5TH YEAR MATHEMATICS – HIGHER LEVEL SOLUTIONS TO REVISION TEST ON TRIGONOMETRY II (a)
sin 4θ . θ →0 3θ
Evaluate lim
In this question we have to replace the 3θ with 4θ in order to produce the trigonometrical identity we are looking for. That means multiplying top and bottom of the original fraction by
⇒
(b)
3 4 and . 3 4
sin 4θ 3 4 sin 4θ 1 4 4 . . = . . . The limit of the result is found to be . 3θ 3 4 4θ 3 1 3
(i) Using cos 2 A = cos 2 A − sin 2 A , or otherwise, prove cos 2 A =
1 (1 + cos 2 A ) . 2
In this question we must use the formula given and substitute it into the equation we are asked to prove. 1 ⇒ cos 2 A = (1 + cos 2 A − sin 2 A ) . 2 From tables p13 we see that sin 2 A = 1 − cos 2 A . 1 ⇒ cos 2 A = 1 + cos 2 A − (1 − cos 2 A ) . 2 1 ⇒ cos 2 A = ( 2 cos 2 A ) which of course is true! 2
(
)
(ii) For the second part we use the result of this proof to make an equation in cosx. If cos 2 A = cos 2 A − sin 2 A , then 1 + cos 2 x = cos x becomes 1 + cos 2 x − sin 2 x = cos x .
⇒ 1 + cos 2 x − (1 − cos 2 x ) = cos x ⇒ 2 cos 2 x − cos x = 0 Since cosx is a common factor to both terms, it can be taken out as a factor: ⇒ ( cos x )( 2 cos x − 1) = 0
⇒ cos x = 0 ⇒ x = 90°; 270°
or
So x = {60°, 90°, 270°,300°}
2 cos x − 1 = 0 1 cos x = 2 x = 60°;300°