits as x approaches any asymptotes. x 0° 30° 45° 60° 90° 120° 135° 150° 180° 210° 225° 240° 270° 300° 315° 330° 360° 2 2 2 2 y – 2 2 1 2 2 – −2 − 2 −  −1 −  −  2 −2 – 3 3 3 3 y 1 y = sin x 0 x 90° 180° 270° 360° −1 y = cosec x The secant function 1 sec x = , so each y value of y = sec x will be the reciprocal of y = cos x. Because cos x = 0 cos x at x = 90°, 270°, 450°, …, y = sec x will have vertical asymptotes at those values. x 0° 30° 45° 60° 90° 120° 135° 150° 180° 210° 225° 240° 270° 300° 315° 330° 360° 2 2 2 2 y 1 2 2 – −2 −  2 −  −1 −  3 − 2 −2 – 2 2 1 3 3 3 y y = sec x 1 y = cos x 0 x 90° 180° 270° 360° −1 ISBN 9780170413299 11. Trigonometric functions 627  The cotangent function 1 cot x = , so each y value of y = cot x will be the reciprocal of y = tan x. Because tan x = 0 tan x at x = 0°, 180°, 360°, …, y = cot x will have vertical asymptotes at those values. Also, because tan x has asymptotes at x = 90°, 270°, 450°, …, y = cot x = 0 and there are x-intercepts at those values. x 0° 30° 45° 60° 90° 120° 135° 150° 180° 210° 225° 240° 270° 300° 315° 330° 360° 1 1 1 1 y – 3 1 0 −  −1 −  3 – 3 1 0 −  −1 −  3 – 3 3 3 3 y = tan x y 1 0 x 90° 180° 270° 360° −1 y = cot x It is more practical to express the trigonometric functions in terms of radians (not degrees), so here are the graphs in radians: y = sin x y = cos x y = tan x y y y 1 1 1 0 x 0 x 0 x π π 3π 2π π π 3π 2π π π 3π 2π −1 2 2 2 2 2 2 −1 −1 Domain (−∞, ∞), range [−1, 1] Domain (−∞, ∞), range [−1, 1] Domain (−∞, ∞) Odd function Even function except for π , 3π , 5π , … 2 2 2 π (odd multiples of ), 2 range (−∞, ∞) Odd function 628 MATHS IN FOCUS 11. Mathematics Extension 1 ISBN 9780170413299 y = cosec x y = sec x y = cot x y y y 1 1 1 0 x 0 x 0 x π π 3π 2π π π 3π 2π π π 3π 2π 2 2 −1 2 2 2 2 −1 −1 Properties of the trigonometric functions All the trigonometric functions have graphs that repeat at regular intervals, so they are called periodic functions. The period is the length of one cycle of a periodic function on the x-axis, before the function repeats itself. The centre of a periodic function is its mean value and is equidistant from the maximum and minimum values. The mean value of y = sin x, y = cos x and y = tan x is 0, represented by the x-axis. The amplitude is the height from the centre of a periodic function to the maximum or minimum values (peaks and troughs of its graph respectively). The range of y = sin x and y = cos x is [-1, 1]. y = sin x has period 2π and amplitude 1. y = cos x has period 2π and amplitude 1. y = tan x has period π and no amplitude. INVESTIGATION TRANSFORMING TRIGONOMETRIC GRAPHS Use a graphics calculator or graphing software to draw the graphs of trigonometric functions with different values. 1 Graphs in the form y = k sin x, y = k cos x and y = k tan x where k = …, −3, −2, −1, 2, 3, … 2 Graphs in the form y = sin ax, y = cos ax and y = tan ax where a = …, −3, −2, −1, 2, 3, … 3 Graphs in the form y = sin x + c, y = cos x + c and y = tan x + c where c = …, −3, −2, −1, 2, 3, … π 4 Graphs in the form y = sin (x + b), y = cos (x + b) and y = tan (x + b) where b = …, ± , π 2 ±π, ± , … 4 Can you see patterns? Could you predict what different graphs look like? ISBN 9780170413299 11. Trigonometric functions 629  Now we shall examine more general trigonometric functions of the form y = k sin ax, y = k cos ax and y = k tan ax, where k and a are constants. Period and amplitude of trigonometric functions 2π y = k sin ax has amplitude k and period. a 2π y = k cos ax has amplitude k and period. a π y = k tan ax has no amplitude and has period. a EXAMPLE 13 a Sketch each function in the domain [0, 2π]. i y = 5 sin x iiy = sin 4x iii y = 5 sin 4x x b Sketch the graph of y = 2 tan for [0, 2π]. 2 Solution a i The graph of y = 5 sin x has y values y 5 that are 5 times as much as y = sin x, so this function has amplitude 5 and y = 5 sin x period 2π. We draw one period of the sine ‘shape’ between ±5. π π 3π 2π x 2 2 −5 ii The graph y = sin 4x has amplitude 1 y 2π π y = sin 4x and period =. 1 4 2 π The curve repeats every , so in x 2 π π 3π π 5π 3π 7π 2π 4 2 4 4 2 4 the domain [0, 2π] there will be −1 4 repetitions. The ‘4’ in sin 4x compresses the graph of y = sin x horizontally. 630 MATHS IN FOCUS 11. Mathematics Extension 1 ISBN 9780170413299  y iii The graph y = 5 sin 4x has amplitud