Show that $(7\cos(x)-\sin(x))^2=A\cos(2x)+B\sin(2x)+C$ for some integers $A,B,C$ How do you solve this question?:$$(7\cos(x)-\sin(x))^2=A\cos(2x)+B\sin(2x)+C$$ is for all $x$. Here $A$, $B$ and $C$ is constants. I need to know $A$, $B$ and $C$ to pass this. They are  integers. 
I got this far:
(LS = left side)
$$LS =  49\cos^2(x)+\sin^2(x)+14\sin(x)\cos(x)$$
And then i follow some steps online and got right side to
$$RS = (A+C)\cos^2(x)+(C-A)\sin^2(x)+2B\sin(x)\cos(x)$$
Only problem is that i get C to 49/2.
 A: $$(7 \cos{(x)}-\sin{(x)})^2=A  \cos{(2x)}+B \sin{(2x)}+C \\ \Rightarrow 49 \cos^2{(x)}-14 \cos{(x)} \sin{(x)}+\sin^2{(x)}=A \cos{(2x)}+B \sin{(2x)}+C \ \ \ (*)$$
Use the formulas:
$$\cos{(2x)}=\cos^2{(x)}-\sin^2{(x)} \text{ and } \sin{(2x)}=2 \sin{(x)} \cos{(x)}$$
A: apply the following identities to your LHS:


*

*$sin^2(x)+cos^2(x)=1$ and 49=48+1;

*$cos^2(x)=1/2+(cos(2x)/2)$

*$sin^2(x)=1/2-(cos(2x)/2)$

*$2sin(x)cos(x)=sin(2x)$


And don't touch the original RHS!


Second method: you can give three different values to $x$ and solve a linear system:
for example set $x=0$ ----> $A+C=49$
$x=\pi/4$ ----->?
A: Anything forall x implies an identity so you can put any * values of x to get C. for e.g.:
$$(7\cos(x)-\sin(x))^2=A\cos(2x)+B\sin(2x)+C$$
$$
\left\{ 
\begin{array}{c}
49=A+C\quad x=0  \\ 
1=-A+C\quad x=\pi/2\\
18=B+C\\ 
\end{array}
\right. 
$$
Solving these we get: 
$$
\left\{ 
\begin{array}{c}
A=24  \\ 
C=25\\
B=-7\\
\end{array}
\right. 
$$

Alternatively:
$$(7\cos(x)-\sin(x))^2=49\cos^2x+\sin^2x-14\cos x\sin x
\\=49\left(\frac{1+\cos{2x}}2\right)+\left(\frac{1-\cos{2x}}2\right)-7\sin 2x
\\=24\cos 2x-7\sin 2x+25$$
So $A=24,B=-7,C=25$

Use:
$$\cos 2x=2\cos^2x-1=1-2\sin^2x,\sin 2x=2\sin x\cos x$$
A: To match coefficients of $\cos^2 x$ and of $\sin^2 x$ you were solving 
$$
\left\{
\begin{array}{c}
C+A = 49 \\
C-A = 1
\end{array}
\right.
$$
This gives $C = 25$ and $A = 24$.
You probably did not notice that $1 \cdot \sin^2 x$ was needed so you used a zero in the second equation, getting $C = A = 49/2$.
