Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
+1 for your work shown in comments to André's answer. – The Chaz 2.0 Jun 15 '12 at 1:26
up vote 5 down vote accepted

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.

share|cite|improve this answer
How did you arrive to $\sqrt{65}$? – εν οίδα ότι ουδέν οίδα Jun 14 '12 at 21:44
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
So in a case like $-2\sin(3x)-5\cos(3x)$ would I use $\sqrt{(29)}$? – εν οίδα ότι ουδέν οίδα Jun 14 '12 at 21:46
Yes.${}{}{}{}{}{}{}{}{}{}{}$ – André Nicolas Jun 14 '12 at 21:47
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

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?

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.