Saturday X-tra X-Sheet: 12 Trigonometry Compound Angles and Transformation Key Concepts In this session we will focus on summarising what you need to know about: Simplifying using compound angles Identities Solving equations When an identity is valid Laws / Key Concepts / Diagram cos( ) cos cos sin sin cos( ) cos cos sin sin sin( ) sin cos cos sin sin( ) sin cos cos sin sin 2 2 sin cos cos 2 cos2 sin 2 cos 2 2 cos2 1 cos 2 1 2 sin 2 cos b b [1 ; 1] cos b k 3600 k Z sin b b [1;1 ] 1 sin 1 b k 3600 tan b b tan 1 b k 3600 or 1800 sin 1 b k 3600 k Z k Z X-ample Questions 1. If cos 200 m , write each of the following in terms of m: a) cos 3400 b) sin110 0 c) 1 sin 70 2 d) cos 40 0 0 e) cos 50 0 Page 1 2. Simplify without using a calculator sin 50 cos15 sin15 cos140 3. Prove the identity cos A [cos A sin 180 A tan A] cos 2 A 0 4. Prove the following identity: tan 1 cos 2 5. a) Prove that sin( A B) sin( A B) 2 cos A sin B sin 2 b) Hence, or otherwise, prove that sin 3 x sin x 2 cos 2 x sin x 6. Prove that tan x 1 cos 2 x , and hence deduce that sin 2 x tan 22 1 o 2 1 2 7. Determine the general solution a) cos 540 cos x sin 540 sin x sin 2 x b) 4 sin x 3cos x 0 c) sin( x 300 ) cos 2 x 0 d) 2 sin( x 30 0 ) 3 cos x e) 8. 3 cos sin For which values of A is the following identity not valid: 2 sin 2 A 1 2 tan A sin 2 A tan A Page 2 X-ercise 1. 2. cos45 x . cos45 x 12 cos 2 x. a) Prove that b) Find the general solution if 1 2 sin 2 x cos 40 cos 2 x cos310 Prove the identity: a) b) 3. For what values of is the identity undefined? Evaluate and simplify: sin 155. cos 25 cos320 4. a) It is given that tan 22 = p. Now, express in terms of p, each of the following: 5. i) tan 338 ii) sin 68 iii) sin 44 Simplify without the use of a calculator: cos330.sin140 (1) sin 160 .tan 405.sin 290 (2) 6. 7. sin 140 cos150 sin 110 sin 340 sin x. sin 2 x cos 2 x 1 cos x sin cos( 30) Determine the general solution: Prove the following identity: Page 3 Answers 1b) x 5 0 k180 0 or x 55 0 k180 0 k Z 2b) x 90 0 k 360 0 or x 120 0 k 360 0 k Z 1 3. 2 4. (1) p 1 ( 2) 1 p2 (3) 2p 1 p2 5. (1) 3 (2) 3 60 0 k180 0 7. k Z Page 4
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