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the relation ${\displaystyle f^2+g^2=1}$ is followed only by trigonometric functions for differentiable functions
Let ${\displaystyle U\subseteq \mathbb{C}}$ be an open, convex and connected subset and let ${\displaystyle f,g:U\to \mathbb{C}}$ be two differentiable functions so that f,g are continuous functions ${\displaystyle f(,g\in C^1)}$
I want to show that if ${\displaystyle f^2+g^2=1}$ over U then there is a function ${\displaystyle z:U\to \mathbb{C}}$ such that ${\displaystyle f\left(t\right)=\sin \left(z\left(t\right)\right)\text{ and }g\left(t\right)=\cos \left(z\left(t\right)\right)}$

Compute ${\displaystyle {\int }_0^{2\pi }\frac{1}{{\sin }^{4}x+{\cos }^{4}x}dx}$

How to evaluate ${\displaystyle \underset{x\to 0}{lim}\frac{5-5\cos \left(2x\right)+\sin \left(4x\right)}{x}}$ without using LHospitals rule?
I thought the best way was to separate it in two limits:
${\displaystyle \underset{x\to 0}{lim}\left(\frac{5-5\cos \left(2x\right)}{x}\right)+\underset{x\to 0}{lim}\left(\frac{\sin \left(4x\right)}{x}\right)}$
Considering that ${\displaystyle \underset{x\to 0}{lim}\left(\frac{\sin x}{x}\right)=1}$ we can easily know that the second limit is 4. I also know the result of the original limit is 4, so the result of the first limit in the second line needs to be 0 (0+4=4).
The issue is that I can not figure out how to remove the x from the denominator so I can avoid the indeterminate form. I already tried replacing ${\displaystyle \cos \left(2x\right)}$ with ${\displaystyle {\cos }^{2}x-{\sin }^{2}x}$, but it seems to be useless.
So, to summarize everything, my problem is how to evaluate this limit without using LHospitals rule:
${\displaystyle \underset{x\to 0}{lim}\left(\frac{5-5\cos \left(2x\right)}{x}\right)}$

Minimum value of ${\displaystyle \frac{b+1}{a+b-2}}$
Attempt:
Then I tried this way: Let a=bk for some real k.
Then I got f(b) in terms of b and k which is minmum when ${\displaystyle b=\frac{2-k}{2(k+1)}}$... then again I got an equation in k which didnt

${\displaystyle \tan \left(a\right)=\frac{3}{4}}$ and ${\displaystyle \tan \left(b\right)=\frac{5}{12}}$, what is ${\displaystyle \cos (a+b)}$
Attempt :
${\displaystyle \cos (a+b)=\cos \left(a\right)\cos \left(b\right)-\sin \left(a\right)\sin \left(b\right)}$
And we can write ${\displaystyle \tan \left(a\right)=\frac{\sin \left(a\right)}{\cos \left(a\right)}=\frac{0.3}{0.4}}$ and ${\displaystyle \frac{\sin \left(b\right)}{\cos \left(b\right)}=\frac{0.05}{0.12}}$ so
${\displaystyle \cos (a+b)=\left(0.4\right)\left(0.12\right)-\left(0.3\right)\left(0.05\right)=\frac{33}{1000}}$
Is this correct?

Evaluating ${\displaystyle {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{\sin \left(x\right)+\cos \left(x\right)}{{\sin }^4\left(x\right)-4}dx}$

Prove ${\displaystyle {\sec }^{2}A+{\csc }^{2}A=4{\csc }^{2}2A}$

Check proof that ${\displaystyle \tan (x+\sin \left(x\right))}$ is periodic and find its period
Ive

