Practice Exercises for Trigonometry

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

where does the ${\displaystyle \cos \left(\theta \right)=1-2{\sin }^2\left(\frac{\theta }{2}\right)}$ come from?

Interval in which ${\displaystyle (\cos p-1)x^2+\cos p.x+\sin p=0}$ has real roots
The solution given in my reference is the interval ${\displaystyle (0,\pi )}$
My Attempt
${\displaystyle \mathrm{\Delta }={\cos }^{2}p-4(\cos p-1)\sin p\ge 0}$
${\displaystyle {\cos }^{2}p-4\sin p\cos p+4\sin p\ge 0}$
${\displaystyle {\mathrm{\Delta }}^{\prime }=16{\sin }^{2}p-16\sin p\le 0\Rightarrow {\sin }^{2}p\le \sin p}$

Why the periodicity of solution for ${\displaystyle \theta +\gamma \theta =0}$ implies ${\displaystyle \sqrt{\gamma }=n\in \mathbb{N}}$
One solution for ${\displaystyle \theta +\gamma \theta =0}$
is, for ${\displaystyle \gamma >0}$
${\displaystyle \mathrm{\Theta }\left(\theta \right)=A\cos \sqrt{\gamma }\theta +B\sin \sqrt{\gamma }\theta }$

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.

Why ${\displaystyle {\int }_0^{2\pi }{\sin }^{2}xdx\ne {\int }_0^{2\pi }\sin x\sin nxdx\to n=1}$
By orthogonality, we know that
${\int }_0^{2\pi }\sin mx\sin nxdx=\pi$ if $m=n$ and 0 otherwise. Nevertheless, when I am calculating it as follows, I get 0.
${\displaystyle {\int }_0^{2\pi }\sin x\sin nxdx=\frac{1}{2}{[\frac{\sin \left[(1-n)x\right]}{1-n}-\frac{\sin \left[(1+n)x\right]}{1+n}]}_0^{2\pi }}$
But the above quantity is 0 even if n=1. Shouldn't it be ${\displaystyle \pi }$?

How can I prove that ${\displaystyle x_{n+1}=x_n\sin \left(x_n\right)}$ converges for any ${\displaystyle x_0}$?

How do I solve this inequality? ${\displaystyle 3\sin \left(2x\right)>\sin \left(x\right)+\cos \left(x\right)+1}$

How to prove ${\displaystyle \frac{\tan \left(A\right)}{\tan \left(A\right)}+\frac{\cot \left(A\right)}{\cot \left(A\right)}=\frac{1}{1-2{\cos \left(A\right)}^2}}$

Two sides of a triangle are $\sqrt{3}+1\text{ and }\sqrt{3}-1$ and the included angle is ${60}^{\circ }$. Find other angles My Attempt Let $a=\sqrt{3}+1,b=\sqrt{3}-1$ and $C=60$ $c^2=a^2+b^2-2.a.b\cos C$ $=(\sqrt{3}+1{)}^2+(\sqrt{3}-1{)}^2-2(\sqrt{3}+1)(\sqrt{3}-1).\frac{1}{2}=8-2=6$ $⟹c=\sqrt{6}=\sqrt{2}\sqrt{3}$ $\frac{a}{\sin A}=\frac{c}{\sin C}⟹\sin A=\frac{\sqrt{3}+1}{\sqrt{2}\sqrt{3}}\frac{\sqrt{3}}{2}=\frac{\sqrt{3}+1}{2\sqrt{2}}$ $A={75}^{\circ }\mathrm{\&}B={45}^{\circ }$ But the solution given in my reference is ${105}^{\circ }$ and ${15}^{\circ }$, what is going wrong with my attempt ?

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.