Trigonometry Modules PDF

BASIC CALCULUS 1| PUP PARAÑAQUE MODULE 4 Trigonometry: Angles in the Unit Circle Sir Kenneth E. Abaja  BASIC CALCULUS...

BASIC CALCULUS 1| PUP PARAÑAQUE MODULE 4 Trigonometry: Angles in the Unit Circle Sir Kenneth E. Abaja  BASIC CALCULUS 2| PUP PARAÑAQUE I. OVERVIEW Trigonometry was invented over 2000 years ago by the Greeks, who needed precise methods for measuring angles and sides of triangles. In fact, the word trigonometry was derived from the two Greek words trigonon (triangle) and metria (measurement). It is the study of the properties of triangles, as the word suggests. This module begins with a discussion of angles and how they are measured. II. MODULE OBJECTIVE At the end of the lesson, the students should be able to: 1. Define an angle. 2. Convert degree to radians and vice versa. 3. Define Coterminal angles and Quadrantal angles. 4. Measure the arc length and area of a sector. III. COURSE MATERIALS MEASURING ANGLE: RADIANS AND DEGREES In geometry an angle is defined as the set of points determined Figure 4.1 by two rays, or half-lines, and, having the same endpoint O. If A and B are points on and, as in Figure 4.1, we refer to angle AOB. An angle may also be considered as two finite line segments with a common endpoint. In trigonometry we often interpret angles as rotations of rays. Start with a fixed ray, having endpoint O, and rotate it about O, in a plane, to a position specified by ray. We call the initial side, the terminal side, and O the vertex of. The amount or direction of rotation is not restricted in any way. Figure 4.2 Figure 4.3 In trigonometry we can also have negative angles aside form positive angles. If an angle 𝜃 is generated by a counter-clockwise rotation, as in Figure 4.2, we call angle 𝜃 a positive angle. On the other hand, if an angle 𝜃 is generated by a clockwise rotation, as in Figure 4.3, we call angle 𝜃 a negative angle. Sir Kenneth E. Abaja  BASIC CALCULUS 3| PUP PARAÑAQUE Angles are often measured in two ways, degrees and radians. A radian, which is short for radius angle, is also based on the concept of a circle. If the arc length of a sector is equal to the radius, then we can say that the angle is 1 radian (Figure 4.4). If the angle is in degrees, we must use the correct symbol ′ ° ′ to show that the angle has been measured in degrees. Otherwise it is assumed that the angle is measured in radians. Often radian is abbreviated, so 1 radian will be abbreviated to 1. The Figure 4.5 below shows that an arc length of 1𝑟 is opposite to (subtends) an angle of 1 radian. Then in Figure 4.6 we can see that an arc Figure 4.4 length of 3𝑟 is opposite to an angle of 3radians. Thus if we think about the circumference of a circle as 𝐶= 2𝜋r, we could say that an arc of 2𝜋r subtends the angle of 2𝜋. In other words, the circumference of a circle subtends a full revolution. This then implies that 2𝜋= 360° ∴ 𝜋= 180°. Figure 4.5 Figure 4.6 CONVERSION FROM DEGREES TO RADIANS, AND VICE VERSA From 𝜋 = 180°, we can use the following identity to: convert from radians to degrees: convert from degrees to radians: 180° 𝜋 𝜋 180° Examples: 1. Convert 3𝜋 to degrees 180° (3𝜋) ( ) = 3(180°) = 540° 𝜋 2. Convert 720° to radians 𝜋 (720°) ( ) = 4𝜋 180° The table above displays the corresponding radian and measure of the special angles. As for your practice, verify if the following conversion is correct. Sir Kenneth E. Abaja  BASIC CALCULUS 4| PUP PARAÑAQUE CONVERSION FROM DECIMAL DEGREE TO DEGREES-MINUTES-SECONDS (DMS), AND VICE VERSA DECIMAL TO DMS 1. For the degrees use the whole number part of the decimal. 2. For the minutes multiply the remaining decimal by 60. Use the whole number part of the answer as minutes. 3. For the seconds multiply the new remaining decimal by 60 Note: For negative angle, just disregard the negative sign and put it back when you have the final answer EXAMPLE: 45.26° The whole number is degrees. So 45.26 gives you 𝟒𝟓°. Multiply the remaining decimal by 60’ 0.26 ∗ 60′ = 15.6′ The whole number is minutes. So 15.6 gives you 15’. Multiply the remaining minutes by 60’’ 0.6 ∗ 60′′ = 𝟑𝟔′′ Therefore the DMS conversion is: 45°15’ 36′′ DMS TO DECIMAL DEGREE Use the formula 𝑀𝑖𝑛𝑢𝑡𝑒𝑠 𝑆𝑒𝑐𝑜𝑛𝑑𝑠 𝐷𝑒𝑔𝑟𝑒𝑒 + + 60′ 3600′′ EXAMPLE: −145°24′52′′ Disregard the sign first, 24′ 52′′ 145° + + 60′ 3600′′ The Decimal degree conversion is 145.4144° Sir Kenneth E. Abaja  BASIC CALCULUS 5| PUP PARAÑAQUE COTERMINAL ANGLES Coterminal Angles are angles in standard position who share the same initial side and terminal sides. An angle is said to be in standard position when its initial side lies along the positive x-axis and the vertex is at the origin. Finding coterminal angles is as simple as adding or subtracting 360° or 2π to each angle, depending on whether the given angle is in degrees or radians. There are an infinite number of coterminal angles that can be found. In trigonometry, we don’t limit the angle to 360°, we can have as high as 43200° or even more and as low as -56200°. In figure 4.7 below, 60° has coterminal angles 420°, 780°, −300° and −660°. As defined, these angle are in standard position, and no matter how many revolution or rotation you take, the terminal sides of each angle are the same. Figure 4.7 To verify if the coterminal angles of 60° are correct, you can add or subtract 360° as much as you want. 60°+360° = 420° 60°+2(360°) = 780° 60° − 360° = −300° 60°-2(360°) = −660° Examples: 1. Find the coterminal angles of −45° −45° + 360= 315° −45° + 2(360)= 675° −45° − 360=−405° −45° − 360=−765° Hence, the coterminal angles of −45° are 315°, 675°, −405° 𝑎𝑛𝑑 =−765° 𝜋 2. Find the coterminal angles of 6 𝜋 13𝜋 + 2𝜋= 6 6 𝜋 25𝜋 + 2(2𝜋)= 6 6 𝜋 −11𝜋 − 2𝜋= 6 6 𝜋 −23𝜋 − 2(2𝜋)= 6 6 𝜋 13𝜋 25𝜋 −11𝜋 −23𝜋 Hence, the coterminal angles of 6 are , 6 , 6 𝑎𝑛𝑑 6 6 The importance of coterminal angle will further be discussed as we go along with the next modules. For now, we can see that if we have a very large number and we cannot visualize at what quadrant where it falls, you can find the coterminal angles to easily visualize the angle. Sir