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IAL 2025 May Q8

A Level / Edexcel / P2

IAL 2025 May Paper · Question 8

题目

Problem

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

(i) Solve, for 0≤x<π0\leq x<\pi, the equation

3tan⁡(2x+π5)=33\tan\left(2x+\frac{\pi}{5}\right)=\sqrt3

giving the answers in radians in the form kπk\pi, where kk is a rational constant to be found.

(3)

(ii) Solve, for 0≤θ<360∘0\leq\theta<360^\circ, the equation

5sin⁡θtan⁡θ=cos⁡θ+45\sin\theta\tan\theta=\cos\theta+4

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

(5)

解答

(i)

解法一

思路

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先令里面的角为 u=2x+π5u=2x+\frac{\pi}{5}。由于 tan⁡u=33\tan u=\frac{\sqrt3}{3}，在一个周期内对应 u=π6u=\frac{\pi}{6}，再加上 π\pi 的周期。

答题过程

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The equation gives

tan⁡(2x+π5)=33.\begin{align*} \tan\left(2x+\frac{\pi}{5}\right)=\frac{\sqrt3}{3}. \end{align*}

Since 0≤x<π0\leq x<\pi,

π5≤2x+π5<11π5.\begin{align*} \frac{\pi}{5}\leq 2x+\frac{\pi}{5}<\frac{11\pi}{5}. \end{align*}

The relevant values are

2x+π5=7π6or2x+π5=13π6.2x+\frac{\pi}{5}=\frac{7\pi}{6} \quad\text{or}\quad 2x+\frac{\pi}{5}=\frac{13\pi}{6}.

So

2x= 7π6−π5=29π30x= 29π60,\begin{align*} 2x=&\,\frac{7\pi}{6}-\frac{\pi}{5} =\frac{29\pi}{30}\\ x=&\,\frac{29\pi}{60}, \end{align*}

or

2x= 13π6−π5=59π30x= 59π60.\begin{align*} 2x=&\,\frac{13\pi}{6}-\frac{\pi}{5} =\frac{59\pi}{30}\\ x=&\,\frac{59\pi}{60}. \end{align*}

Therefore,

x=29π60,59π60.\begin{align*} x=\frac{29\pi}{60},\quad \frac{59\pi}{60}. \end{align*}

解法二

思路

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也可以用 compound angle formula，把 tan⁡(2x+π5)\tan\left(2x+\frac{\pi}{5}\right) 展开成 tan⁡2x\tan2x 的方程。关键是把解出的 tan⁡2x\tan2x 重新看成一个特殊角的正切值。

答题过程

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From

tan⁡(2x+π5)=33,\begin{align*} \tan\left(2x+\frac{\pi}{5}\right)=\frac{\sqrt3}{3}, \end{align*}

use

tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B.\begin{align*} \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}. \end{align*}

Let

T=tan⁡π5.\begin{align*} T=\tan\frac{\pi}{5}. \end{align*}

This gives

tan⁡2x+tan⁡π51−tan⁡2xtan⁡π5=33.\frac{\tan2x+\tan\frac{\pi}{5}} {1-\tan2x\tan\frac{\pi}{5}} =\frac{\sqrt3}{3}.

So

tan⁡2x+T1−Ttan⁡2x= 333(tan⁡2x+T)= 3(1−Ttan⁡2x)tan⁡2x(3+3T)= 3−3Ttan⁡2x= 3−3T3+3T.\begin{align*} \frac{\tan2x+T}{1-T\tan2x} =&\,\frac{\sqrt3}{3}\\ 3(\tan2x+T) =&\,\sqrt3(1-T\tan2x)\\ \tan2x(3+\sqrt3T) =&\,\sqrt3-3T\\ \tan2x =&\,\frac{\sqrt3-3T}{3+\sqrt3T}. \end{align*}

Since 33=tan⁡π6\frac{\sqrt3}{3}=\tan\frac{\pi}{6},

3−3T3+3T=tan⁡π6−tan⁡π51+tan⁡π6tan⁡π5.\frac{\sqrt3-3T}{3+\sqrt3T} =\frac{\tan\frac{\pi}{6}-\tan\frac{\pi}{5}} {1+\tan\frac{\pi}{6}\tan\frac{\pi}{5}}.

Therefore,

tan⁡2x=tan⁡(π6−π5)=tan⁡(−π30).\tan2x=\tan\left(\frac{\pi}{6}-\frac{\pi}{5}\right) =\tan\left(-\frac{\pi}{30}\right).

So

2x=−π30+nπ.\begin{align*} 2x=-\frac{\pi}{30}+n\pi. \end{align*}

Since 0≤x<π0\leq x<\pi, we have 0≤2x<2π0\leq2x<2\pi.

Thus the possible values are

2x=29π30or2x=59π30.2x=\frac{29\pi}{30} \quad\text{or}\quad 2x=\frac{59\pi}{30}.

Therefore,

x=29π60,59π60.\begin{align*} x=\frac{29\pi}{60},\quad \frac{59\pi}{60}. \end{align*}

(ii)

解法一

思路

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把 tan⁡θ\tan\theta 写成 sin⁡θcos⁡θ\frac{\sin\theta}{\cos\theta}，然后乘以 cos⁡θ\cos\theta，再用 sin⁡2θ=1−cos⁡2θ\sin^2\theta=1-\cos^2\theta，就能得到关于 cos⁡θ\cos\theta 的二次方程。

答题过程

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Using

tan⁡θ=sin⁡θcos⁡θ,\begin{align*} \tan\theta=\frac{\sin\theta}{\cos\theta}, \end{align*}

the equation becomes

5sin⁡θ⋅sin⁡θcos⁡θ=cos⁡θ+4.5\sin\theta\cdot\frac{\sin\theta}{\cos\theta} =\cos\theta+4.

Multiply by cos⁡θ\cos\theta:

5sin⁡2θ=cos⁡2θ+4cos⁡θ.\begin{align*} 5\sin^2\theta=\cos^2\theta+4\cos\theta. \end{align*}

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

5(1−cos⁡2θ)= cos⁡2θ+4cos⁡θ5−5cos⁡2θ= cos⁡2θ+4cos⁡θ6cos⁡2θ+4cos⁡θ−5= 0.\begin{align*} 5(1-\cos^2\theta)=&\,\cos^2\theta+4\cos\theta\\ 5-5\cos^2\theta=&\,\cos^2\theta+4\cos\theta\\ 6\cos^2\theta+4\cos\theta-5=&\,0. \end{align*}

Solve the quadratic:

cos⁡θ= −4±42−4(6)(−5)2(6)= −4±13612= −2±346.\begin{align*} \cos\theta =&\,\frac{-4\pm\sqrt{4^2-4(6)(-5)}}{2(6)}\\ =&\,\frac{-4\pm\sqrt{136}}{12}\\ =&\,\frac{-2\pm\sqrt{34}}{6}. \end{align*}

Only

cos⁡θ=−2+346\begin{align*} \cos\theta=\frac{-2+\sqrt{34}}{6} \end{align*}

is in the range −1≤cos⁡θ≤1-1\leq\cos\theta\leq1.

Therefore,

θ=50.3∘,309.7∘\begin{align*} \theta=50.3^\circ,\quad 309.7^\circ \end{align*}

to one decimal place.