# Solving $\tan{4\theta} = \dfrac{\cos{\theta} - \sin{\theta}}{\cos{\theta} + \sin{\theta}}$

Solve: $\tan{4\theta} = \dfrac{\cos{\theta} - \sin{\theta}}{\cos{\theta} + \sin{\theta}}$ for acute angle $\theta$

I need help solving that problem. I have tried to do both side, but no result yet.

What I have done. $\dfrac{\sin{4\theta}}{\cos{4\theta}} = \dfrac{\cos{2\theta}}{1 + \sin{2\theta}}$

I got that by multiply the RHS with $\dfrac{\cos{\theta} + \sin{\theta}}{\cos{\theta} + \sin{\theta}}$

• Don't post a picture. Write up your question. And if you must post a picture, orient it so that I don't get a crick in my neck when I read it. – Doug M Jun 28 '18 at 1:20
• You only need $1$ picture also rotate it $90$ degrees anticlockwise. – Chris2018 Jun 28 '18 at 1:21
• How are we supposed to read this? – Ryan Jun 28 '18 at 1:21
• I read it by pressing the downward pointing arrow on the upper left ofvthe question. -1. – Oscar Lanzi Jun 28 '18 at 1:23
• I am sorry. I use mobile phone, its hard to write equation. I will edit my question later – Muhamad Abdul Rosid Jun 28 '18 at 1:32

## 2 Answers

We need to solve $$\tan4\theta=\frac{\frac{1}{\sqrt2}\cos\theta-\frac{1}{\sqrt2}\sin\theta}{\frac{1}{\sqrt2}\cos\theta+\frac{1}{\sqrt2}\sin\theta}$$ or $$\tan4\theta=\frac{\sin(45^{\circ}-\theta)}{\cos(45^{\circ}-\theta)}$$ or $$\tan4\theta=\tan(45^{\circ}-\theta)$$ or $$4\theta=45^{\circ}-\theta+180^{\circ},k$$ where $k$ is an integer number, or $$\theta=9^{\circ}+36^{\circ}k,$$ which gives the answer: $$\left\{9^{\circ}, 45 ^{\circ}, 81^{\circ}\right\}$$ By the way, your idea is also works.

Indeed, we need to solve that $$\frac{\sin4\theta}{\cos4\theta}=\frac{\cos2\theta}{1+\sin2\theta}$$ or $$\frac{2\sin2\theta\cos2\theta}{\cos4\theta}=\frac{\cos2\theta}{1+\sin2\theta}.$$ Now, $\cos2\theta=0$ gives $$2\theta=90^{\circ}+180^{\circ}k,$$ where $k$ is an integer number, which gives $\theta=45^{\circ}$.

Also, $$\frac{2\sin2\theta}{\cos4\theta}=\frac{1}{1+\sin2\theta}$$ or $$4\sin^22\theta+2\sin\theta-1=0$$ or $$\sin2\theta=\frac{-1+\sqrt5}{4}$$ or $$\sin2\theta=\sin18^{\circ},$$ which gives $$2\theta=18^{\circ}$$ or $$2\theta=162^{\circ}$$ and we get the same answer.

• Thank you. I prefer the first one. Honestly, I get the answer myself with different steps. – Muhamad Abdul Rosid Jun 28 '18 at 5:41
• You are welcome! – Michael Rozenberg Jun 28 '18 at 5:42
• I think the problem is solved now. – Muhamad Abdul Rosid Jun 28 '18 at 5:44

(Remark: This answer is much like Michael Rozenberg's, which I didn't see until posting. It differs enough, I think, to be worth keeping.)

Since $\sin(\pi/4)=\cos(\pi/4)$, we have

$${\cos\theta-\sin\theta\over\cos\theta+\sin\theta}={\sin(\pi/4)\cos\theta-\cos(\pi/4)\sin\theta\over\cos(\pi/4)\cos\theta+\sin(\pi/4)\sin\theta}={\sin(\pi/4-\theta)\over\cos(\pi/4-\theta)}=\tan(\pi/4-\theta)$$

Writing $\phi=\pi/4-\theta$, we get

$$\tan\phi=\tan(\pi/4-\theta)={\cos\theta-\sin\theta\over\cos\theta+\sin\theta}=\tan(4\theta)=\tan(\pi-4\phi)=-\tan4\phi$$

It's clear that $\phi=0$, which corresponds to $\theta=\pi/4$, is a solution, and it's also easy to see that we get a solution from $\phi=\pi-4\phi$, i.e., $\phi=\pi/5$, which corresponds to $\theta=\pi/4-\pi/5=\pi/20$. To round this out, the symmetry of $\tan\phi=-\tan4\phi$ for $\phi$ and $-\phi$ tells us $\phi=-\pi/5$ is also a solution, corresponding to $\theta=\pi/4+\pi/5=9\pi/20$.

It remains to see if there are any other solutions in the interval $-\pi/4\le\phi\le\pi/4$, which corresponds to acute angles $0\le\theta\le\pi/2$. But the curves $y=\tan\phi$ and $y=-\tan4\phi$ are straightforward to sketch: the first is stricly increasing, and the latter is strictly decreasing (because of the minus sign) in each interval where it's defined, so it's easy to see the two curves intersect exactly three times. Thus $\phi=0$ and $\phi=\pm\pi/5$ are the only three solutions corresponding to acute angles $\theta$, which are

$$\theta=\pi/20,\pi/4,\text{and }9\pi/20$$