Kenneth E. Abaja  BASIC CALCULUS 6| PUP PARAÑAQUE QUADRANTAL ANGLES Quadrantal Angles are angles in standard position where terminal side of the angle falls in a certain quadrant. Figure 4.8 In Figure 4.8, all angles are in standard position and all of their terminal side lies along quadrant. In radians, quadrantal angles are measured in π increments (π, 2π, 3π, 4π, 5π, etc.) Take a look at figure 4.9 to see the quadrantal angles in radian measure. Figure 4.9 ARC LENGTH AND AREA OF A SECTOR FOMULA FOR THE LENGTH OF A CIRCULAR ARC 𝜃 is in radian measure: 𝜃 is in degree: 𝑆 = 𝑟𝜃 𝑟𝜃𝜋 𝑆= 180° Where: Where: S= Arc length S= Arc length r=radius r=radius 𝜃=angle (in radians) 𝜃=angle (in degrees) Sir Kenneth E. Abaja  BASIC CALCULUS 7| PUP PARAÑAQUE PROOF. A typical arc of length S and the corresponding central angle are shown in Figure 4.10(a). Figure 4.10(b) shows an arc of length and central angle. If radian measure is used, then, from plane geometry, the ratio of the lengths of the arcs is the same as the ratio of the angular measures; that is, Figure 4.10 𝑆 𝜃 𝜃 = 𝑜𝑟 𝑆= 𝑆 𝑆1 𝜃1 𝜃1 1 If we consider the special case in which 𝜃1 has radian measure 1, then, from the definition of radian, 𝑆1 = 𝑟 and the last equation becomes 𝜃 𝑆= 𝑟 = 𝑟𝜃 1 FOMULA FOR THE AREA OF A CIRCULAR SECTOR 𝜃 is in radian measure: 𝜃 is in degree: 1 2 1 2 𝐴= 𝑟 𝜃 𝐴= 𝑟 𝜃𝜋 2 360 Where: Where: A=Area A=Area r=radius r=radius 𝜃=angle (in radians) 𝜃=angle (in degrees) PROOF. If 𝐴 and 𝐴1 are the areas of the sectors in Figure 4.11(a) and Figure 4.11(b), respectively, from plane geometry, 𝐴 𝜃 𝜃 = 𝑜𝑟 𝐴= 𝐴 𝐴1 𝜃1 𝜃1 1 If we consider the special case in which 𝜃1 = 2𝜋, then 𝐴1 = 𝜋𝑟 2 and Figure 4.11 𝜃 1 𝐴= 𝜋𝑟 2 = 𝑟 2 𝜃 2𝜋 2 Sir Kenneth E. Abaja  BASIC CALCULUS 8| PUP PARAÑAQUE Examples: 2𝜋 1. Find the arc length and area of a sector of a circle of radius 6 cm and the central angle. 5 Solution: 𝑆 = 𝑟𝜃 1 2 𝐴= 𝑟 𝜃 2 1 2 𝑆 = (6 𝑐𝑚)𝐴 = 𝑟 𝜃 1 2𝜋 2 𝐴= (6𝑐𝑚)2 ( ) 2 5 22𝜋 𝑆= 𝑐𝑚. 5 36𝜋 𝐴= 𝑐𝑚2 5 2. Find the arc length and area of a sector of a circle of radius 4 m and the central angle 30°. Solution: 𝑟𝜃𝜋 1 2 𝑆= 𝐴= 𝑟 𝜃𝜋 180° 360 (4𝑚)(30°)𝜋 1 𝑆= 𝐴= (4 𝑚)2 (30°)𝜋 180° 360 2𝜋 𝑆= 𝑚. 3 4𝜋 2 𝐴= 𝑚 3 3. The radius of the Earth is about 3960 miles. If you travel 500 miles due north, how many degrees of latitude will you traverse? Solution: Going back to the formula of the arc length 𝑆 = 𝑟𝜃 In the given, S=500mi and r=3960mi. 500mi = 3960mi 𝜃 500mi 3960mi = 𝜃 3960mi 3960mi 𝜃 = 0.1263 𝑟𝑎𝑑 𝑟𝜃𝜋 If you want the answer to be in degrees, convert the answer or, from the start use 𝑆 = 180° 3960𝑚𝑖 𝜃𝜋 500mi = 180° θ = 7.23° Sir Kenneth E. Abaja  BASIC CALCULUS 9| PUP PARAÑAQUE I. ACTIVITIE/ASSESSMENT 1. Verify the conversion below, show your complete solution. 2. Find 5 coterminals angles on each given and sketch it. a. θ = 50° b. θ = −120° 4𝜋 c. θ = 5 5𝜋 d. θ = − 4 7𝜋 e. θ = −2 3. Convert from Decimal Degree to DMS. a. 63.75° b. 200.325° c. −317.06° d. 179.999° e. −261.56° 4. Convert from DMS to Decimal Degree a. 63°45′ 24′′ b. 125°50′ c. −32°10′ 12′′ d. 502°12′′ e. −314°12′ 40′′ 5. Compute for the length of the arc and the area of the sector of the following: 𝜋 a. 𝜃 = 6 , 𝑟 = 12𝑐𝑚. b. 𝜃 = 𝜋, 𝑟 = 1𝑚. 5𝜋 c. 𝜃 = 4 , 𝑟 = 100𝑖𝑛. d. 𝜃 = 240°, 𝑟 = 5𝑐𝑚. e. 𝜃 = 330°, 𝑟 = 9.3𝑚. f. 𝜃 = 1°, 𝑟 = 117𝑖𝑛. 6. The distance around the face of a clock from 12 to 1 is 25cm. What is the radius of the clock? WORKS CITED Stitz, Carl and Zeager, Jeff. College Algebra and Trigonometry. 2010 Swokowski, Earl W., Cole, Jeffert A. Algebra and Trigonomnetry with Abakythic Geometry. Thomson Learning, Inc. 2008 Sir Kenneth E. Abaja  BASIC CALCULUS 10| PUP PARAÑAQUE MODULE 5 The Unit Circle: Sine and Cosine Sir Kenneth E. Abaja  BASIC CALCULUS 11| PUP PARAÑAQUE I. OVERVIEW This module will focus on two major trigonometric functions: sine and cosine, and its derivation from the unit circle. The Unit circle is a special type of circle where center lies on the origin with radius 1. The standard equation of a unit circle is: 𝑥2 + 𝑦 2 = 1 We will also discuss the reference angle and its importance. These are all preparations to tackle the other 4 trigonometric functions on the next modules. II. MODULE OBJECTIVE At the end of the lesson, the students should be able to: 1. Derive the sine and cosine values of the special angles; 2. Use the Reference Angle Theorem in finding cosine and sine values. III. COURSE MATERIALS SINE AND COSINE Consider an angle θ in standard position and let P denote the point where the terminal side of θ intersects the Unit Circle. By associating the point P with the angle θ, we are assigning a position on the Unit Circle to the angle θ. The x-coordinate of P is called the cosine of θ, written cos(θ), while the y-coordinate of P is called the sine of θ, written sin(θ). The reader is encouraged to verify that these rules used to match an angle with its cosine and sine do, in fact, satisfy the definition of a function. That is, for each angle θ, there is only one associated value of cos(θ) and only one associated value of sin(θ). In simple words, in (x,y), x is cosine value and y is sine value. Figure 5.1 Sir Kenneth E. Abaja  BASIC CALCULUS 12| PUP PARAÑAQUE Examples: Find the cosine and sine of the following angles. 1. 𝜃 = 270° To find 𝑐𝑜𝑠 (270°) and 𝑠𝑖𝑛 (270°), we plot the angle 𝜃 = 270° in standard position and find the point on the terminal side of 𝜃 which lies along the Unit Circle (refer to figure 5.2). Since 270° is a quadrantal angles that lies along the negative y-axis, the point we seek is (0,-1) so that 𝑐𝑜𝑠 (270°) = 0 and 𝑠𝑖𝑛 (270°) = −1 Figure 5.2 2. 𝜃 = −𝜋 The angle θ = −π represents one half of a clockwise revolution so its terminal side lies on the negative x-axis (Refer to Figure 5.3). The point on the Unit Circle that lies on the negative x-axis is (−1, 0) which means cos(−π) = −1 and sin(−π) = 0. Figure 5.3 3. 