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P2 2025 May Q8

A Level / Edexcel / P2

IAL 2025 May Paper · Question 8

题目

Problem

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

3tan⁡(2x+π5)=33\tan\bigg(2x+\frac{\pi}{5}\bigg)=\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)
题目中文翻译

(i) 在 0≤x<π0\leq x<\pi 范围内解方程

3tan⁡(2x+π5)=3.3\tan\bigg(2x+\frac{\pi}{5}\bigg)=\sqrt3.

答案以弧度表示，并写成 kπk\pi 的形式，其中 kk 是待求的有理常数。（3分）

(ii) 在 0≤θ<360∘0\leq\theta<360^\circ 范围内解方程

5sin⁡θtan⁡θ=cos⁡θ+4.5\sin\theta\tan\theta=\cos\theta+4.

答案以度为单位，保留至小数点后一位。（5分）

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\bigg(2x+\frac{\pi}{5}\bigg)=\frac{\sqrt3}{3}. \end{align*}

Since 0≤x<π0\leq x<\pi,

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

The relevant values are

u= 7π6oru= 13π6.\begin{align*} u=&\,\frac{7\pi}{6}\\ \text{or}\quad u=&\,\frac{13\pi}{6}. \end{align*}

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\bigg(2x+\frac{\pi}{5}\bigg) 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\bigg(2x+\frac{\pi}{5}\bigg)=\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+T1−Ttan⁡2x=33.\frac{\tan2x+T}{1-T\tan2x}=\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−T1+Ttan⁡π6= tan⁡(π6−π5).\begin{align*} &\,\frac{\sqrt3-3T}{3+\sqrt3T}\\ =&\,\frac{\tan\frac{\pi}{6}-T}{1+T\tan\frac{\pi}{6}}\\ =&\,\tan\bigg(\frac{\pi}{6}-\frac{\pi}{5}\bigg). \end{align*}

Therefore,

tan⁡2x= tan⁡(π6−π5)= tan⁡(−π30).\begin{align*} \tan2x=&\,\tan\bigg(\frac{\pi}{6}-\frac{\pi}{5}\bigg)\\ =&\,\tan\bigg(-\frac{\pi}{30}\bigg). \end{align*}

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*}

解法三

思路

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将正切写成正弦除以余弦，再平方并用 sin⁡2u+cos⁡2u=1\sin^2u+\cos^2u=1。平方会引入多余角，因此必须用原来的正切符号筛选。

答题过程

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Let u=2x+π5u=2x+\frac{\pi}{5}. The range is

π5≤u<11π5.\frac{\pi}{5}\leq u<\frac{11\pi}{5}.

Since tan⁡u\tan u is defined, cos⁡u≠0\cos u\ne0. Thus

3tan⁡u= 33sin⁡u= 3cos⁡u.\begin{align*} 3\tan u=&\,\sqrt3\\ 3\sin u=&\,\sqrt3\cos u. \end{align*}

Squaring gives

9sin⁡2u= 3cos⁡2u9sin⁡2u= 3(1−sin⁡2u)12sin⁡2u= 3sin⁡2u= 14sin⁡u= ±12.\begin{align*} 9\sin^2u=&\,3\cos^2u\\ 9\sin^2u=&\,3(1-\sin^2u)\\ 12\sin^2u=&\,3\\ \sin^2u=&\,\frac14\\ \sin u=&\,\pm\frac12. \end{align*}

In the required range, the candidate angles are

u= 5π6, 7π6, 11π6, 13π6.\begin{align*} u=&\,\frac{5\pi}{6},\ \frac{7\pi}{6},\\ &\,\hspace{2pt}\frac{11\pi}{6},\ \frac{13\pi}{6}. \end{align*}

The original equation requires tan⁡u=33>0\tan u=\frac{\sqrt3}{3}>0.

At u=5π6u=\frac{5\pi}{6} and u=11π6u=\frac{11\pi}{6}, tan⁡u<0\tan u<0, so these are rejected. The other two angles satisfy the original equation.

For u=7π6u=\frac{7\pi}{6},

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*}

For u=13π6u=\frac{13\pi}{6},

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*}

Both lie in 0≤x<π0\leq x<\pi. Hence x=29π60,59π60x=\frac{29\pi}{60},\frac{59\pi}{60}.

(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.

The original equation contains tan⁡θ\tan\theta, so cos⁡θ≠0\cos\theta\ne0. Multiplying by cos⁡θ\cos\theta is therefore valid:

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.

The positive cosine gives angles in quadrants I and IV. Let

α= arccos⁡(−2+346)≈ 50.320537876∘.\begin{align*} \alpha=&\,\arccos\bigg(\frac{-2+\sqrt{34}}{6}\bigg)\\ \approx&\,50.320537876^\circ. \end{align*}

Thus θ=α\theta=\alpha or 360∘−α360^\circ-\alpha. Therefore,

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

to one decimal place.