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# app-a - Exercise 4 b cos θ = – 2 3 = x r and θ is an...

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Unformatted text preview: Exercise 4. b. cos θ = – } 2 3 } = } x r } and θ is an angle in the third quadrant. Since x 2 + y 2 = r 2 , 4 + y 2 = 9 y = – Ï 5 w . Hence, sin θ = – } Ï 3 5 w } and tan θ = } Ï 2 5 w } . c. tan θ = –2 = } y x } and θ is an angle in the fourth quadrant. Since x 2 + y 2 = r 2 , 1 + 4 = r 2 r = Ï 5 w . Hence, sin θ = – } Ï 2 5 w } and cos θ = } Ï 1 5 w } . 7. a. tan x + cot x = sec x csc x L.S. = tan x + cot x = } c s o in s x x } + } c s o in s x x } = } c si o n s 2 x x + + c s o in s 2 x x } = } cos x 1 + sin x } R.S. = sec x csc x = } co 1 s x } ? } sin 1 x } = } cos x 1 sin x } Therefore, tan x + cot x = sec s csc x . b. } 1 s – in si x n x } tan x + sec x L.S. = } 1 s – in si x n 2 x } = sin } co x x 2 x } R.S. = tan x sec x = } c s o in s x x } ? } co 1 s x } = } c s o in s 2 x x } Therefore, } 1 s – in si x n 2 x } = tan x sec x. c. sin 4 x – cos 4 x = 1 – 2cos 2 x L.S. = sin 4 x – cos 4 x = (sin 2 x + cos 2 x )(sin 2 x – cos 2 x ) = (1)(1 – cos 2 x – cos 2 x ) = 1 – 2cos 2 x Therefore, sin 4 x – cos 4 x = 1 – 2cos 2 x. d. } 1 + 1 sin x } = sec 2 x – } c ta o n s x x } R.S. = sec 2 x – } c ta o n s x x } = } co 1 s 2 x } – = } 1 c – o s s i 2 n x x } = } 1 1 – – s s i i n n 2 x x } = = } 1 + 1 sin x } Therefore, } 1 + 1 sin x } = sec 2 x – } c ta o n s x x } . 8. a. 6 sin x – 3 = 1 – 2 sin x 8 sin x = 4 sin x = } 1 2 } x = } π 6 } , } 5 6 π } . b. cos 2 x – cos x = 0 cos x (cos x – 1) = 0 cos x = 0 or cos x = 1 x = } π 2 } , } 3 2 π } or x = 0, 2 π 1 – sin x } (1 – sin x )(1 + sin x ) } c s o in s x x } } cos x Appendix A 259 Appendix A c. 2 sin x cos x = 0 sin 2 x = 0 where 0 ≤ 2 x ≤ 4 π 2 x = 0, π , 2 π , 3 π , 4 π x = 0, } π 2 } , π , } 3 2 π } , 2 π d. 2 sin 2 x – sin x – 1 = 0 (2 sin x + 1)(sin x – 1) = 0 sin x = – } 1 2 } or sin x = 1 x = } 7 6 π } , } 11 6 π } or x = } π 2 } e. sin x + Ï 3 w cos x = 0 sin x = – Ï 3 w cos x tan x = – Ï 3 w x = } 2 3 π } , } 5 3 π } f. 2 sin 2 x – 3 cos x = 0 2(1 – cos 2 x ) – 3 cos x = 0 2 – 3 cos x – 2 cos 2 x = 0 (2 + cos x )(1 – 2 cos x ) = 0 cos x = –2 or cos x = } 1 2 } no solutions or x = } π 3 } , } 5 3 π } Exercise A1 3. a. sin( W + T ) = sin W cos T + cos W sin T = } 3 5 } ? } 1 1 2 3 } + } 4 5 } ? } 1 5 3 } = } 36 6 + 5 20 } = } 5 6 6 5 } b. cos( W – T ) < sin ( W + T ) cos( W – T ) = cos W cos T + sin W sin T = } 4 5 } ? } 1 1 2 3 } + } 3 5 } ? } 1 5 3 } = } 48 6 + 5 15 } = } 6 6 3 5 } 4. sin( A – B ) = sin A cos B – cos A sin B sin( A – B ) = sin( A + (– B )) = sin A cos(– B ) + cos A sin(– B ) = sin A cos B – cos A sin B 5. a. cos 2 A = cos 2 A – sin 2 A cos 2 A = cos( A + A ) = cos A cos A – sin A sin A = cos 2 A – sin 2 A b. cos 2 A = 2 cos 2 A – 1 cos 2 A = cos 2 A – sin 2 A = cos 2 A – (1 – cos 2 A ) = 2cos 2 A – 1 c. cos 2 A = cos 2 A – sin 2 A = 1 – sin 2 A – sin 2 A = 1 – 2sin 2 A 7. a. cos 75° = cos (45° + 30°)...
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app-a - Exercise 4 b cos θ = – 2 3 = x r and θ is an...

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