𝜃 = 45° When we sketch θ = 45° in standard position, we see that its terminal does not lie along any of the coordinate axes which makes our job of finding the cosine and sine values a bit more difficult. Let P(x,y) denote the point on the terminal side of θ which lies on the Unit Circle. By definition, x = cos (45° ) and y = sin (45°). If we drop a perpendicular line segment from P to the x-axis, we obtain a 45° − 45°− 90° right triangle whose legs have lengths x and y units. Kindly refer to figure 5.4. Figure 5.4 From Geometry, we get y=x. Since P(x,y) lies on the Unit Circle, we have 𝑥 2 + 𝑦 2 = 1. Substituting y = x into this equation yields 𝑥2 + 𝑥2 = 1 2𝑥 2 = 1 1 𝑥 = ±√ 2 1 𝑥=± √2 By the process of rationalization, we obtain: √2 𝑥=± 2 Sir Kenneth E. Abaja  BASIC CALCULUS 13| PUP PARAÑAQUE Since x=y, then √2 𝑦=± 2 45° lies in Quadrant 1, where x and y are both positive. √2 √2 Therefore, cos (45°) = and sin (45°) = 2 2 𝜋 4. 𝜃 = 6 𝜋 As what we have in question 3, the terminal side of 𝜃 = 6 does not lie on any of the coordinate axes, so we proceed using a triangle approach. Letting P(x, y) denote the point on the terminal side of θ which lies on the Unit Circle, we drop a perpendicular line segment from P to the x-axis to form a 30° − 60° − 90° right triangle. Refer to the figure 5.5. Figure 5.5 1 After a bit of Geometry, we find 𝑦 = 2. Since P(x, y) lies on the Unit Circle, we substitute 1 𝑦 = 2 to the equation of the unit circle. Thus, 1 2 𝑥2 + ( ) = 1 2 Solve for the value of x 1 𝑥2 + =1 4 1 3 𝑥2 = 1 − = 4 4 3 √𝑥 2 = √ 4 √3 𝑥=± 2 𝜋 lies in Quadrant 1, where x and y are both positive. 6 𝜋 √3 𝜋 1 Therefore, cos ( 6 ) = and sin ( 6 ) = 2 2 Sir Kenneth E. Abaja  BASIC CALCULUS 14| PUP PARAÑAQUE 5. 𝜃 = 60° 𝜋 Plotting 𝜃 = 6 in standard positon, we find that it’s not quadrantal angle and set abut using the triangle approach. Once again, we get a 30° − 60° − 90° right triangle. Refer to the figure 5.6. Figure 5.6 1 Unlike in question 4, here 𝑥 =. 2 1 2 ( ) + 𝑦2 = 1 2 Solve for the value of y 1 + 𝑦2 = 1 4 1 3 𝑦2 = 1 − = 4 4 3 √𝑦 2 = √ 4 √3 𝑦=± 2 60° lies in Quadrant 1, where x and y are both positive. 1 √3 Therefore, cos (60°) = and sin (60°) = 2 2 PYTHAGOREAN IDENTITY Pythagorean Identity using the Pythagorean Theorem with respect to the figure 5.7 is 𝑥2 + 𝑦2 = 𝑟2 Substituting 𝑥 = 𝑐𝑜𝑠 𝜃 and 𝑦 = 𝑠𝑖𝑛 𝜃, we have 𝑐𝑜𝑠 2 𝜃 + 𝑠𝑖𝑛2 𝜃 = 1 Figure 5.7 We will use this identity to solve for the examples below but before that, let’s discuss first the sign in each quadrant. Recall in the previous module that Sir Kenneth E. Abaja  BASIC CALCULUS 15| PUP PARAÑAQUE Quadrant II Quadrant I (cos 𝜃, sin 𝜃) (cos 𝜃, sin 𝜃) (−, +) (+, +) Sine is only positive All are positive Quadrant III Quadrant IV (cos 𝜃, sin 𝜃) (cos 𝜃, sin 𝜃) (−, −) (+, −) Tangent is only positive Cosine is only positive Examples: Using the given information about θ, find the indicated value 3 1. If θ is a Quadrant II angle with sin(θ) = 5, find cos(θ) Use the Pythagorean identity: 𝑐𝑜𝑠 2 𝜃 + 𝑠𝑖𝑛2 𝜃 = 1 3 2 Thus, 𝑐𝑜𝑠 2 𝜃 + (5) = 1 Solve for 𝑐𝑜𝑠 𝜃, 9 𝑐𝑜𝑠 2 𝜃 = 1 − 25 2 9 𝑐𝑜𝑠 𝜃 = 1 − 25 16 √𝑐𝑜𝑠 2 𝜃 = ±√ 25 4 𝑐𝑜𝑠 𝜃 = ± 5 Since we are at Quadrant II, where sine is the only positive, then the value of 4 cos 𝜃 = − 5 3𝜋 √5 2. If 𝜋 < 𝜃 < with 𝑐𝑜𝑠 𝜃 = − , find 𝑠in 𝜃. 2 5 Use the Pythagorean identity: 𝑐𝑜𝑠 2 𝜃 + 𝑠𝑖𝑛2 𝜃 = 1 2 √5 Thus, (− 5 ) + 𝑠𝑖𝑛2 𝜃 = 1 Sir Kenneth E. Abaja  BASIC CALCULUS 16| PUP PARAÑAQUE Solve for 𝑠𝑖𝑛 𝜃, 5 𝑠𝑖𝑛2 𝜃 + =1 25 5 𝑠𝑖𝑛2 𝜃 = 1 − 25 20 √𝑠𝑖𝑛2 𝜃 = ±√ 25 2√5 𝑠𝑖𝑛 𝜃 = ± 5 3𝜋 Since 𝜃 in 𝜋 < 𝜃 < lies in Quadrant III, where tangent is the only positive, then 2 c 2√5 sin 𝜃 = − 5 3. If sin(θ) = 1, find cos(θ). 𝑐𝑜𝑠 2 𝜃 + (1)2 = 1 𝑐𝑜𝑠 2 𝜃 = 1 − 1 = 0 𝑐𝑜𝑠 𝜃 = 0 THE REFERENCE ANGLE THEOREM Definition: Reference Angle for a non-quadrantal angle θ, the reference angle for θ (usually denoted α) is the acute angle made between the terminal side of θ and the x-axis. If θ is a Quadrant I or IV angle, α is the angle between the terminal side of θ and the positive x - axis; if θ is a Quadrant II or III angle, α is the angle between the terminal side of θ and the negative x-axis. If we let P denote the point (𝑐𝑜𝑠(𝜃), 𝑠𝑖𝑛(𝜃)), then P lies on the Unit Circle. Since the Unit Circle possesses symmetry with respect to the x-axis, y-axis and origin, regardless of where the terminal side of θ lies, there is a point Q symmetric with P which determines θ’s reference angle, α as seen below Sir Kenneth E. Abaja  BASIC CALCULUS 17| PUP PARAÑAQUE Example of the reference angle: 1. 2. 3. 4. 5. 6. -325 Sir Kenneth E. Abaja  BASIC CALCULUS 18| PUP PARAÑAQUE Answer: The reference angles are: 1. 𝛼 = 45° The angle is referred to the nearest x-axis, which is 180°, so 180° − 135° = 45° 2. 𝛼 = 30° The angle is referred to the nearest x-axis, which is 180°, so 240° − 180° = 30° 3. 𝛼 = 60° The angle is referred to the nearest x-axis, which is 360°, so 360° − 300° = 60° 4. 𝛼 = 60° The angle is referred to the nearest x-axis, which is −180°, so −180° − (−240°) = 60° 5. 𝛼 = 30° The angle is referred to the nearest x-axis, which is −180°, so −150° − (−180°) = 30° 6. 𝛼 = 35° The angle is referred to the nearest x-axis, which is −360°, so −325° − (−360°) = 35° Note: If you are confused with the reference angle, just look at it as an angle formed by the original angle to the x-xis. HINT to easily find the reference angle in radian measure 𝜋 𝜋 𝜋 30° = 45° = 60° = 6 4 3 Angles with 2 as denominator or has no Angles that have Angles that have Angles that have denominator is a denominator of 6 has a denominator of 4 has a denominator of 6 has a quadrantal angle. They reference angle of reference angle of reference angle of don’t have reference 𝜋 𝜋 𝜋 30°𝑜𝑟 45° 𝑜𝑟 60° 𝑜𝑟 angle. 6 4 3 We’ll use this reference angle to find the cosine and sine value. That is what we call, Reference Angle Theorem. Reference Angle Theorem. Suppose α is the reference angle for θ. Then cos(θ) = ± cos(α) and sin(θ) = ± sin(α), where the choice of the (±) depends on the quadrant in which the terminal side of θ lies. Examples: Find the cosine and sine values of the following angle. 1. 