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IAL 2026 Jan A Q7

A Level / Edexcel / P3

IAL 2026 Jan A Paper · Question 7

题目

Problem

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

Solutions relying entirely on calculator technology are not acceptable.

(a) Express 5cos⁡θ−2sin⁡θ\sqrt5\cos\theta-2\sin\theta in the form Rcos⁡(θ+α)R\cos(\theta+\alpha), where R>0R>0 and 0<α<π20<\alpha<\dfrac{\pi}{2}.

State the value of RR and give the value of α\alpha to 4 significant figures.

(3)

(b) Solve, for −π<θ<π-\pi<\theta<\pi,

5cos⁡θ−2sin⁡θ=0.5\sqrt5\cos\theta-2\sin\theta=0.5

giving your answers to 3 significant figures.

(4)
f(θ)=A(5cos⁡θ−2sin⁡θ)+Bθ∈Rf(\theta)=A\left(\sqrt5\cos\theta-2\sin\theta\right)+B\qquad \theta\in\mathbb{R}

where AA and BB are constants.

Given that the range of ff is

−15≤f(θ)≤33-15\le f(\theta)\le 33

(c) find the value of BB and the possible values of AA.

(4)
题目中文翻译

本题必须写出全部解题步骤。

不接受完全依赖计算器技术的解法。

(a) 将 5cos⁡θ−2sin⁡θ\sqrt5\cos\theta-2\sin\theta 写成 Rcos⁡(θ+α)R\cos(\theta+\alpha) 的形式，其中 R>0R>0，且 0<α<π20<\alpha<\dfrac{\pi}{2}。

写出 RR 的值，并将 α\alpha 的值保留 4 位有效数字。

(b) 在 −π<θ<π-\pi<\theta<\pi 内解方程

5cos⁡θ−2sin⁡θ=0.5\sqrt5\cos\theta-2\sin\theta=0.5

答案保留 3 位有效数字。

已知

f(θ)=A(5cos⁡θ−2sin⁡θ)+Bθ∈Rf(\theta)=A\left(\sqrt5\cos\theta-2\sin\theta\right)+B\qquad \theta\in\mathbb{R}

其中 A,BA,B 为常数。

若 ff 的值域为

−15≤f(θ)≤33-15\le f(\theta)\le 33

(c) 求 BB 的值以及 AA 的可能取值。

解答

(a)

解法一

思路

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把

5cos⁡θ−2sin⁡θ\sqrt5\cos\theta-2\sin\theta

写成 Rcos⁡(θ+α)R\cos(\theta+\alpha)。展开右边：

Rcos⁡(θ+α)=Rcos⁡θcos⁡α−Rsin⁡θsin⁡αR\cos(\theta+\alpha) =R\cos\theta\cos\alpha-R\sin\theta\sin\alpha

然后比较 cos⁡θ\cos\theta 和 sin⁡θ\sin\theta 的系数。

答题过程

展开

Let

5cos⁡θ−2sin⁡θ≡Rcos⁡(θ+α)\sqrt5\cos\theta-2\sin\theta \equiv R\cos(\theta+\alpha)

Expand the right-hand side.

Rcos⁡(θ+α)= R(cos⁡θcos⁡α−sin⁡θsin⁡α)= Rcos⁡αcos⁡θ−Rsin⁡αsin⁡θ\begin{align*} R\cos(\theta+\alpha) =&\,R(\cos\theta\cos\alpha-\sin\theta\sin\alpha) \\[2mm] =&\,R\cos\alpha\cos\theta-R\sin\alpha\sin\theta \end{align*}

Comparing coefficients,

Rcos⁡α=5,Rsin⁡α=2R\cos\alpha=\sqrt5, \qquad R\sin\alpha=2

Hence

R2= (5)2+22= 5+4= 9\begin{align*} R^2 =&\,(\sqrt5)^2+2^2 \\[2mm] =&\,5+4 \\[2mm] =&\,9 \end{align*}

Since R>0R>0,

R=3R=3

Also,

tan⁡α=Rsin⁡αRcos⁡α=25\tan\alpha =\frac{R\sin\alpha}{R\cos\alpha} =\frac{2}{\sqrt5}

Therefore

α=tan⁡−1(25)=0.7297\alpha=\tan^{-1}\left(\frac{2}{\sqrt5}\right)=0.7297

to 4 significant figures.

Thus

5cos⁡θ−2sin⁡θ=3cos⁡(θ+0.7297)\boxed{\sqrt5\cos\theta-2\sin\theta =3\cos(\theta+0.7297)}

(b)

解法一

思路

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利用 (a) 的结果，把方程化为

3cos⁡(θ+α)=0.53\cos(\theta+\alpha)=0.5

然后在 −π<θ<π-\pi<\theta<\pi 内找出所有解。注意本题用弧度。

答题过程

展开

From part (a),

5cos⁡θ−2sin⁡θ=3cos⁡(θ+α)\sqrt5\cos\theta-2\sin\theta =3\cos(\theta+\alpha)

where

α=0.7297…\alpha=0.7297\ldots

So

3cos⁡(θ+α)=0.53\cos(\theta+\alpha)=0.5

Hence

cos⁡(θ+α)=16\cos(\theta+\alpha)=\frac{1}{6}

The principal angle is

cos⁡−1(16)=1.4033…\cos^{-1}\left(\frac16\right)=1.4033\ldots

Therefore

θ+α=1.4033…orθ+α=−1.4033…\theta+\alpha=1.4033\ldots \quad\text{or}\quad \theta+\alpha=-1.4033\ldots

Using α=0.7297…\alpha=0.7297\ldots,

θ= 1.4033…−0.7297…= 0.6736…\begin{align*} \theta=&\,1.4033\ldots-0.7297\ldots \\[2mm] =&\,0.6736\ldots \end{align*}

or

θ= −1.4033…−0.7297…= −2.1330…\begin{align*} \theta=&\,-1.4033\ldots-0.7297\ldots \\[2mm] =&\,-2.1330\ldots \end{align*}

Both values lie in −π<θ<π-\pi<\theta<\pi. Therefore

θ=0.674, −2.13\boxed{\theta=0.674,\ -2.13}

to 3 significant figures.

(c)

解法一

思路

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由 (a)，

5cos⁡θ−2sin⁡θ\sqrt5\cos\theta-2\sin\theta

的范围是 [−3,3][-3,3]。所以

A(5cos⁡θ−2sin⁡θ)+BA(\sqrt5\cos\theta-2\sin\theta)+B

的中间值是 BB，振幅是 3∣A∣3|A|。

答题过程

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Since

5cos⁡θ−2sin⁡θ=3cos⁡(θ+α)\sqrt5\cos\theta-2\sin\theta=3\cos(\theta+\alpha)

we know

−3≤5cos⁡θ−2sin⁡θ≤3-3\leq \sqrt5\cos\theta-2\sin\theta\leq3

The range of ff is given as

−15≤f(θ)≤33-15\leq f(\theta)\leq33

The midpoint of the range is

−15+332=9\frac{-15+33}{2}=9

so

B=9B=9

The half-width of the range is

33−(−15)2=24\frac{33-(-15)}{2}=24

Since the half-width is 3∣A∣3|A|,

3∣A∣=243|A|=24

Therefore

∣A∣=8|A|=8

Thus

B=9,A=8 or A=−8\boxed{B=9,\qquad A=8\text{ or }A=-8}