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Find the period of $2\sin <!--\; \; -->(2x){\cos }^2<!--\; \; -->(x)$

How do I convert to only cos?
$y=\cos <!--\; \; -->(t)-<!--\; -\; -->\sin <!--\; \; -->(t)\to <!--\; \to \; -->y=A\cos <!--\; \; -->(\mathrm{\omega <!--\; \omega \; -->}t-<!--\; -\; -->\mathrm{\varphi <!--\; \varphi \; -->})$

Prove $\cos <!--\; \; -->x+\cos <!--\; \; -->(2x)+\cdots <!--\; \cdots \; -->+\cos <!--\; \; -->(nx)=\frac{\sin <!--\; \; -->(\frac{nx}{2})\cos <!--\; \; -->\frac{(n+1)x}{2}}{\sin <!--\; \; -->(\frac{x}{2})}.(1)$

Sum to infinity of trignometry inverse: $\underset{r=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\arctan <!--\; \; -->\left(\frac{4}{r^2+3}\right)$

Why is $2{\cos }^2<!--\; \; -->{10}^{\circ <!--\; \circ \; -->}-<!--\; -\; -->2{\cos }^2<!--\; \; -->{20}^{\circ <!--\; \circ \; -->}=\sin <!--\; \; -->{10}^{\circ <!--\; \circ \; -->}$?

The question gives us 3 vowels (A,E,O) and 4 consonants (B,C,D,F).
a) How many ways can you make a 7 letter word if each letter can only be used once. (Word doesn't have to be real). This was easy, as the answer is just 7!
b) If the vowels have to be together and the consonants have to be together?
c) If the vowels have to be together?
d) If B and C have to be together, but no other vowels or consonants can be together?

If $\frac{{\sin }^3<!--\; \; -->x-<!--\; -\; -->{\cos }^3<!--\; \; -->x}{\sin <!--\; \; -->x-<!--\; -\; -->\cos <!--\; \; -->x}-<!--\; -\; -->\frac{\cos <!--\; \; -->x}{\sqrt{1+{\cot }^2<!--\; \; -->x}}-<!--\; -\; -->2\tan <!--\; \; -->x\cot <!--\; \; -->x=-<!--\; -\; -->1$ where $x\in <!--\; \in \; -->[0,2\mathrm{\pi <!--\; \pi \; -->}]$

You have 2 boxes that contains 100 balls each, and the balls can be white, blue, red, or half red and blue.
Box A contains: 90 white balls, 10 red balls
Box B contains: 86 white balls, 4 red balls, 9 blue balls, 1 red and blue ball
If you don't know which box you pick a ball from (the probability is equal for what box you chose), what is the probability for pulling exactly 2 white balls and 1 red ball, if you don't put it back and don't take a look at the balls until all 3 are pulled from the box you chose? (You only pick 3 balls).
I thought this would be pretty straight forward by using combinatorics:
P(1 red and 2 white balls) = $\frac{(\genfrac{}{}{0ex}{}{176}{2})}{(\genfrac{}{}{0ex}{}{200}{3})}$

If $f(\mathrm{\theta <!--\; \theta \; -->})=sin(\mathrm{\theta <!--\; \theta \; -->})=0.1$, then find $f(\mathrm{\theta <!--\; \theta \; -->}+\mathrm{\pi <!--\; \pi \; -->})=sin(\mathrm{\theta <!--\; \theta \; -->}+\mathrm{\pi <!--\; \pi \; -->})$

Solve, in the interval $0^{\circ <!--\; \circ \; -->}\le <!--\; \le \; -->\mathrm{\theta <!--\; \theta \; -->}\le <!--\; \le \; -->{360}^{\circ <!--\; \circ \; -->}$
$\cos <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}+{25}^{\circ <!--\; \circ \; -->})+\sin <!--\; \; -->(\mathrm{\theta <!--\; \theta \; -->}+{65}^{\circ <!--\; \circ \; -->})=1$

Suppose $S\subset <!--\; \subset \; -->\mathbb{Z}$ (set of integers) such that
1) $|S|=15$
2) $\mathrm{\forall <!--\; \forall \; -->}s\in <!--\; \in \; -->S,\mathrm{\exists <!--\; \exists \; -->}a,b\in <!--\; \in \; -->S$ such that $s=a+b$
Show that for every such $S$, there is a non-empty subset $T$ of $S$ such that the sum of elements of $T$ is zero and $|T|\le <!--\; \le \; -->7$.