𝜽 = 𝟐𝟐𝟓° The reference angle of 𝜃 is 𝛼 = 45° (refer to figure 5.8) We’ll use the 45° instead of 225°. √2 √2 We all know that 𝑐𝑜𝑠 45° = and 𝑠𝑖𝑛 45° = Figure 5.8 2 2 Sir Kenneth E. Abaja  BASIC CALCULUS 19| PUP PARAÑAQUE From the Reference Angle Theorem, that is also the value of 𝑐𝑜𝑠 225° and 𝑠𝑖𝑛 225°. However, we need to respect the sign of its quadrant. Since 𝜃 = 225° lies in Quadrant III, where Tangent is the only positive, therefore, √2 𝑐𝑜𝑠 225° = − 2 √2 𝑠𝑖𝑛 225° = − 2 𝟏𝟏𝝅 2. 𝜽 = 𝟔 The since 𝜃 has denominator 6, then the reference angle is 𝛼 = 𝜋 (refer to figure 5.9) 6 𝜋 We’ll use the 6 𝑜𝑟 30° 𝜋 √3 𝜋 1 We all know that 𝑐𝑜𝑠 ( 6 ) = and 𝑠𝑖𝑛 ( 6 ) = 2 Figure 5.9 2 With respect to the quadrant and from the Reference Angle Theorem, 11𝜋 √3 𝑐𝑜𝑠 ( )= 6 2 11𝜋 1 𝑠𝑖𝑛 ( )=− 6 2 −𝟓𝝅 3. 𝜽 = 𝟒 The since 𝜃 has denominator 4, then the reference angle is 𝜋 𝛼 = 4 (refer to figure 5.9) 𝜋 We’ll use the 4 𝑜𝑟 45° 𝜋 √2 𝜋 √2 We all know that 𝑐𝑜𝑠 ( ) = and 𝑠𝑖𝑛 ( ) = 4 2 4 2 Figure 5.10 With respect to the quadrant and from the Reference Angle Theorem, −5𝜋 √2 𝑐𝑜𝑠 ( )=− 4 2 −5𝜋 √2 𝑠𝑖𝑛 ( )= 4 2 Sir Kenneth E. Abaja  BASIC CALCULUS 20| PUP PARAÑAQUE IMPORTANT POINTS ON THE UNIT CIRCLE The next example summarizes all of the important ideas discussed thus far in the section. 5 Example: Suppose 𝛼 is an acute angle with cos(𝛼) = 13. 1. Find sin(𝛼) and use this to plot 𝛼 in standard position. 2. Find the sine and cosine of the following angles: 𝜋 (𝑎) 𝜃 = 𝜋 + 𝛼 (𝑏) 𝜃 = 2𝜋 − 𝛼 (𝑐) 𝜃 = 3𝜋 − 𝛼 (𝑑) 𝜃 = +𝛼 2 Solution: 5 1. We substitute cos(α) = into 𝑐𝑜𝑠 2 (𝛼) + 𝑠𝑖𝑛2 (𝛼) = 1 and using Pythagorean Identity, 13 12 sin(𝛼) = ± 13. Since α is an acute (and therefore Quadrant I) angle, sin(α) is positive. 12 Hence, sin(α) =. To plot α in standard position, we begin our rotation on the positive 13 5 12 x-axis to the ray which contains the point (cos(α),sin(α)) = (13 , 13 ). Sir Kenneth E. Abaja  BASIC CALCULUS 21| PUP PARAÑAQUE 2. (a) To find the cosine and sine of θ = π + α, we first plot θ in standard position. We can imagine the sum of the angles π+α as a sequence of two rotations: a rotation of π radians followed by a rotation of α radians. We see that α is the reference angle for θ, 5 so by The Reference Angle Theorem, cos(θ) = ± cos(α) = ± ± 13 and sin(θ) = ± sin(α) = 12 ± 13. Since the terminal side of θ falls in Quadrant III, both cos(θ) and sin(θ) are 5 12 negative, hence, cos(θ) = −13, and sin(θ) = − 13. (b) Rewriting θ = 2π − α as θ = 2π + (−α), we can plot θ by visualizing one complete revolution counter-clockwise followed by a clockwise revolution, or ‘backing up,’ of α radians. We see that α is θ’s reference angle, and since θ is a Quadrant IV angle, the 5 12 Reference Angle Theorem gives: cos(θ) = 13 and sin(θ) = − 13. Sir Kenneth E. Abaja  BASIC CALCULUS 22| PUP PARAÑAQUE (c) Taking a cue from the previous problem, we rewrite θ = 3π − α as θ = 3π + (−α). The angle 3π represents one and a half revolutions counter-clockwise, so that when we ‘back up’ α radians, we end up in Quadrant II. Using the Reference Angle Theorem, 5 12 we get cos(θ) = −13 and sin(θ) = 13. 𝜋 (d) To plot θ = π 2 +α, we first rotate 2 , radians and follow up with α radians. The reference angle here is not α, so The Reference Angle Theorem is not immediately applicable. (It’s important that you see why this is the case. Take a moment to think about this before reading on.) Let Q(x, y) be the point on the terminal side of θ which lies on the Unit Circle so that x = cos(θ) and y = sin(θ). Once we graph α in standard position, we use the fact that equal angles subtend equal chords to show that the dotted 12 lines in the figure below are equal. Hence, x = cos(θ) = − 13. Similarly, we find y = sin(θ) 5 = 13 BEYOND THE UNIT CIRCLE We began the section with a quest to describe the position of a particle experiencing circular motion. In defining the cosine and sine functions, we assigned to each angle a position on the Unit Circle. In this subsection, we broaden our scope to include circles of radius r centered at the origin. Consider for the moment the acute angle θ drawn below in standard position. Let Q(x, y) be the point on the terminal side of θ which lies on the circle 𝑥 2 + 𝑦 2 = 𝑟 2 , and let 𝑃(𝑥 ′ , 𝑦 ′ ) be the point on the terminal side of θ which lies on the Unit Circle. Now consider dropping perpendiculars from P and Q to create two right triangles, ∆OPA and ∆OQB. These triangles are 𝑥 𝑟 similar, thus it follows that 𝑥′ = 1 = 𝑟, 𝑠𝑜 𝑥 = 𝑟𝑥 ′ and, similarly, we find 𝑦 = 𝑟𝑦′. Since, by definition, 𝑥 ′ = cos(𝜃) 𝑎𝑛𝑑 𝑦 ′ = sin(𝜃), we get the coordinates of Q to be x = rcos(θ) and y = rsin(θ). By Sir Kenneth E. Abaja  BASIC CALCULUS 23| PUP PARAÑAQUE reflecting these points through the x-axis, y-axis and origin, we obtain the result for all non- quadrantal angles θ, and we leave it to the reader to verify these formulas hold for the quadrantal angles. Not only can we describe the coordinates of Q in terms of cos(θ) and sin(θ) but since the radius of the circle is 𝑟 = √𝑥 2 + 𝑦 2 , we can also express cos(θ) and sin(θ) in terms of the coordinates of Q. These results are summarized in the following theorem.\ Theorem. If 𝑄(𝑥, 𝑦) is the point on the terminal side of an angle θ, plotted in standard position, which lies on the circle 𝑥 2 + 𝑦 2 = 𝑟 2 then 𝑥 = 𝑟 𝑐𝑜𝑠(𝜃) 𝑎𝑛𝑑 𝑦 = 𝑟 𝑠𝑖𝑛(𝜃). Moreover, 𝑥 𝑥 𝑦 𝑟 cos(𝜃) = = 𝑎𝑛𝑑 sin(𝜃) = = 𝑟 √𝑥 + 𝑦 2 2 𝑟 √𝑥 + 𝑦 2 2 Note that in the case of the Unit Circle we have 𝑟 = √𝑥 2 + 𝑦 2 = 1, so Theorem reduces to our definitions of cos(θ) and sin(θ). Examples: Suppose that the terminal side of an angle θ, when plotted in standard position, contains the point Q(4, −2). Find sin(θ) and cos(θ). Solution: Using Theorem above with x = 4 and y = −2, we find 𝑟 = √𝑥 2 + 𝑦 2 = √(4)2 + (−2)2 = 𝑥 4 2√5 𝑦 −2 −√5 √20 = 2√5 so that cos(𝜃) = 𝑟 = 2√5 = 5 𝑎𝑛𝑑 sin(𝜃) = 𝑟 = 2√5 = 5. Consider the generic right triangle below with corresponding acute angle θ. The side with length a is called the side of the triangle adjacent to θ; the side with length b is called the side of the triangle opposite θ; and the remaining side of length c (the side opposite the right angle) is called the hypotenuse. We now imagine drawing this triangle in Quadrant I so that the angle θ is in standard position with the adjacent side to θ lying along the positive x-axis. Sir Kenneth