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Chapter 11: Trigonometric Identities and Equations 11.1 Trigonometric Identities 11.2 Addition and Subtraction Formulas 11.3 Double-Angle, Half-Angle, and Product-Sum Formulas 11.4 Inverse Trigonometric Functions 11.5 Trigonometric Equations 11.3 Double-Angle, Half-Angle, and Product-Sum Formulas Double-Angle Identities E.g. cos 2A = cos(A + A) = cos A cos A – sin A sin A = cos² A – sin² A Other forms for cos 2A are obtained by substituting either cos² A = 1 – sin² A or sin² A = 1 – cos² A to get cos 2A = 1 – 2 sin² A or cos 2A = 2 cos² A – 1. cos 2 A cos A sin A cos 2 A 1 2 sin A cos 2 A 2 cos A 1 sin 2 A 2 sin A cos A 2 2 2 2 2 tan A tan 2 A 1 tan 2 A 11.3 Finding Function Values of 2 Example Given cos 53 and sin < 0, find a) sin 2, b) cos 2, and c) tan 2. Solution To find sin 2, we must find sin . sin 2 cos 2 1 2 a) sin 2 2 sin cos 4 3 2 5 5 24 25 3 sin 1 5 9 sin 1 25 4 Choose the negative sin square root since sin < 0. 5 2 So cos 2 is… 8 5 94% 3% a. .. 0% he 3% of t 0% No ne 1. 4 2. 5 24 3. 7 7 4. 25 5. None of the above 11.3 Finding Function Values of 2 b) cos 2 cos 2 sin 2 2 3 4 5 5 2 7 25 2 tan sin 54 4 c) tan 2 , where tan 3 1 tan 2 cos 3 5 2 43 1 43 2 24 OR 7 sin 2 tan 2 cos 2 24 257 25 24 7 11.3 Simplifying Expressions Using Double-Number Identities Example Simplify each expression. (a) cos² 7x – sin² 7x (b) sin 15° cos 15° Solution (a) cos 2A = cos² A – sin² A. Substituting 7x in for A gives cos² 7x – sin² 7x = cos 2(7x) = cos 14x. (b) Apply sin 2A = 2 sin A cos A directly. 1 sin 15 cos15 (2) sin 15 cos15 2 1 sin( 2 15 ) 1 sin 30 1 2 2 4 11.3 Product-to-Sum Identities • Product-to-sum identities are used in calculus to find integrals of functions that are products of trigonometric functions. • Adding identities for cos(A + B) and cos(A – B) gives 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 B) 2 cos A cos B 1 cos A cos B [cos( A B) cos( A B)]. 2 11.3 Product-to-Sum Identities • Similarly, subtracting and adding the sum and difference identities of sine and cosine, we may derive the identities in the following table. Product-to-Sum Identities cos A cos B 12 [cos( A B) cos( A B)] sin A sin B 12 [cos( A B) cos( A B)] sin A cos B 12 [sin( A B) sin( A B)] cos A sin B 12 [sin( A B) sin( A B)] 11.3 Using a Product-to-Sum Identity Example Rewrite cos 2 sin as the sum or difference of two functions. Solution By the identity for cos A sin A, with 2 = A and = B, cos 2 sin 1 [sin( 2 ) sin( 2 )] 2 1 1 sin 3 sin . 2 2 Since, cos A sin B 12 [sin( A B) sin( A B)] 11.3 Sum-to-Product Identities • From the previous identities, we can derive another group of identities that are used to rewrite sums of trigonometric functions as products. Sum-to-Product Identities sin A sin B 2 sin A2 B cos A2 B sin A sin B 2 cos A2 B sin A2 B cos A cos B 2 cos A2 B cos A2 B cos A cos B 2 sin A2 B sin A2 B 11.3 Using a Sum-to-Product Identity Example Write sin 2 – sin 4 as a product of two functions. Solution Use the identity for sin A – sin B, with 2 = A and 4 = B. sin 2 sin 4 2 4 2 4 Since, sin A sin B 2 cos A B sin AB 2 2 2 cos sin 2 2 6 2 2 cos sin 2 2 2 cos 3 sin( ) 2 cos 3 sin Since sine is an odd function 11.3 Half-Number Identities • Half-angle identities for sine and cosine are used in calculus when eliminating the xy-term from an equation of the form Ax² + Bxy + Cy² + Dx + Ey + F = 0, so the type of conic it represents can be determined. • From the alternative forms of the identity for cos 2A, we can derive three additional identities, e.g. sin A . 2 cos 2x 1 2sin 2 x 2sin 2 x 1 cos 2x 1cos 2 x A sin x Let 2x A so that x . 2 2 1cos A Choose the sign ± depending on A sin the quadrant of the angle A/2. 2 2 11.3 Half-Number Identities Half-Number Identities A 1 cos A cos 2 2 A 1 cos A sin 2 2 A 1 cos A A sin A tan tan 2 1 cos A 2 1 cos A A 1 cos A tan 2 sin A 11.3 Using a Half-Number Identity to Find an Exact Value Example Find the exact value of cos . 12 Solution 3 6 cos 1 2 cos 12 2 2 22 1 cos 6 2 2 3 3 2 1 2 2 11.3 Finding Function Values of x/2 Example Given cos x 23 , with 3 2 x 2 , find cos 2x , sin 2x , and tan 2x . Solution The half-angle terminates in quadrant II since 3 3 x x 2 2 4 2 . 2 x 1 3 a) sin 2 2 x 1 23 b) cos 2 2 x c) tan 2 sin cos x 2 x 2 1 6 6 6 5 30 6 6 6 6 30 6 5 5