Find ${\displaystyle S=\left({\tan 1}^{\circ }\right)+\left({\tan 2}^{\circ }\right)+\left({\tan 3}^{\circ }\right)+\left({\tan 4}^{\circ }\right)+\left({\tan 5}^{\circ }\right)+\left({\tan 6}^o\{\circ \}\right)+\left({\tan 7}^{\circ }\right)+\left({\tan 8}^{\circ }\right)+\left({\tan 9}^{\circ }\right)+\left({\tan 10}^{\circ }\right)}$
My attempt:
${\displaystyle \cos x+\cos 2x+\cos 3x+...+\cos nx=\frac{\sin (n+\frac{1}{2})x-\frac{\sin x}{2}}{2\frac{\sin x}{2}}}$
${\displaystyle \sin x+\sin 2x+\sin 3x+...+\sin nx=\frac{\frac{\cos x}{2}-\cos (n+\frac{1}{2})x}{2\frac{\sin x}{2}}}$
Since angles are small we use term by term division which gives:
${\displaystyle \tan x+\tan 2x+\tan 3x+...+\tan nx=\frac{\frac{\cos x}{2}-\cos (n+\frac{1}{2})x}{\sin (n+\frac{1}{2})x-\frac{\sin x}{2}}}$
Putting x=1 and n=10 gives s=0.9629..
Wolfram alpha gives s=0.9653..
Is my calculations correct?

Solve ${\displaystyle \cos \left(x\right)-\sin 3x=\cos 2x}$
I have tried to rewrite as follows:
${\displaystyle \cos \left(2x\right)-\cos \left(x\right)=-\sin 3x}$
${\displaystyle =-2\sin \left(\frac{3}{2}x\right)\sin x=-3\sin x+4{\sin }^{3}x}$
Is it correct? Is it possible to proceed from here or I need to transform it differently?
Transform ${\displaystyle \cos 2x}$ into ${\displaystyle 1-2{\sin }^{2}x}$ is not useful here either.

Prove that ${\displaystyle \sin \left(x\right)\sin (\frac{\pi }{3}+x)\sin (\frac{\pi }{3}-x)=\frac{1}{4}\sin 3x}$

Evaluation of ${\displaystyle {\int }_0^{\frac{\pi }{2}}\frac{1}{{(a{\cos }^2\left(x\right)+b{\sin }^2\left(x\right))}^n}dx}$
${\displaystyle n=1,2,3,\dots }$
I thought about using ${\displaystyle u=\tan x\text{ or }u=\frac{\pi }{2}-x}$ but did not work.

Let ${\displaystyle z=p{e}^{i\theta }}$ be a complex number. What is the argument of ${\displaystyle 1-\bar{z}}$ in terms of ${\displaystyle \theta }$? Is it ${\displaystyle \pi -\theta }$ because the argument of ${\displaystyle -\bar{z}\text{ is }\pi -\theta }$?
Follow-up: Is there a way to simplify the argument: ${\displaystyle \arctan \left(\frac{\rho \sin \theta }{1-\rho \cos \theta }\right)}$?

Finf the critical points of f(x,y)= e^xy +3 and use the second derivative test

What is the value of: ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\sum _{r=1}^{n-1}\frac{{\cot }^2\left(r\frac{\pi }{n}\right)}{n^2}}$

How do I show that ${\displaystyle \cos \left(t\right)+1=2{\cos }^2\left(\frac{t}{2}\right)}$
I have been trying to show that ${\displaystyle \cos \left(t\right)+1}$ is equal to ${\displaystyle 2{\cos }^2\left(\frac{t}{2}\right)}$ Heres

How many solutions has the equation ${\displaystyle \tan \left(3x\right)=1}$?
Ive

Show that when written in terms of t, where ${\displaystyle t=\tan \left(\frac{x}{2}\right)}$ the expression ${\displaystyle 2(1+\cos \left(x\right))(5\sin \left(x\right)+12\cos \left(x\right)+13)}$ is a perfect square.

Idea for ${\displaystyle \underset{x\to \frac{\pi }{2}}{lim}{\left(\tan \left(\frac{\pi }{4}\sin x\right)\right)}^{\frac{1}{\tan \left(\pi \sin x\right)}}}$
Need help for this. I tried using trigonometric identities but nothing seems to fit. I'm tired and clueless.