Two questions on trigonometry

Question

Excuse me please. I cannot solve two tasks on trigonometry.

i) Prove the inequality $$ \sin x (\sin x-2)+\cos^2 (x-1)>0. $$

I've reduced it to $$ (1-\sin x)^2-\sin^2 (x-1)>0, $$ or, equivalently, $$ 1-\sin x>|\sin (x-1)|. $$

But I cannot prove the last inequality.

ii) Solve the equation $$ 4\tan\frac{x}{2}+2\tan\frac{x}{4}+\tan\frac{x}{8}=\tan\frac{x}{12}-\frac{8}{\tan x}. $$

The answer says that it is equivalent to $$ \cos\frac{5x}{12}=-1. $$

But how to get it?

Thank you very much in advance!

Answer 1

Concerning the inequality, as already said in the comments, it is false. If you plot the function $$f(x)=\sin x (\sin x-2)+\cos^2 (x-1)$$ you will identify that it shows $x$ intercepts at $x_1=1.10620$, $x_2=3.03539$, $x_3=7.38938$, $x_4=9.31858$, ... Between $x_1$ and $x_2$, $f(x) \lt 0$ as well as between $x_3$ and $x_4$ and so on.

Answer 2

I assume you meant $~\sin x\cdot\sin(x-2)+\cos^2(x-1)>0$, which is indeed true, since it can be simplified to $(\cos1)^2>0$ using angle addition formulas in conjunction with $\sin^2x+\cos^2x=1$.