E. Abaja  BASIC CALCULUS 24| PUP PARAÑAQUE According to the Pythagorean Theorem, 𝑎 2 + 𝑏2 = 𝑐 2 , so that the point P(a, b) lies on a 𝑎 𝑏 circle of radius c. Theorem above tells us that cos(𝜃) = 𝑐 and sin(𝜃) = 𝑐 so we have determined the cosine and sine of θ in terms of the lengths of the sides of the right triangle. Thus we have the following theorem. Theorem. Suppose θ is an acute angle residing in a right triangle. If the length of the side adjacent to θ is a, the length of the side opposite θ is b, and the length of the hypotenuse is c, 𝑎 𝑏 then cos(𝜃) = 𝑐 and sin(𝜃) = 𝑐. Example: Find the measure of the missing angle and the lengths of the missing sides of: Solution: The first and easiest task is to find the measure of the missing angle. Since the sum of angles of a triangle is 180°, we know that the missing angle has measure 180° − 30°− 90°= 60°. We now proceed to find the lengths of the remaining two sides of the triangle. Let c denote the 7 7 length of the hypotenuse of the triangle. By Theorem above, we have cos (30°)= 𝑐 , or c = cos(30°) √3 14√3 cos(30°). Since cos (30° ) = 2 , we have, after the usual fraction gymnastics, c = 3. At this point, we have two ways to proceed to find the length of the side opposite the 30° angle, which we’ll denote b. We know the length of the adjacent side is 7 and the length of the hypotenuse is 14√3 , so we could use the Pythagorean Theorem to find the missing side and solve 72 + 𝑏2 = 3 2 14√3 𝑏 ( ) for b. Alternatively, we could use the theorem above, namely that sin (30°)=. Choosing 3 𝑐 14√3 1 7√3 the latter, we find 𝑏 = 𝑐 sin(30°) = ∙2 =. The triangle with all of its data is recorded below. 3 3 Sir Kenneth E. Abaja  BASIC CALCULUS 25| PUP PARAÑAQUE IV. ACTIVITIES/ ASSESSMENTS 1. Find the exact value of the cosine and sine of the given angle. Show your complete solution. 13𝜋 a. θ = − 2 23𝜋 b. θ = 6 c. θ = 117𝜋 7𝜋 d. θ = 6 10𝜋 e. θ = 3 5𝜋 f. θ= −4 2. Use the results developed throughout the section to find the requested value. 7 a. If sin(θ) = − 25 with θ in Quadrant IV, what is cos(θ)? 4 b. If cos(θ) = with θ in Quadrant I, what is sin(θ)? 9 5 c. If sin(θ) = with θ in Quadrant II, what is cos(θ)? 13 2 d. If cos(θ) = − 11 with θ in Quadrant III, what is sin(θ)? 2 e. If sin(θ) = − 3 with θ in Quadrant III, what is cos(θ)? 2√5 𝜋 f. If sin(θ) = with θ in < 𝜃 < 𝜋, what is cos(θ)? 5 2 √10 5𝜋 g. If cos(θ) = with θ in 2𝜋 < 𝜃 < , what is sin(θ)? 10 2 3. Find θ, a, and c 4. Assume that θ is an acute angle in a right triangle and use Theorem 10.4 to find the requested side. a. If θ = 12° and the side adjacent to θ has length 4, how long is the hypotenuse? b. If θ = 78.123° and the hypotenuse has length 5280, how long is the side adjacent to θ? c. If θ = 37.5° and the side opposite θ has length 306, how long is the side adjacent to θ? d. If θ = 5° and the hypotenuse has length 10, how long is the side adjacent to θ? e. If θ = 59° and the side opposite θ has length 117.42, how long is the hypotenuse? WORKS CITED Stitz, Carl and Zeager, Jeff. College Algebra and Trigonometry. 2010 Sir Kenneth E. Abaja  BASIC CALCULUS 26| PUP PARAÑAQUE MODULE 6 6 Circular Trigonometric Functions and Identities Sir Kenneth E. Abaja  BASIC CALCULUS 27| PUP PARAÑAQUE I. OVERVIEW In the previous module, we defined cos(θ) and sin(θ) for angles θ using the coordinate values of points on the Unit Circle. As such, these functions earn the moniker circular functions. It turns out that cosine and sine are just two of the six commonly used circular functions which we define below. II. MODULE OBJECTIVE At the end of the lesson, the students should be able to: 1. Introduce the Sum and Difference Identity, Double Angle and Half Angle Identities. 2. Use the Identities properly. III. COURSE MATERIALS THE SIX CIRCULAR FUNCTIONS AND FUNDAMENTAL IDENTITES Definition. The Circular Functions: Suppose θ is an angle plotted in standard position and P(x, y) is the point on the terminal side of θ which lies on the Unit Circle. The cosine of θ, denoted cos(θ), is defined by cos(θ) = x. The sine of θ, denoted sin(θ), is defined by sin(θ) = y. 1 The secant of θ, denoted sec(θ), is defined by sec(θ) = 𝑥, provided x≠0. 1 The cosecant of θ, denoted csc(θ), is defined by csc(θ) = 𝑦, provided y≠0. 𝑦 The tangent of θ, denoted tan(θ), is defined by tan(θ) = 𝑥 , provided x≠0. 𝑥 The cotangent of θ, denoted cot(θ), is defined by cot(θ) = 𝑦, provided y≠0. Consider the acute angle θ below in standard position. Let P(x, y) denote, as usual, the point on the terminal side of θ which lies on the Unit Circle and let Q(1, y’) denote the point on the terminal side of θ which lies on the vertical line x = 1. The word ‘tangent’ comes from the Latin meaning ‘to touch,’ and for this reason, the line x = 1 is called a tangent line to the Unit Circle since it intersects, or ‘touches’, the circle at only one point, namely (1, 0). Dropping perpendiculars from P and Q creates a pair of similar triangles 𝑦′ 1 𝑦 ∆OPA and ∆OQB. Thus 𝑦 = 𝑥 which y’= 𝑥 = tan (𝜃), where this last equality comes from applying Sir Kenneth E. Abaja  BASIC CALCULUS 28| PUP PARAÑAQUE the definition above. We have just shown that for acute angles θ, tan(θ) is the y-coordinate of the point on the terminal side of θ which lies on the line x = 1 which is tangent to the Unit Circle. Now the word ‘secant’ means ‘to cut’, so a secant line is any line that ‘cuts through’ a circle at two points.2 The line containing the terminal side of θ is a secant line since it intersects the Unit Circle in Quadrants I and III. With the point P lying on the Unit Circle, the length of the hypotenuse of ∆OPA is 1. If we let h denote the length of the hypotenuse of ∆OQB, we have from similar ℎ 1 1 triangles that 1 = 𝑥 𝑜𝑟 𝑥 = 𝑥 = sec(𝜃). Hence for an acute angle θ, sec(θ) is the length of the line segment which lies on the secant line determined by the terminal side of θ and ‘cuts off’ the tangent line x = 1. Not only do these observations help explain the names of these functions, they serve as the basis for a fundamental inequality needed for Calculus which we’ll explore in the Exercises. Of the six circular functions, only cosine and sine are defined for all angles. Since cos(θ) = x and sin(θ) = y in the definition above, it is customary to rephrase the remaining four circular functions in terms of cosine and sine. The following theorem is a result of simply replacing x with cos(θ) and y with sin(θ). Theorem. Reciprocal and Quotient Identities: 1 ▪ sec(θ) = cos(𝜃), provided cos(θ) ≠ 0; if cos(θ) = 0, sec(θ) is undefined. 