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

Question

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 $\dfrac{\cos{\theta} + \sin{\theta}}{\cos{\theta} + \sin{\theta}}$

Answer 1

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.

Answer 2

(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

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