Practice Exercises for Trigonometry

Solve ${\displaystyle 2{\cos }^{2}x+\sin x=1}$ for all possible x

Solving ${\displaystyle \tan 2x=1+2\sin 4x}$

What is the period of the ${\displaystyle f\left(x\right)=\sin x+\sin 3x}$?
${\displaystyle f\left(x\right)=\sin x+\sin 3x=2\frac{3x+x}{2}\cos \frac{x-3x}{2}=2\sin 2x\cos x=4\sin x{\cos }^{2}x}$
${\displaystyle f(x+T)=4\sin (x+T){\cos }^2(x+T)=4\sin x{\cos }^{2}x}$

What is ${\displaystyle \arctan \left(z_1\right)\pm \arctan \left(z_2\right)}$ with ${\displaystyle z_1,z_2\in \mathbb{C}}$
On wikipedia, there is the following identity:
${\displaystyle \arctan \left(u\right)\pm \arctan \left(v\right)=\arctan \left(\frac{u\pm v}{1\mp uv}\right)}$
However when I try some ${\displaystyle u,v\in \mathbb{C}}$ to check, the formula does not hold. Is there an equivalent formula for complex numbers?

What is the value of ${\displaystyle \tan \theta =\frac{1}{-1}}$?

If ${\displaystyle a_k=\tan (\sqrt{2}+\frac{k\pi }{2011})}$ , then evaluate ${\displaystyle \frac{a_1+a_2+\dots +a_{2011}}{a_1{a}_2\dots a_{2011}}}$

Express ${\displaystyle \cos \left(\frac{\alpha }{3}\right)}$ as a function of ${\displaystyle \cos \left(\alpha \right)}$
Give an expression for ${\displaystyle \cos \left(\frac{\alpha }{3}\right)}$ one can do the following trick:
${\displaystyle \cos \left(3x\right)=\cos (2x+x)=\cos \left(2x\right)\cos \left(x\right)-\sin \left(2x\right)\sin \left(x\right)}$
${\displaystyle =[{\cos }^2\left(x\right)-{\sin }^2\left(x\right)]\cos -2{\sin }^2\left(x\right)\cos \left(x\right)=4{\cos }^3\left(x\right)-3\cos \left(x\right)}$
By defining ${\displaystyle \alpha =3x}$, we have
${\displaystyle 4{\cos }^3\left(\frac{\alpha }{3}\right)-3\cos \left(\frac{\alpha }{3}\right)=\cos \left(\alpha \right)}$
However, I have not found any way to obtain an expression for ${\displaystyle \cos \left(\frac{\alpha }{3}\right)}$ solely as a function of ${\displaystyle \cos \left(\alpha \right)}$, as one can instead do with ${\displaystyle \cos \left(\frac{\alpha }{2}\right)=\pm \frac{1}{\sqrt{2}}\sqrt{1+\cos \left(\alpha \right)}}$
Is it simply impossible or is there a way to obtain the desired result?

Simplifying ${\displaystyle \frac{{\cos }^{2}x-{\sin }^{2}x}{\sin 2x}}$

How to evaluate limit without LHospitals rule, but using ${\displaystyle \underset{x\to 0}{lim}\frac{\sin x}{x}=1}$
I am having hard time evaluating this limit:
${\displaystyle \underset{x\to 0}{lim}\frac{\cos 3x-1}{\cos 2x-1}}$
I know, that this fact can help
${\displaystyle \underset{x\to 0}{lim}\frac{\sin x}{x}=1}$
But how? We can transform the first expression to
${\displaystyle \underset{x\to 0}{lim}\frac{\cos 3x-1}{-2{\sin }^{2}x}}$
but what should I do next? I cannot use LHopitals rule for solving this.

Simplify ${\displaystyle \frac{1}{1-z}}$, given ${\displaystyle z=\cos \vartheta +i\sin \vartheta }$
Im

How can I rewrite ${\displaystyle {\sin }^3\left(x\right)+{\csc }^3\left(x\right)}$?
If ${\displaystyle m=\sin x+\csc x}$ compute:
${\displaystyle R={\sin }^{3}x+{\csc }^{3}x}$
how can I rewrite this expression

Finding the range of ${\displaystyle f\left(x\right)=\frac{{\sin }^{2}x+\sin x-1}{{\sin }^{2}x-\sin x+2}}$
I am attempting to find the range of the function ${\displaystyle f\to \mathbb{R}\to \mathbb{R}}$ which is given by the expression as follows.
${\displaystyle f\left(x\right)=\frac{{\sin }^{2}x+\sin x-1}{{\sin }^{2}x-\sin x+2}}$
This is rational function in ${\displaystyle \sin x}$, so it will be periodic. Usually when there is a rational function with quadratic expressions, differentiation can be used or, x is rewritten in terms of y, so that gives.
${\displaystyle \sin x=\frac{(y+1)\pm \sqrt{10-2y-3y^2}}{2(y-1)}\in [-1,1]}$
This is a tedious task and so is using differentiation to find minima and maxima. Any hints to go about this problem using a different technique are appreciated.

how is this solved? ${\displaystyle \underset{x\to 0}{lim}\frac{1}{x^2}(\left({\tan (x+\frac{\pi }{4})}^{\frac{1}{x}}-e^2\right)}$

Finding the ${\displaystyle \underset{x\to 0}{lim}\frac{1-\cos x}{\sin x\ln (1+x)}}$ using Taylors

Evaluate ${\displaystyle \int {\cos }^2\left(x\right){\tan }^3\left(x\right)dx}$ using trigonometric substitution

Evaluate
${\displaystyle \cos \theta \cos 2\theta \cos 3\theta +\cos 2\theta \cos 3\theta \cos 4\theta +\dots }$
upto n terms

If ${\displaystyle \sin \alpha +\cos \alpha =-\frac{\sqrt{7}}{2}}$, then ${\displaystyle \alpha }$ is the angle of which quadrant?
My attempts:
${\displaystyle \sqrt{2}\sin (\frac{\pi }{4}+\alpha )=-\frac{\sqrt{7}}{2}}$
${\displaystyle \sin (\frac{\pi }{4}+\alpha )=-\frac{\sqrt{14}}{4}}$

Can I put ${\displaystyle x=\frac{\pi }{2}}$ in ${\displaystyle \tan 2x=\frac{2\tan x}{1-{\tan }^{2}x}}$?

How would I go about finding the roots of ${\displaystyle (e^x-1)-k\arctan \left(x\right)=0}$?

For ${\displaystyle \alpha \in (0^{\circ };{90}^{\circ })}$ simplify ${\displaystyle {\sin }^2\alpha +{\tan }^2\alpha +{\sin }^2\alpha \cdot {\cos }^2\alpha +{\cos }^4\alpha }$
My try: ${\displaystyle {\sin }^2\alpha +{\tan }^2\alpha +{\sin }^2\alpha \cdot {\cos }^2\alpha +{\cos }^4\alpha ={\sin }^2\alpha +\frac{{\sin }^2\alpha }{{\cos }^2\alpha }+{\sin }^2\alpha \cdot {\cos }^2\alpha +{\cos }^4\alpha }$
${\displaystyle =\frac{{\sin }^2\alpha \cdot {\cos }^2\alpha +{\sin }^2+{\sin }^2\alpha \cdot {\cos }^4\alpha +{\cos }^6\alpha }{{\cos }^2\alpha }}$