1 ▪ csc(θ) = sin(𝜃), provided sin(θ) ≠ 0; if sin(θ) = 0, csc(θ) is undefined. sin(𝜃) ▪ tan(θ) = cos(𝜃), provided cos(θ) ≠ 0; if cos(θ) = 0, tan(θ) is undefined. cos(𝜃) ▪ cot(θ) = , provided sin(θ) ≠ 0; if sin(θ) = 0, cot(θ) is undefined. sin(𝜃) Example: Find the indicated value, if it exists. 1. 𝑠𝑒𝑐 (60°) 7𝜋 2. 𝑐𝑠𝑐 ( 4 ) 3𝜋 3. 𝑡𝑎𝑛 (𝜃), where θ is any angle coterminal with. 2 4. 𝑐𝑜𝑠 (𝜃), 𝑤ℎ𝑒𝑟𝑒 𝑐𝑠𝑐(𝜃) = −√ 5 𝑎𝑛𝑑 𝜃 𝑖𝑠 𝑎 𝑄𝑢𝑎𝑑𝑟𝑎𝑛𝑡 𝐼𝑉 𝑎𝑛𝑔𝑙𝑒. 5. 𝑠𝑖𝑛 (𝜃), 𝑤ℎ𝑒𝑟𝑒 𝑡𝑎𝑛(𝜃) = 3 𝑎𝑛𝑑 𝜋 < 𝜃 < 3 Solution: 1 1 1. According to reciprocal identity, 𝑠𝑒𝑐 (60°) = = 1⁄. Hence, 𝑠𝑒𝑐 (60°) = 2. 𝑐𝑜𝑠(60°) 2 7𝜋 √2 7𝜋 1 2 2. Since 𝑠𝑖𝑛 ( 4 ) = − 2 , 𝑐𝑠𝑐 ( 4 ) = − √2 = − − √2. √2 2 3𝜋 3𝜋 3𝜋 3. If θ is coterminal with , then 𝑐𝑜𝑠(𝜃) = 𝑐𝑜𝑠 ( ) = 0 and 𝑠𝑖𝑛(𝜃) = 𝑠𝑖𝑛 ( ) = −1. 2 2 2 𝑠𝑖𝑛(𝜃) −1 Attempting to compute 𝑡𝑎𝑛(𝜃) = 𝑐𝑜𝑠(𝜃) results in 0 , so 𝑡𝑎𝑛(𝜃) is undefined. 1 1 √5 4. We are given that 𝑐𝑠𝑐(𝜃) = = − √5 𝑠𝑜 𝑠𝑖𝑛(𝜃) = − =−. We can use the 𝑠𝑖𝑛(𝜃) √5 5 Pythagorean Identity, 𝑐𝑜𝑠 (𝜃) + 𝑠𝑖𝑛 (𝜃) = 1, to find cos(θ) by knowing sin(θ). 2 2 Sir Kenneth E. Abaja  BASIC CALCULUS 29| PUP PARAÑAQUE 2 √5 4 2√5 Substituting, we get 𝑐𝑜𝑠 2 (𝜃) + ( 5 ) = 1, which gives 𝑐𝑜𝑠 2 (𝜃) = 5, or 𝑐𝑜𝑠 (𝜃) = ± 5. Since 2√5 θ is a Quadrant IV angle, cos(𝜃) > 0, so 𝑐𝑜𝑠 (𝜃) = 5. 𝑠𝑖𝑛(𝜃) 5. If 𝑡𝑎𝑛(𝜃) = 3, then 3. Be careful - this does NOT mean we can take 𝑠𝑖𝑛(𝜃) = 𝑐𝑜𝑠(𝜃) 𝑠𝑖𝑛(𝜃) 3 and 𝑐𝑜𝑠(𝜃) = 1. Instead, from 𝑐𝑜𝑠(𝜃) = 3. we get: 𝑠𝑖𝑛 (𝜃) = 3 𝑐𝑜𝑠 (𝜃). To relate 𝑐𝑜𝑠(𝜃) and 𝑠𝑖𝑛(𝜃), we once again employ the Pythagorean Identity, 𝑐𝑜𝑠 2 (𝜃) + 𝑠𝑖𝑛2 (𝜃) = 1. = 1. 1 Solving 𝑠𝑖𝑛 (𝜃) = 3 𝑐𝑜𝑠 (𝜃) for 𝑐𝑜𝑠 (𝜃), we find 𝑐𝑜𝑠 (𝜃) = sin(𝜃). Substituting this into 3 1 2 9 the Pythagorean Identity, we find (3 sin(𝜃)) + 𝑠𝑖𝑛 (𝜃) = 1. Solving, we get 𝑠𝑖𝑛2 (𝜃) = 10 2 3√10 3𝜋 so 𝑠𝑖𝑛(𝜃) = ± 10. Since 𝜋 < 𝜃 < 2 , θ is a Quadrant III angle. This means 𝑠𝑖𝑛(𝜃) < 0, 3√10 so our final answer is 𝑠𝑖𝑛(𝜃) = 10. Theorem. Generalized Reference Angle Theorem: The values of the circular functions of an angle, if they exist, are the same, up to a sign, of the corresponding circular functions of its reference angle. More specifically, if α is the reference angle for θ, then: 𝑐𝑜𝑠(𝜃) = ± 𝑐𝑜𝑠(𝛼), 𝑠𝑖𝑛(𝜃) = ± 𝑠𝑖𝑛(𝛼), 𝑠𝑒𝑐(𝜃) = ± 𝑠𝑒𝑐(𝛼), 𝑐𝑠𝑐(𝜃) = ± 𝑐𝑠𝑐(𝛼), 𝑡𝑎𝑛(𝜃) = ± 𝑡𝑎𝑛(𝛼) 𝑎𝑛𝑑 𝑐𝑜𝑡(𝜃) = ± 𝑐𝑜𝑡(𝛼). The choice of the (±) depends on the quadrant in which the terminal side of θ lies. Example: Find the indicated value, if it exists. 1. sec(𝜃) = 2 2. 𝑡𝑎𝑛 (𝜃) = √3. 3. 𝑐𝑜𝑡 (𝜃) = −1 Solution: 1 1 1. To solve 𝑠𝑒𝑐(𝜃) = 2, we convert to cosines and get = 2 𝑜𝑟 𝑐𝑜𝑠(𝜃) =. So we know 𝑐𝑜𝑠(𝜃) 2 𝜋 5𝜋 the answer is: 𝜃 = + 2𝜋𝑘 𝑜𝑟 𝜃 = + 2𝜋𝑘 for integers k. 3 3 Sir Kenneth E. Abaja  BASIC CALCULUS 30| PUP PARAÑAQUE 𝜋 2. From the table of common values, we see 𝑡𝑎𝑛 ( 3 ) = √3. According to the Reference Angle Theorem, we know the solutions to 𝑡𝑎𝑛(𝜃) = √3. must, therefore, have a reference 𝜋 angle of 3. Our next task is to determine in which quadrants the solutions to this equation 𝑦 lie. Since tangent is defined as the ratio 𝑥 of points (𝑥, 𝑦) on the Unit Circle with 𝑥 ≠ 0, tangent is positive when 𝑥 and 𝑦 have the same sign (i.e., when they are both positive or both negative.) This happens in Quadrants I and III. In Quadrant I, we get the solutions: 𝜋 4𝜋 𝜃 = 3 + 2𝜋𝑘 for integers k, and for Quadrant III, we get 𝜃 = 3 + 2𝜋𝑘 for integers k. While these descriptions of the solutions are correct, they can be combined into one list as 𝜃 = 𝜋 + 𝜋𝑘 for integers k. The latter form of the solution is best understood looking at the 3 geometry of the situation in the diagram below. 𝜋 3. From the table of common values, we see that has a cotangent of 1, which means the 4 𝜋 solutions to 𝑐𝑜𝑡(𝜃) = −1 have a reference angle of 4. To find the quadrants in which our 𝑥 solutions lie, we note that 𝑐𝑜𝑡(𝜃) = \for a point (𝑥, 𝑦) on the Unit Circle where 𝑦 ≠ 0. If 𝑦 𝑐𝑜𝑡(𝜃) is negative, then x and y must have different signs (i.e., one positive and one negative.) Hence, our solutions lie in Quadrants II and IV. Our Quadrant II solution is 𝜃 = 3𝜋 7𝜋 + 2𝜋𝑘, and for Quadrant IV, we get 𝜃 = 4 + 2𝜋𝑘 for integers k. Can these lists be 4 3𝜋 combined? Indeed they can - one such way to capture all the solutions is: 𝜃 = + 𝜋𝑘 4 for integers k. We have already seen the importance of identities in trigonometry. Our next task is to use the Reciprocal and Quotient Identities coupled with the Pythagorean Identity to derive new Pythagorean-like identities for the remaining four circular functions. Assuming cos(𝜃) ≠ 0, we may 𝑠𝑖𝑛2 (𝜃) 1 start with 𝑐𝑜𝑠 2 (𝜃) + 𝑠𝑖𝑛2 (𝜃) = 1 and divide both sides by 𝑐𝑜𝑠 2 (𝜃) to obtain 1 + 𝑐𝑜𝑠 2(𝜃) = 𝑐𝑜𝑠 2(𝜃). Using properties of exponents along with the Reciprocal and Quotient Identities, this reduces to 1 + 𝑡𝑎𝑛2 (𝜃) = 𝑠𝑒𝑐 2 (𝜃). If 𝑠𝑖𝑛(𝜃) ≠ 0, we can divide both sides of the identity 𝑐𝑜𝑠 2 (𝜃) + 𝑠𝑖𝑛2 (𝜃) = 1 by Sir Kenneth E. Abaja  BASIC CALCULUS 31| PUP PARAÑAQUE 𝑠𝑖𝑛2 (𝜃), apply Reciprocal and Quotient Identities once again, and obtain 𝑐𝑜𝑡 2 (𝜃) + 1 = 𝑐𝑠𝑐 2 (𝜃). These three Pythagorean Identities are worth memorizing and they, along with some of their other common forms, are summarized in the following theorem. Theorem. The Pythagorean Identities: 1. 𝑐𝑜𝑠 2 (𝜃) + 𝑠𝑖𝑛2 (𝜃) = 1 Common Alternate Forms: ▪ 𝑐𝑜𝑠 2 (𝜃) = 1 − 𝑠𝑖𝑛2 (𝜃) ▪ 𝑠𝑖𝑛2 (𝜃) = 1 − 𝑐𝑜𝑠 2 (𝜃) 2. 1 + 𝑡𝑎𝑛2 (𝜃) = 𝑠𝑒𝑐 2 (𝜃), provided that 𝑐𝑜𝑠(𝜃) ≠ 0. Common Alternate Forms: ▪ 𝑠𝑒𝑐 2 (𝜃) − 𝑡𝑎𝑛2 (𝜃) = 1 ▪ 𝑠𝑒𝑐 2 (𝜃) − 1 = 𝑡𝑎𝑛2 (𝜃) 3. 1 + 𝑐𝑜𝑡 2 (𝜃) = 𝑐𝑠𝑐 2 (𝜃), provided that 𝑠𝑖𝑛(𝜃) ≠ 0. Common Alternate Forms: ▪ 𝑐𝑠𝑐 2 (𝜃) − 𝑐𝑜𝑡 2 (𝜃) = 1 ▪ 𝑐𝑠𝑐 2 (𝜃) − 1 = 𝑐𝑜𝑡 2 (𝜃) Examples: Verify the following identities. Assume that all quantities are defined. 1. 𝑡𝑎𝑛(𝜃) = 𝑠𝑖𝑛(𝜃) 𝑠𝑒𝑐(𝜃) 2. (𝑠𝑒𝑐(𝜃) − 𝑡𝑎𝑛(𝜃))(𝑠𝑒𝑐(𝜃) + 𝑡𝑎𝑛(𝜃)) = 1 𝑠𝑒𝑐(𝜃) 1 3. 