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I am asked as a part of a question to express $\sin(2x)-8\cos(2x)$ as a single sine function.

I know it has something to do with the trigonometric identity $$\sin(a-b)=\sin(a) \cos(b)-\cos(a)\sin(b)$$ but I can't get my head around it because of that $8$ in front of $\cos2x$.

Any tips on how I can move on?

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+1 for your work shown in comments to André's answer. –  The Chaz 2.0 Jun 15 '12 at 1:26
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3 Answers 3

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Hint: Our function is $$\sqrt{65}\left(\frac{1}{\sqrt{65}}\sin 2x -\frac{8}{\sqrt{65}}\cos 2x\right).$$ Let $b$ be an angle whose cosine is $\frac{1}{\sqrt{65}}$ and whose sine is $\frac{8}{\sqrt{65}}$, and use the identity you quoted.

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How did you arrive to $\sqrt{65}$? –  Panayiotis Jun 14 '12 at 21:44
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It is the square root of $1^2+8^2$. By dividing by it, we transform our numbers $1$ and $8$ into numbers that can be a cosine and sine of something. Recall that $\cos^2 b+\sin^2 b$ must be $1$. –  André Nicolas Jun 14 '12 at 21:45
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So in a case like $-2\sin(3x)-5\cos(3x)$ would I use $\sqrt{(29)}$? –  Panayiotis Jun 14 '12 at 21:46
    
Yes.${}{}{}{}{}{}{}{}{}{}{}$ –  André Nicolas Jun 14 '12 at 21:47
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I called it $b$. And it would be $\sin(2x-b)$ although later we can change that. We want somebody whose cosine is $1/\sqrt{65}$and whose sine is $\dots$. If you press the arccos button you will have it. Or if you don't want numbers, can't really do better than $\arccos(1/\sqrt{65})$. –  André Nicolas Jun 14 '12 at 21:54
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I believe the answer you'll be looking for will be in the form

$$k\sin(2x-b)=k\sin 2x\cos b-k\cos 2x\sin b=\sin 2x-8\cos 2x$$

Equating coefficients, we get

$$k\cos b=1$$ $$k\sin b=8$$

Next, use the trig identity $\sin^2x+\cos^2x=1$ to solve for $k$. Can you take it from here?

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This is another way of getting to Andre's answer, which is probably easier to remember.

Let $u$ be so that $\tan(u)=8$. Then

$$\sin(2x)-8\cos(2x) = \sin(2x)-\tan(u)\cos(2x)= \sin(2x)-\frac{\sin(u)}{\cos(u)}\cos(2x) = \frac{\sin(2x)\cos(u)-\sin(u)\cos(2x)}{\cos(u)}$$

Now, knowing that $\tan(u)=8 \Rightarrow \cos(u)=...$, you recover that answer.

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