Given $n+1$ points $(x_i,f_i)$ with $x_i\ne <!--\; \ne \; -->x_j$ for $i\ne <!--\; \ne \; -->j$, define $L_k(x)$ to be
$\frac{(x-<!--\; -\; -->x_0)(x-<!--\; -\; -->x_1)\cdots <!--\; \cdots \; -->(x-<!--\; -\; -->x_{k-<!--\; -\; -->1})(x-<!--\; -\; -->x_{k+1})\cdots <!--\; \cdots \; -->(x-<!--\; -\; -->x_n)}{(x_k-<!--\; -\; -->x_0)(x_k-<!--\; -\; -->x_1)\cdots <!--\; \cdots \; -->(x_k-<!--\; -\; -->x_{k-<!--\; -\; -->1})(x_k-<!--\; -\; -->x_{k+1})\cdots <!--\; \cdots \; -->(x_k-<!--\; -\; -->x_n)}\cdots <!--\; \cdots \; -->(1)$
Show that $L_k(x_j)$ is a polynomial of degree n, that $L_k(x_j)=0$ for $j\ne <!--\; \ne \; -->k$ and $L_k(x_k)=1$

Simplify $\frac{\cos <!--\; \; -->(2x)}{\cot <!--\; \; -->(x)-<!--\; -\; -->1}-<!--\; -\; -->\frac{\sin <!--\; \; -->(2x)}{2}$

Using equation $9x^2+8xy+7y^2=6$ to find maximum possible value of $7x+5y+12xy$

Minimizing ${({\sin }^2<!--\; \; -->(x)+\frac{1}{{\sin }^2<!--\; \; -->(x)})}^2+{({\cos }^2<!--\; \; -->(x)+\frac{1}{{\cos }^2<!--\; \; -->(x)})}^2$

There are 4 rows and 3 columns in a table, and each slot is painted with black or white with equal probability, and I wish to find the probability that one, and only one of the rows is painted black. What I did was:
$P=4\cdot <!--\; \cdot \; -->(1/2{)}^3\ast <!--\; \ast \; -->(7/8{)}^3=\frac{343}{1024}$
and it indeed seems to be the correct answer. I thought I should also try solving it using combinatorics, but that is something I fail to do. My try is:
$\frac{4\cdot <!--\; \cdot \; -->1\cdot <!--\; \cdot \; -->(2^9-<!--\; -\; -->7)}{2^{12}}=\frac{505}{1024}\ne <!--\; \ne \; -->\frac{343}{1024}$
what have I done wrong?

Determine the remainder when $(x^4-<!--\; -\; -->1)(x^2-<!--\; -\; -->1)$ is divided by $1+x+x^2$

Question: Find the general solution to $\sin <!--\; \; -->(4x)-<!--\; -\; -->\cos <!--\; \; -->(x)=0$

I need to find the limit for $x\in <!--\; \in \; -->\mathbb{R}$
$\underset{n\to <!--\; \backslash rightarrow\; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}|e^{ixk/n}-<!--\; -\; -->e^{ix(k-<!--\; -\; -->1)/n}|$
Using Euler's formula and cosine of a sum this simplifies to
$\underset{n\to <!--\; \backslash rightarrow\; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}n\sqrt{2(1-<!--\; -\; -->\cos <!--\; \; -->(x/n))}$

There are 9 boys and 7 girls sitting around a table. If a boy sits next to a girl we call it a switch. Example: In BBBBBGGGBBGBGGG there are 6 switches.(If a boy sits next to 2 girls, there are 2 switches counted.) What is the expected number of switches?