# 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.

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Do you want a polar equation? $r=2\cos2\theta+3\sin2\theta$? – David Mitra May 9 '12 at 0:51
Or is it $y=2\cos2x+3\sin2x$? – David Mitra May 9 '12 at 0:56
I think, it is almost the same? or totally different? – Sb Sangpi May 9 '12 at 0:58
They are totally different. – David Mitra May 9 '12 at 0:58
I presume you want the graph of $y=\cos 2x+3\sin2x$ for $x\in[-\pi,\pi]$. – David Mitra May 9 '12 at 1:00

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.

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thx for that! Does your answer look the same as my graph. $y=3sin(2\theta)$ its not in the graph. thx – Sb Sangpi May 9 '12 at 4:31
I didn't graph it myself; I know how to do it, and I don't have to turn it in, so there was no point in my drawing it. The graph you added seems to be the graph of something else, so I do not see how it is relevant. – Arturo Magidin May 9 '12 at 4:36
yea.when I read your answer, I understand it but compare to the answer graph.It make me confuse, the answer its the graph. thx@Arturo Magidin – Sb Sangpi May 9 '12 at 5:17
@SbSangpi: No. The graph you posted is the graph of $y=\sin\theta + 2\cos\theta$, just like it says, not of $y=2\cos(2\theta)+3\sin(2\theta)$. The dashed red line is the graph of $y=2\cos\theta$, the dashed yellow line is the graph of $y=\sin\theta$, and the solid blue line is the graph of the sum. There is nothing to compare it to, because (i) I have not posted a graph; and (ii) the graph you posted is completely and utterly irrelevant to your question. What makes you think it is related, when it is so clearly and prominently labeled differently? – Arturo Magidin May 9 '12 at 5:20
sorry. You are right.It is totally different! thx – Sb Sangpi May 9 '12 at 5:34

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.

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It is like this.

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yea, but can you please show me the working out. Sorry I'll be able to work out other questions as well. thx so much – Sb Sangpi May 9 '12 at 1:13