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Triple-Angle Formulas and Linear Combinations Bradley Hughes Larry Ottman Lori Jordan Mara Landers Andrea Hayes Brenda Meery Art Fortgang Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) To access a customizable version of this book, as well as other interactive content, visit www.ck12.org CK-12 Foundation is a non-profit organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook®, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. 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Printed: November 16, 2013 AUTHORS Bradley Hughes Larry Ottman Lori Jordan Mara Landers Andrea Hayes Brenda Meery Art Fortgang www.ck12.org C ONCEPT Concept 1. Triple-Angle Formulas and Linear Combinations 1 Triple-Angle Formulas and Linear Combinations Here you’ll learn to derive equations for formulas with triple angles using existing trig identities, as well as to construct linear combinations of trig functions. In other Concepts you’ve dealt with double angle formulas. This was useful for finding the value of an angle that was double your well known value. Now consider the idea of a "triple angle formula". If someone gave you a problem like this: sin 135◦ Could you compute its value? Keep reading, and at the end of this Concept you’ll know how to simplify equations such as this using the triple angle formula. Watch This MEDIA Click image to the left for more content. Deriving a Triple Angle Formula Guidance Double angle formulas are great for computing the value of a trig function in certain cases. However, sometimes different multiples than two times and angle are desired. For example, it might be desirable to have three times the value of an angle to use as the argument of a trig function. By combining the sum formula and the double angle formula, formulas for triple angles and more can be found. Here, we take an equation which takes a linear combination of sine and cosine and converts it into a simpler cosine function. p A cos x + B sin x = C cos(x − D), where C = A2 + B2 , cos D = CA and sin D = CB . You can also use the TI-83 to solve trigonometric equations. It is sometimes easier than solving the equation algebraically. Just be careful with the directions and make sure your final answer is in the form that is called for. You calculator cannot put radians in terms of π. Example A Find the formula for sin 3x Solution: Use both the double angle formula and the sum formula. 1 www.ck12.org sin 3x = sin(2x + x) = sin(2x) cos x + cos(2x) sin x = (2 sin x cos x) cos x + (cos2 x − sin2 x) sin x = 2 sin x cos2 x + cos2 x sin x − sin3 x = 3 sin x cos2 x − sin3 x = 3 sin x(1 − sin2 x) − sin3 x = 3 sin x − 4 sin3 x Example B Transform 3 cos 2x − 4 sin 2x into the form C cos(2x − D) q Solution: A = 3 and B = −4, so C = 32 + (−4)2 = 5. Therefore cos D = 35 and sin D = − 54 which makes the reference angle is −53.1◦ or −0.927 radians. since cosine is positive and sine is negative, the angle must be a fourth quadrant angle. D must therefore be 306.9◦ or 5.36 radians.The final answer is 3 cos 2x − 4 sin 2x = 5 cos(2x − 5.36). Example C Solve sin x = 2 cos x such that 0 ≤ x ≤ 2π using a graphing calculator. Solution: In y =, graph y1 = sin x and y2 = 2 cos x. Next, use CALC to find the intersection points of the graphs. Vocabulary Linear Combination: A linear combination is a set of terms that are added or subtracted from each other with a multiplicative constant in front of each term. Triple Angle Identity: A triple angle identity relates the a trigonometric function of three times an argument to a set of trigonometric functions, each containing the original argument. 2 www.ck12.org Concept 1. Triple-Angle Formulas and Linear Combinations Guided Practice 1. Transform 5 cos x − 5 sin x to the form C cos(x − D) 2. Transform −15 cos 3x − 8 sin 3x to the form C cos(x − D) 3. Derive a formula for tan 4x. Solutions: √ 5 = √1 = 1. If 5 cos x − 5 sin x, then A = 5 and B = −5. By the Pythagorean Theorem, C = 5 2 and cos D = √ 5 2 2 √ 2 . So, because B is negative, D is in Quadrant IV. Therefore, D = 7π . Our final answer is 5 √2 cos x − 7π . 2 4 4 2. If −15 cos 3x − 8 sin 3x, then A = −15 and B = −8. By the Pythagorean Theorem, C = 17. Because A and B are both negative, D is in Quadrant III, which means D = cos−1 15 = 0.49 + π = 3.63 rad. Our final answer is 17 17 cos 3(x − 3.63). 3. tan 4x = tan(2x + 2x) tan 2x + tan 2x = 1 − tan 2x tan 2x 2 tan 2x = 1 − tan2 2x 2 tan x 2 · 1−tan 2x = 2 2 tan x 1 − 1−tan 2x 4 tan x (1 − tan2 x)2 − 4 tan2 x ÷ 1 − tan2 x (1 − tan2 x)2 4 tan x 1 − 2 tan2 x + tan4 x − 4 tan2 x = ÷ 1 − tan2 x (1 − tan2 x)2 4 tan x (1 − tan2 x)2 = · 1 − tan2 x 1 − 6 tan2 x + tan4 x 4 tan x − 4 tan3 x = 1 − 6 tan2 x + tan4 x = Concept Problem Solution Using the triple angle formula we learned in this Concept for the sine function, we can break the angle down into three times a well known angle: sin 3x = 3 sin x − 4 sin3 x we can solve this problem. 3 www.ck12.org sin(3 × 45◦ ) = 3 sin 45◦ − 4 sin3 45◦ √ √ !3 2 2 =3 −4 2 2 ! √ 2 4(2)2/3 =3 − 2 8 √ √ 3 2−2 2 = 2 √ 2 = 2 Practice Transform each expression to the form C cos(x − D). 1. 2. 3. 4. 5. 6. 7. 3 cos x − 2 sin x 2 cos x − sin x −4 cos x + 5 sin x 7 cos x − 6 sin x 11 cos x + 9 sin x 14 cos x + 2 sin x −2 cos x − 4 sin x Derive a formula for each expression. 8. 9. 10. 11. 12. sin 4x cos 6x cos 4x csc 2x cot 2x Find all solutions to each equation in the interval [0, 2π). 13. cos x + cos 3x = 0 14. sin 2x = cos 3x 15. cos 2x + cos 4x = 0 4