Simplify $2 \sin(x) \cos(7x) + \sin(6x)$

Question

I was doing a problem and in my chain of computations I arrived at a seemingly complicated function $$2 \sin(x) \cos(7x) + \sin(6x)$$

However, I typed it into Wolfram and was surprised to find $$2 \sin(x) \cos(7x) + \sin(6x) = \sin(8x)$$

Is there a way I could have seen this beforehand? