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
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(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°
(c) The dotted line in the queston reveals a rectangle and a right angled triangle. These are used to solve the question.
Since two sides of the triangle are given, we use the theorem of Pythagoras to solve the remaining one.
⇒ h2 = a2 + b2 ⇒ 122 = 62 + b 2 ⇒ 144 = 36 + b 2 ⇒ b = 108 = 36 × 3 = 6 3 Area of rectangle = 3 × 6 3 = 18 3u 2 1 1 Area of triangle = base× ⊥ height = 6 3 ( 6 ) = 18 3u 2 2 2 2 Total area = 36 3u
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Shaded region = Area of quadrilateral – Combined area of two sectors In order to calculate the area of the sectors we need to know the angle at the centre of each sector. We can do this using SOHCAHTOA. 6 ⇒ α = 60° . Since all angles in a quadrilateral add 12 to 360° , the angle in the small sector is 120° . In large sector cos α =
1 1 ⇒ 36 3 − (9)(9) sin 60 + (3)(3) sin120 2 2 1 3 1 3 ⇒ 36 3 − (81)( ) + (9)( ) 2 2 2 2 90 3 ⇒ 36 3 − 4 = 144 3 − 90 3 = 54 3u 2