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IAL 2022 Jan Q2

A Level / Edexcel / P3

IAL 2022 Jan Paper · Question 2

题目

Problem

(a) Show that the equation

8cos⁡θ=3cosec⁡θ8\cos\theta=3\cosec\theta

can be written in the form

sin⁡2θ=k\sin 2\theta=k

where kk is a constant to be found.

(3)

(b) Hence find the smallest positive solution of the equation

8cos⁡θ=3cosec⁡θ8\cos\theta=3\cosec\theta

giving your answer, in degrees, to one decimal place.

(2)
题目中文翻译

(a) 证明方程

8cos⁡θ=3cosec⁡θ8\cos\theta=3\cosec\theta

可以写成

sin⁡2θ=k\sin 2\theta=k

的形式，其中 kk 是需要求出的常数。

(b) 由此求方程

8cos⁡θ=3cosec⁡θ8\cos\theta=3\cosec\theta

的最小正解，并以角度制表示，精确到小数点后 1 位。

解答

(a)

Start with

8cos⁡θ=3cosec⁡θ8\cos\theta=3\cosec\theta

Since

cosec⁡θ=1sin⁡θ\cosec\theta=\frac{1}{\sin\theta}

we have

8cos⁡θ=3sin⁡θ8\cos\theta=\frac{3}{\sin\theta}

Multiply both sides by sin⁡θ\sin\theta:

8sin⁡θcos⁡θ=38\sin\theta\cos\theta=3

Using

sin⁡2θ=2sin⁡θcos⁡θ\sin2\theta=2\sin\theta\cos\theta

we get

4sin⁡2θ=34\sin2\theta=3

Therefore

sin⁡2θ=34\sin2\theta=\frac34

So the equation can be written in the form

sin⁡2θ=k\sin2\theta=k

where

k=34\boxed{k=\frac34}

(b)

From part (a),

sin⁡2θ=34\sin2\theta=\frac34

The smallest positive solution occurs when

2θ=sin⁡−1342\theta=\sin^{-1}\frac34

Thus

θ=12sin⁡−134\theta=\frac12\sin^{-1}\frac34

So

θ=24.295…∘\theta=24.295\ldots^\circ

Therefore, to one decimal place,

θ=24.3∘\boxed{\theta=24.3^\circ}