A geometry problem by A Former Brilliant Member

$cos(\theta)=sin(\theta)(1+\sqrt{2})$

$\frac{sin(\theta)}{cos(\theta)}=\frac{1}{1+\sqrt{2}}$

$tan(\theta)=\frac{1}{1+\sqrt{2}}$

set $x=\frac{1}{1+\sqrt{2}}$

$\theta=tan^{-1}(\frac{1}{1+\sqrt{2}})$ so by standard identities $sin(\theta)=\frac{x}{\sqrt{1+x^2}}$ and $cos(\theta)=\frac{1}{\sqrt{1+x^2}}$ . So $sin(\theta)+cos(\theta)=\frac{1+x}{\sqrt{1+x^2}}$

Then by noting that $x=\frac{1}{1+\sqrt{2}}=\sqrt{2}-1$

we get $1-\sqrt{2}+x=0$

so $\frac{1+x-\sqrt{2}}{\sqrt{1+x^2}}=0$

from which it follows

$sin(\theta)+cos(\theta)=\frac{1+x}{\sqrt{1+x^2}}=\frac{\sqrt{2}}{\sqrt{1+x^2}}=\sqrt{2}cos(\theta)$

By the harmonic addition formula, $\cos\theta-\sin\theta=\sqrt{2}\cos(\theta+\pi/4)=\sqrt{2}\sin(\pi/4-\theta).$

So, $\sqrt{2}\sin(\pi/4-\theta)=\sqrt{2}\sin\theta$

$\Rightarrow \pi/4-\theta+2k\pi=\theta$

$\Rightarrow \theta = \pi/8+k\pi.$

Apply the harmonic addition formula again to obtain:

$\cos\theta+\sin\theta=\sqrt{2}\cos(\theta-\pi/4)$

$=\sqrt{2}\cos(\pi/8+k\pi-\pi/4)$

$=\sqrt{2}\cos(-\pi/8+k\pi)$

$=\sqrt{2}\cos(\pi/8-k\pi).$

$=\sqrt{2}\cos(\pi/8-k\pi+2k\pi)$

$=\sqrt{2}\cos(\pi/8+k\pi)$

$=\sqrt{2}\cos\theta.$