How to sketch $y=2\cos\,2x+3\sin\,2x$ , $x$ for $[-\pi,\pi]$. 
Use addition of ordinate to sketch the graph of  $y=2\cos\,2x+3\sin\,2x$  , $x$ for $[-\pi,\pi]$.

I know that there will be three line in graph from the example it show that
$x=0$, $x=\frac{\pi}{4}$, $x=\frac{\pi}{2}$ and something like that I haven't no clue how to do. Can you please explain in step by step, so that I'll be able to do other questions.
Answer look like this.
Thanks.!
 A: You probably know the graph of $y=\cos(\theta)$ and of $y=\sin(\theta)$ on $[-2\pi,2\pi]$.
The graph of $y=\cos(2\theta)$ on $[-\pi,\pi]$ is obtained from the graph of $y=\cos(\theta)$ on $[-2\pi,2\pi]$ by performing a horizontal compression by a factor of $2$ (we are making the change from $y=f(x)$ to $y=f(2x)$). 
Likewise, the graph of $y=\sin(2\theta)$ on $[-\pi,\pi]$ is the result of compressing horizontally by a factor of 2 the graph of $y=\sin(\theta)$ on $[-2\pi,2\pi]$.
The graph of $y=2\cos(2\theta)$ is obtained from the graph of $y=\cos(2\theta)$ by performing a vertical stretch by a factor of $2$. The graph of $y=3\sin(2\theta)$ is obtained from the graph of $y=\sin(2\theta)$ by performing a vertical stretch by a factor of $3$.
Once you have the graphs of both $y=2\cos(2\theta)$ and $y=3\sin(2\theta)$ (obtained by the simple geometric operations described above) you obtain the graph of
$$y= 2\cos(2\theta) + 3\sin(2\theta)$$
by "addition of ordinate". You want to imagine that you are graphing $y=3\sin(2\theta)$ "on top of" the graph of $y=2\cos(2\theta)$, so that you end up adding the values. You can get a fairly reasonable geometric approximation by doing this.
A: I assuming you meant $\theta \in [-\pi,\pi]$, rather than $x$.
You might notice that $y = \sqrt{13} (\frac{2}{\sqrt{13}} \cos 2 \theta+\frac{3}{\sqrt{13}} \sin 2 \theta)$. Let $\alpha$ be the angle such that $\sin(\alpha) = \frac{2}{\sqrt{13}}$, and $\cos(\alpha) = \frac{3}{\sqrt{13}}$ (you should convince yourself that such an angle exists). Then the sum-product formulae for $\sin, \cos$ gives:
$$y = \sqrt{13} (\sin\alpha \cos 2 \theta+\cos\alpha \sin 2 \theta) = \sqrt{13} \sin(\alpha+2\theta).$$
You can compute $\alpha \approx 33.69^{\circ}$, plotting the function should be straightforward after that.
A: 

It is like this. 
