$\cos \frac{7 \pi}{8}+\cos \frac{\pi}{4}+\cos \left(\frac{-\pi}{8}\right)-1=$

$\sin ^3 10^{\circ}+\sin ^3 50^{\circ}-\sin ^3 70^{\circ}=$

$\sum_{k=1}^3 \cos^2 \left((2k-1) \frac{\pi}{12}\right)$ is equal to

For non-negative integers $n$,let $f(n) = \frac{\sum_{k=0}^n \sin \left(\frac{k+1}{n+2} \pi\right) \sin \left(\frac{k+2}{n+2} \pi\right)}{\sum_{k=0}^n \sin ^2\left(\frac{k+1}{n+2} \pi\right)}$. Assuming $\cos ^{-1} x$ takes values in $[0, \pi]$,which of the following options is/are correct?
$(1)$ $\sin \left(7 \cos ^{-1} f(5)\right)=0$
$(2)$ $f(4)=\frac{\sqrt{3}}{2}$
$(3)$ $\lim _{n \rightarrow \infty} f(n)=\frac{1}{2}$
$(4)$ If $\alpha=\tan \left(\cos ^{-1} f(6)\right)$,then $\alpha^2+2 \alpha-1=0$

$\text{If } \sin(\alpha+\beta)=1, \sin(\alpha-\beta)=\frac{1}{2}, \alpha, \beta \in [0, \frac{\pi}{2}], \text{ then } \tan(\alpha+2\beta) \cdot \tan(2\alpha+\beta) = ?$

Prove that: $(\sin 3x + \sin x) \sin x + (\cos 3x - \cos x) \cos x = 0$