1−𝑡𝑎𝑛(𝜃) = 𝑐𝑜𝑠(𝜃)−𝑠𝑖𝑛(𝜃) 3 3 4. 6𝑠𝑒𝑐(𝜃)𝑡𝑎𝑛(𝜃) = − 1−𝑠𝑖𝑛(𝜃) 1+𝑠𝑖𝑛(𝜃) Solution: In verifying identities, we typically start with the more complicated side of the equation and use known identities to transform it into the other side of the equation. 1 1. Starting with the right hand side of the equation, we use sec(𝜃) = 𝑐𝑜𝑠(𝜃) and find: 1 𝑠𝑖𝑛(𝜃) 𝑠𝑖𝑛(𝜃) 𝑠𝑒𝑐(𝜃) = 𝑠𝑖𝑛(𝜃) = = tan(𝜃) 𝑐𝑜𝑠(𝜃) 𝑐𝑜𝑠(𝜃) 2. Expanding the left hand side of the equation gives: (𝑠𝑒𝑐(𝜃) − 𝑡𝑎𝑛(𝜃))(𝑠𝑒𝑐(𝜃) + 𝑡𝑎𝑛(𝜃)) = 𝑠𝑒𝑐 2 (𝜃) − 𝑡𝑎𝑛2 (𝜃). According to Pythagorean Identity, 𝑠𝑒𝑐 2 (𝜃) − 𝑡𝑎𝑛2 (𝜃) = 1. We have verified the identity. 3. While both sides of our last identity contain fractions, the left side affords us more 1 𝑠𝑖𝑛(𝜃) opportunities to use our identities. Substituting sec(𝜃) = 𝑐𝑜𝑠(𝜃) and tan(𝜃) = 𝑐𝑜𝑠(𝜃), we get: 1 1 𝑠𝑒𝑐(𝜃) 𝑐𝑜𝑠(𝜃) 𝑐𝑜𝑠(𝜃) 1 = = = 1 − 𝑡𝑎𝑛(𝜃) 𝑠𝑖𝑛(𝜃) 𝑐𝑜𝑠(𝜃) − 𝑠𝑖𝑛(𝜃) 𝑐𝑜𝑠(𝜃) − 𝑠𝑖𝑛(𝜃) 1− 𝑐𝑜𝑠(𝜃) 𝑐𝑜𝑠(𝜃) 4. The right hand side of the equation seems to hold more promise. We get common denominators and add: Sir Kenneth E. Abaja  BASIC CALCULUS 32| PUP PARAÑAQUE 3 3 3(1 + 𝑠𝑖𝑛(𝜃)) − 3(1 − 𝑠𝑖𝑛(𝜃)) − = 1 − 𝑠𝑖𝑛(𝜃) 1 + 𝑠𝑖𝑛(𝜃) (1 − 𝑠𝑖𝑛(𝜃))(1 + 𝑠𝑖𝑛(𝜃)) 3 + 3𝑠𝑖𝑛(𝜃) − (3 − 3𝑠𝑖𝑛(𝜃)) = 1 − 𝑠𝑖𝑛2 (𝜃) 3 + 3𝑠𝑖𝑛(𝜃) − 3 + 3𝑠𝑖𝑛(𝜃)) = 1 − 𝑠𝑖𝑛2 (𝜃) 6𝑠𝑖𝑛(𝜃) = 1 − 𝑠𝑖𝑛2 (𝜃) 6𝑠𝑖𝑛(𝜃) 6𝑠𝑖𝑛(𝜃) 6𝑠𝑖𝑛(𝜃) From Pythagorean Identity, 1 − 𝑠𝑖𝑛2 (𝜃) = 𝑐𝑜𝑠 2 (𝜃). Thus, 1−𝑠𝑖𝑛2(𝜃) = 𝑐𝑜𝑠 2 (𝜃) = 𝑐𝑜𝑠(𝜃) cos (𝜃). The 𝑠𝑖𝑛(𝜃) 6𝑠𝑖𝑛(𝜃) 6𝑡𝑎𝑛(𝜃) 1 Qoutient Identity says cos (𝜃) = tan (𝜃). Making 𝑐𝑜𝑠(𝜃) cos (𝜃) = cos (𝜃). From the reciprocal identity, cos (𝜃) = 6𝑠𝑖𝑛(𝜃) sec(𝜃). Therefore, = 6𝑠𝑒𝑐(𝜃)𝑡𝑎𝑛(𝜃). 1−𝑠𝑖𝑛2(𝜃) Strategies for Verifying Identities ▪ Try working on the more complicated side of the identity. ▪ Use the Reciprocal and Quotient Identities to write functions on one side of the identity in terms of the functions on the other side of the identity. Simplify the resulting complex fractions. ▪ Add rational expressions with unlike denominators by obtaining common denominators. ▪ Use the Pythagorean Identities to ‘exchange’ sines and cosines, secants and tangents, cosecants and cotangents, and simplify sums or differences of squares to one term. ▪ Multiply numerator and denominator by Pythagorean Conjugates in order to take advantage of the Pythagorean Identities. ▪ If you find yourself stuck working with one side of the identity, try starting with the other side of the identity and see if you can find a way to bridge the two parts of your work. Theorem. Sum and Difference Identities: For all possible angles 𝛼 and 𝛽. ▪ 𝑐𝑜𝑠(𝛼 ± 𝛽) = 𝑐𝑜𝑠(𝛼) 𝑐𝑜𝑠(𝛽) ∓ 𝑠𝑖𝑛(𝛼) 𝑠𝑖𝑛(𝛽) ▪ 𝑠𝑖𝑛(𝛼 ± 𝛽) = 𝑠𝑖𝑛(𝛼) 𝑐𝑜𝑠(𝛽) ± 𝑐𝑜𝑠(𝛼) 𝑠𝑖𝑛(𝛽) tan(𝛼) ± tan(β) ▪ tan(𝛼 ± 𝛽) = 1 ∓(tan(𝛼) tan(β)) Example: Find the exact value of 𝑐𝑜𝑠(𝜃), 𝑠𝑖𝑛 (𝜃) and 𝑡𝑎𝑛 (𝜃) if 𝜃 = 15° a. 𝑐𝑜𝑠(15°) = cos(45° − 30°) = 𝑐𝑜𝑠(45°) 𝑐𝑜𝑠(30°) + 𝑠𝑖𝑛(45°) 𝑠𝑖𝑛(30°) √2 √3 √2 1 = ( )( )+ ( )( ) 2 2 2 2 √6 + √2 = 2 Sir Kenneth E. Abaja  BASIC CALCULUS 33| PUP PARAÑAQUE b. 𝑠𝑖𝑛(15°) = cos(45° − 30°) = 𝑠𝑖𝑛(45°) 𝑐𝑜𝑠(30°) − 𝑐𝑜𝑠(45°) 𝑠𝑖𝑛(30°) √2 √3 √2 1 = ( )( )+ ( )( ) 2 2 2 2 √6 − √2 = 2 tan(45°) − tan(30°) c. 𝑡𝑎𝑛(15°) = 𝑡𝑎𝑛(45° − 30°) = 1+(tan(45°) tan(30°)) √3 3 − √3 1− 3 = = 3 √3 √3 1 + (1 × 3 ) 1 + 3 3 − √3 3 3 − √3 = = 3 + √3 3 + √3 3 Perform the rationalization process by getting the conjugate of the denominator. 2 3 − √3 3 − √3 (3 − √3) = × = 2 3 + √3 3 − √3 (3)2 − (√3) 9 − 6√3 + 3 12 − 6√3 = = 9−3 6 𝑡𝑎𝑛(15°) = 2 + √3 Theorem. Double Angle Identities: For all possible angles 𝜃. 𝑐𝑜𝑠 2 (𝜃) − 𝑠𝑖𝑛2 (𝜃) ▪ 𝑐𝑜𝑠(2𝜃) = { 2𝑐𝑜𝑠 2 (𝜃) − 1 1 − 2𝑠𝑖𝑛2 (𝜃) ▪ 𝑠𝑖𝑛(2𝜃) = 2 𝑠𝑖𝑛(𝜃) 𝑐𝑜𝑠(𝜃) 2 tan(θ) ▪ tan(2𝜃) = 1−tan2(𝛼) Example: 1. Suppose 𝑃(−3, 4) lies on the terminal side of 𝜃 when 𝜃 is plotted in standard position. Find 𝑐𝑜𝑠(2𝜃) and 𝑠𝑖𝑛(2𝜃) and determine the quadrant in which the terminal side of the angle 2𝜃 lies when it is plotted in standard position. 𝜋 𝜋 2. If 𝑠𝑖𝑛(𝜃) = 𝑥 for – 2 ≤ 𝜃 ≤ 2 , find an expression for 𝑠𝑖𝑛(2𝜃) in terms of 𝑥. 2 𝑡𝑎𝑛(𝜃) 3. Verify the identity: 𝑠𝑖𝑛(2𝜃) = 1+𝑡𝑎𝑛2(𝜃). Solution: 1. With 𝑥 = −3 and 𝑦 = 4, we find 𝑟 = √𝑥 2 + 𝑦 2 = √(−3)2 + (4)2 = 5. Hence, 𝑐𝑜𝑠(𝜃) = 3 4 − and sin(θ) =. Applying the Double Angle Identity, we get 𝑐𝑜𝑠(2𝜃) = 𝑐𝑜𝑠 2 (𝜃) − 5 5 Sir Kenneth E. Abaja  BASIC CALCULUS 34| PUP PARAÑAQUE 3 2 4 2 7 4 3 24 𝑠𝑖𝑛2 (𝜃) = (− 5 ) − ( 5 ) = 5, and 𝑠𝑖𝑛(2𝜃) = 2 𝑠𝑖𝑛(𝜃) 𝑐𝑜𝑠(𝜃) = 2 ( 5 ) (− 5 ) = − 5. Since both cosine and sine of 2𝜃 are negative, the terminal side of 2𝜃, when plotted in standard position, lies in Quadrant III. 2. Here, ‘𝑥’ is just a variable - it does not necessarily represent the x-coordinate of the point on The Unit Circle which lies on the terminal side of 𝜃, assuming 𝜃 is drawn in standard position. Here, 𝑥 represents the quantity 𝑠𝑖𝑛(2𝜃), and what we wish to know is how to express 𝑠𝑖𝑛(2𝜃) in terms of 𝑥. Since 𝑠𝑖𝑛(2𝜃) = 2 𝑠𝑖𝑛(𝜃) 𝑐𝑜𝑠(𝜃), we need to write 𝑐𝑜𝑠(𝜃) in terms of 𝑥 to finish the problem. We substitute 𝑥 = 𝑠𝑖𝑛(𝜃) into the Pythagorean Identity, 𝜋 𝜋 𝑐𝑜𝑠 2 (𝜃) + 𝑠𝑖𝑛2 (𝜃) = 1, to get 𝑐𝑜𝑠 2 (𝜃) + 𝑥 2 = 1, or 𝑐𝑜𝑠(𝜃) = ±√1 − 𝑥 2. Since – 2 ≤ 𝜃 ≤ 2 , 𝑐𝑜𝑠(𝜃) 0, and thus 𝑐𝑜𝑠(𝜃) = √1 − 𝑥 2. Our final answer is 𝑠𝑖𝑛(2𝜃) = 2 𝑠𝑖𝑛(𝜃) 𝑐𝑜𝑠(𝜃) = 2𝑥 𝑠𝑖𝑛(𝜃)√1 − 𝑥 2. 3. We start with the right hand side of the identity and note that 1 + 𝑡𝑎𝑛2 (𝜃) = 𝑠𝑒𝑐 2 (𝜃). From this point, we use the Reciprocal and Quotient Identities to rewrite 𝑡𝑎𝑛(𝜃) and 𝑠𝑒𝑐(𝜃) in terms of 𝑐𝑜𝑠(𝜃) and 𝑠𝑖𝑛(𝜃): 𝑠𝑖𝑛(𝜃) 2 𝑡𝑎𝑛(𝜃) 2 𝑡𝑎𝑛(𝜃) 2 𝑐𝑜𝑠(𝜃) = = = 2 sin(𝜃) 𝑐𝑜𝑠(𝜃) = 2 sin(2𝜃) 1 + 𝑡𝑎𝑛2 (𝜃) 𝑠𝑒𝑐 2 (𝜃) 1 𝑐𝑜𝑠 2 (𝜃) Theorem. Power Reduction Formula: For all angles 𝜃. 1+𝑐𝑜𝑠(2𝜃) ▪ 𝑐𝑜𝑠 2 (𝜃) = 2 1−𝑐𝑜𝑠(2𝜃) ▪ 𝑠𝑖𝑛2 (𝜃) = 2 Example: Rewrite 𝑠𝑖𝑛2 (𝜃) 𝑐𝑜𝑠 2 (𝜃) as a sum and difference of cosines to the first power. Solution: We begin with a straightforward application of the power Reduction Formula 1−𝑐𝑜𝑠(2𝜃) 1+𝑐𝑜𝑠(2𝜃) 𝑠𝑖𝑛2 (𝜃) 𝑐𝑜𝑠 2 (𝜃) = ( )( ) 2 2 1 1 1 = (1 − 𝑐𝑜𝑠 2 (2𝜃)) = − 𝑐𝑜𝑠 2 (2𝜃) 4 4 4 Next, we apply the power reduction formula to 𝑐𝑜𝑠 2 (𝜃) to finish the reduction 1 1 𝑠𝑖𝑛2 (𝜃) 𝑐𝑜𝑠 2 (𝜃) = 4 − 4 𝑐𝑜𝑠 2 (𝜃) 1 1 1 + 𝑐𝑜𝑠(2(2𝜃)) = − ( ) 4 4 2 1 1 1 = − − cos(4𝜃) 4 8 8 1 1 = − cos(4𝜃) 8 8 Sir Kenneth E. Abaja  BASIC CALCULUS 35| PUP PARAÑAQUE Theorem. Half Angle Identities: For all applicable angles 𝜃. 