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IAL 2022 Oct Q9

A Level / Edexcel / P3

IAL 2022 Oct Paper · Question 9

题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Given that cos⁡2θ−sin⁡3θ≠0\cos 2\theta-\sin 3\theta\ne 0

(a) prove that

cos⁡2θcos⁡2θ−sin⁡3θ≡1+sin⁡θ1−2sin⁡θ−4sin⁡2θ\frac{\cos^2\theta}{\cos 2\theta-\sin 3\theta}\equiv \frac{1+\sin\theta}{1-2\sin\theta-4\sin^2\theta}
(4)

(b) Hence solve, for 0<θ≤360∘0<\theta\le 360^\circ,

cos⁡2θcos⁡2θ−sin⁡3θ=2cosec⁡θ\frac{\cos^2\theta}{\cos 2\theta-\sin 3\theta}=2\cosec\theta

Give your answers to one decimal place.

(5)
题目中文翻译

本题中你必须写出解题过程的所有步骤。

不接受完全依赖计算器技术的解法。

已知 cos⁡2θ−sin⁡3θ≠0\cos 2\theta-\sin 3\theta\ne 0。

(a) 证明

cos⁡2θcos⁡2θ−sin⁡3θ≡1+sin⁡θ1−2sin⁡θ−4sin⁡2θ\frac{\cos^2\theta}{\cos 2\theta-\sin 3\theta}\equiv \frac{1+\sin\theta}{1-2\sin\theta-4\sin^2\theta}

(b) 由此在 0<θ≤360∘0<\theta\le 360^\circ 内解方程

cos⁡2θcos⁡2θ−sin⁡3θ=2cosec⁡θ\frac{\cos^2\theta}{\cos 2\theta-\sin 3\theta}=2\cosec\theta

答案精确到小数点后 1 位。

解答

(a)

Start with

cos⁡2θcos⁡2θ−sin⁡3θ\frac{\cos^2\theta}{\cos 2\theta-\sin 3\theta}

Using

sin⁡3θ=sin⁡2θcos⁡θ+cos⁡2θsin⁡θ\sin 3\theta=\sin 2\theta\cos\theta+\cos 2\theta\sin\theta

we get

cos⁡2θcos⁡2θ−sin⁡3θ=cos⁡2θcos⁡2θ−sin⁡2θcos⁡θ−cos⁡2θsin⁡θ\frac{\cos^2\theta}{\cos 2\theta-\sin 3\theta} = \frac{\cos^2\theta} {\cos 2\theta-\sin 2\theta\cos\theta-\cos 2\theta\sin\theta}

Now use

cos⁡2θ=1−2sin⁡2θ\cos2\theta=1-2\sin^2\theta

and

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

Then the denominator becomes

1−2sin⁡2θ−2sin⁡θcos⁡2θ−sin⁡θ(1−2sin⁡2θ)1-2\sin^2\theta-2\sin\theta\cos^2\theta-\sin\theta(1-2\sin^2\theta)

Use

cos⁡2θ=1−sin⁡2θ\cos^2\theta=1-\sin^2\theta

So the denominator is

1−2sin⁡2θ−2sin⁡θ(1−sin⁡2θ)−sin⁡θ(1−2sin⁡2θ)=1−2sin⁡2θ−2sin⁡θ+2sin⁡3θ−sin⁡θ+2sin⁡3θ=1−3sin⁡θ−2sin⁡2θ+4sin⁡3θ\begin{aligned} &1-2\sin^2\theta-2\sin\theta(1-\sin^2\theta) -\sin\theta(1-2\sin^2\theta) \\ &=1-2\sin^2\theta-2\sin\theta+2\sin^3\theta -\sin\theta+2\sin^3\theta \\ &=1-3\sin\theta-2\sin^2\theta+4\sin^3\theta \end{aligned}

Factorise this:

1−3sin⁡θ−2sin⁡2θ+4sin⁡3θ=(1−sin⁡θ)(1−2sin⁡θ−4sin⁡2θ)1-3\sin\theta-2\sin^2\theta+4\sin^3\theta =(1-\sin\theta)(1-2\sin\theta-4\sin^2\theta)

Also,

cos⁡2θ=1−sin⁡2θ=(1−sin⁡θ)(1+sin⁡θ)\cos^2\theta=1-\sin^2\theta=(1-\sin\theta)(1+\sin\theta)

Therefore

cos⁡2θcos⁡2θ−sin⁡3θ=(1−sin⁡θ)(1+sin⁡θ)(1−sin⁡θ)(1−2sin⁡θ−4sin⁡2θ)\frac{\cos^2\theta}{\cos 2\theta-\sin 3\theta} = \frac{(1-\sin\theta)(1+\sin\theta)} {(1-\sin\theta)(1-2\sin\theta-4\sin^2\theta)}

Cancel the common factor:

cos⁡2θcos⁡2θ−sin⁡3θ≡1+sin⁡θ1−2sin⁡θ−4sin⁡2θ\boxed{ \frac{\cos^2\theta}{\cos 2\theta-\sin 3\theta} \equiv \frac{1+\sin\theta}{1-2\sin\theta-4\sin^2\theta} }

as required.

(b)

Using part (a), the equation becomes

1+sin⁡θ1−2sin⁡θ−4sin⁡2θ=2cosec⁡θ\frac{1+\sin\theta}{1-2\sin\theta-4\sin^2\theta} =2\cosec\theta

Since

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

we have

1+sin⁡θ1−2sin⁡θ−4sin⁡2θ=2sin⁡θ\frac{1+\sin\theta}{1-2\sin\theta-4\sin^2\theta} =\frac{2}{\sin\theta}

Cross multiply:

sin⁡θ(1+sin⁡θ)=2(1−2sin⁡θ−4sin⁡2θ)\sin\theta(1+\sin\theta) =2(1-2\sin\theta-4\sin^2\theta)

Expand:

sin⁡θ+sin⁡2θ=2−4sin⁡θ−8sin⁡2θ\sin\theta+\sin^2\theta =2-4\sin\theta-8\sin^2\theta

Bring all terms to one side:

9sin⁡2θ+5sin⁡θ−2=09\sin^2\theta+5\sin\theta-2=0

Using the quadratic formula,

sin⁡θ=−5±52−4(9)(−2)2(9)\sin\theta =\frac{-5\pm\sqrt{5^2-4(9)(-2)}}{2(9)}

So

sin⁡θ=−5±9718\sin\theta=\frac{-5\pm\sqrt{97}}{18}

Thus

sin⁡θ=0.2693…\sin\theta=0.2693\ldots

or

sin⁡θ=−0.8249…\sin\theta=-0.8249\ldots

For

0<θ≤360∘0<\theta\le360^\circ

the solutions are

θ=15.6∘, 164.4∘, 235.6∘, 304.4∘\theta=15.6^\circ,\ 164.4^\circ,\ 235.6^\circ,\ 304.4^\circ

Therefore

θ=15.6∘, 164.4∘, 235.6∘, 304.4∘\boxed{\theta=15.6^\circ,\ 164.4^\circ,\ 235.6^\circ,\ 304.4^\circ}