𝜃 1+𝑐𝑜𝑠(𝜃) ▪ 𝑐𝑜𝑠 ( 2 ) = ±√ 2 𝜃 1−𝑐𝑜𝑠(𝜃) ▪ 𝑠𝑖𝑛 ( 2 ) = ±√ 2 𝜃 1−𝑐𝑜𝑠(𝜃) ▪ 𝑐𝑜𝑠 ( 2 ) = ±√1+𝑐𝑜𝑠(𝜃) Example: 1. Use a half angle formula to find the exact value of 𝑐𝑜𝑠 (15°). 3 𝜃 2. Suppose −𝜋 ≤ 𝜃 ≤ 0 with 𝑐𝑜𝑠(𝜃) = − 5. Find 𝑠𝑖𝑛 ( 2 ).. Solution: 30° 1. To use the half angle formula, we note that 15° = and since 15° is a Quadrant I angle, 2 its cosine is positive. Thus we have √3 1 + 𝑐𝑜𝑠(30°) √1 + 2 𝑐𝑜𝑠 (15°) = √ = 2 2 √3 √1 + 2 2 √2 + √3 √2 + √3 = ∙ = = 2 2 4 2 Back in Sum and Difference Identity, we found 𝑐𝑜𝑠 (15°) by using the difference formula for cosine. In that case, we determined 𝑐𝑜𝑠 (15°) = √6 + √2. The student is encouraged to prove that these two expressions are equal. 𝜋 𝜃 𝜃 2. If −𝜋 ≤ 𝜃 ≤ 0, − 2 ≤ 2 ≤ 0, which means 𝑖𝑛 ( 2 ) < 0. 3 𝜃 √ 1 − 𝑐𝑜𝑠(𝜃) √1 − (− 5) 𝑖𝑛 ( ) = − =− 2 2 2 3 1+ 5 =− √ 5 ∙ = −√ 8 = 2√5 2 5 10 5 1−𝑐𝑜𝑠(2𝜃) 1+𝑐𝑜𝑠(2𝜃) 𝑠𝑖𝑛2 (𝜃) 𝑐𝑜𝑠 2 (𝜃) = ( )( ) 2 2 1 1 1 = (1 − 𝑐𝑜𝑠 2 (2𝜃)) = − 𝑐𝑜𝑠 2 (2𝜃) 4 4 4 Sir Kenneth E. Abaja  BASIC CALCULUS 36| PUP PARAÑAQUE IV. ACTIVITIES/ ASSESSMENTS 1. Use the Sum and Difference Identities to find the exact value. You may have need of the Quotient, Reciprocal or Even / Odd Identities as well. a. 𝑐𝑜𝑠(75°) b. 𝑐𝑠𝑐(195°) c. 𝑠𝑖𝑛(105°) d. 𝑡𝑎𝑛(375°) e. 𝑠𝑒𝑐(165°) f. 𝑐𝑜𝑠(−105°) 𝜋 g. 𝑠𝑖𝑛 (12) 17𝜋 h. 𝑡𝑎𝑛 ( 12 ) 7𝜋 i. 𝑐𝑜𝑠 ( 12 ) 11𝜋 j. 𝑐𝑜𝑡 ( 12 ) √5 √10 𝜋 2. If 𝛼 is a Quadrant IV angle with 𝑐𝑜𝑠(𝛼) = , and 𝑠𝑖𝑛(𝛽) = , where 2 < 𝛽 < 𝜋, find 5 10 a. 𝑐𝑜𝑠(𝛼 + 𝛽) b. 𝑐𝑜𝑠(𝛼 − 𝛽) c. 𝑠𝑖𝑛(𝛼 + 𝛽) d. 𝑠𝑖𝑛(𝛼 − 𝛽) e. 𝑡𝑎𝑛(𝛼 + 𝛽) f. 𝑡𝑎𝑛(𝛼 − 𝛽) 3. Verify the Identity. a. 𝑐𝑜𝑠(𝜃 − 𝜋) = − 𝑐𝑜𝑠(𝜃) b. 𝑠𝑖𝑛(𝛼 + 𝛽) − 𝑠𝑖𝑛(𝛼 − 𝛽) = 2 𝑐𝑜𝑠(𝛼) 𝑠𝑖𝑛(𝛽) 𝜋 c. 𝑡𝑎𝑛 (𝜃 + 2 ) = − 𝑐𝑜𝑡(𝜃) 𝑡𝑎𝑛(𝑡 + ℎ) − 𝑡𝑎𝑛(𝑡) 𝑡𝑎𝑛(ℎ) 𝑠𝑒𝑐 2 (𝑡) d. =( ) ( ) ℎ ℎ 1 − 𝑡𝑎𝑛(𝑡) 𝑡𝑎𝑛(ℎ) e. (𝑐𝑜𝑠(𝜃) + sin (𝜃))2 = 1 + sin (2𝜃) 1 1 2 cos(θ) f. cos(θ) − sin(θ) + cos(θ)+ sin(θ) = cos(2θ) 4. Use the Half Angle Formulas to find the exact value. You may have need of the Quotient or Reciprocal as well. a. 𝑐𝑜𝑠 (75°) b. 𝑐𝑜𝑠(67.5°) c. 𝑠𝑖𝑛(105°) d. 𝑠𝑖𝑛(157.5°) e. 𝑡𝑎𝑛(112.5°) 𝜋 f. 𝑠𝑖𝑛 (12) 7𝜋 g. 𝑡𝑎𝑛 ( 8 ) 7𝜋 h. 𝑐𝑜𝑠 ( 12 ) Sir Kenneth E. Abaja  BASIC CALCULUS 37| PUP PARAÑAQUE 5. use the given information about θ to find the exact values of 𝑠𝑖𝑛(2𝜃), 𝑐𝑜𝑠(2𝜃), 𝑡𝑎𝑛(2𝜃), 𝜃 𝜃 𝜃 𝑐𝑜𝑠 ( 2 ) , 𝑠𝑖𝑛 ( 2 ) , 𝑎𝑛𝑑 𝑡𝑎𝑛 ( 2 ). 7 3𝜋 a. 𝑠𝑖𝑛(𝜃) = − 𝑤ℎ𝑒𝑟𝑒 < 𝜃 < 2𝜋 25 2 12 3𝜋 b. 𝑡𝑎𝑛(𝜃) = 𝑤ℎ𝑒𝑟𝑒 𝜋 < 𝜃 < 5 2 28 𝜋 c. 𝑐𝑜𝑠(𝜃) = 𝑤ℎ𝑒𝑟𝑒 0 < 𝜃 < 53 2 5 𝜋 d. 𝑠𝑖𝑛(𝜃) = − 𝑤ℎ𝑒𝑟𝑒 < 𝜃 < 𝜋 13 2 𝜋 e. 𝑡𝑎𝑛(𝜃) = − 2 𝑤ℎ𝑒𝑟𝑒 < 𝜃 < 𝜋 2 WORKS CITED Stitz, Carl and Zeager, Jeff. College Algebra and Trigonometry. 2010 Sir Kenneth E. Abaja  BASIC CALCULUS 38| PUP PARAÑAQUE MODULE 7 Inverse Trigonometric Functions Sir Kenneth E. Abaja  BASIC CALCULUS 39| PUP PARAÑAQUE I. OVERVIEW Every mathematical function, from the simplest to the most complex, has an inverse. In mathematics, inverse usually means opposite. For addition, the inverse is subtraction. For multiplication, it's division. And for trigonometric functions, it's the inverse trigonometric functions. Trigonometric functions are the functions of an angle. The term function is used to describe the relationship between two sets of numbers or variables. In modern mathematics, there are six basic trigonometric functions: sine, cosine, tangent, secant, cosecant, and cotangent. The inverse of these functions are inverse sine, inverse cosine, inverse tangent, inverse secant, inverse cosecant, and inverse cotangent. II. MODULE OBJECTIVE At the end of the lesson, the students should be able to: 1. Discuss different Inverse Trigonometric Functions. 2. Apply the inverse trigonometric function. III. COURSE MATERIALS THE IINVERSE TRIGONOMETRIC FUNCTIONS We already know about inverse operations. For example, addition and subtraction are inverse operations, and multiplication and division are inverse operations. Each operation does the opposite of its inverse. The idea is the same in trigonometry. Inverse trig functions do the opposite of the “regular” trig functions. For example: ▪ Inverse sine (𝑠𝑖𝑛−1 ) does the opposite of the sine. ▪ Inverse cosine (𝑐𝑜𝑠 −1 ) does the opposite of the cosine. ▪ Inverse tangent (𝑡𝑎𝑛−1 ) does the opposite of the tangent. In general, if you know the trigonometric ratio but not the angle, you can use the corresponding inverse trigonometric function to find the angle. TRIGONOMETRIC RATIO: SOH-CAH-TOA I’ve already introduce you from the previous module to the trigonometric ratio. Recall the lecture of sine and cosine beyond the unit circle. The trigonometric functions can all be defined as ratios of the sides of a right triangle. Since all right triangles conform to the Pythagorean Theorem, as long as the angles of two right triangles are the same, their sides will be proportional. Because of this, the ratios of one side to another will always be the same. Take a look at this example at the right. These triangles have the same angle measures, so their sides are proportional. Any ratio 6 3 of one side to another will be the same for both triangles. Hence, 10 = 5. Sir Kenneth E. Abaja  BASIC CALCULUS 40| PUP PARAÑAQUE By discovering that these ratios are the same for any sized right triangle (as long as they have the same angle measure), the trigonometric functions were discovered. These functions relate one angle of a triangle to the ratio of two of its sides. Because of these ratios, when an angle (other than the right angle) of a right triangle and at least one side are known, you can determine the length of the other sides using these ratios. And inversely, when the lengths of two sides are known, the angle measure can be determined. Since memorizing these ratios can prove to be difficult, there is a mnemonic that helps keep them straight. SOH CAH TOA is a helpful device to remember which ratio goes with which function. 𝑶𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑺𝑖𝑛𝑒 = 𝑯𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑨𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑪𝑜𝑠𝑖𝑛𝑒 = 𝑯𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑶𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑻𝑎𝑛𝑔𝑒𝑛𝑡 = 𝑨𝑑𝑗𝑎𝑐𝑒𝑛𝑡 Sine Inverse or Arcsin Follow the link for discussion Lectur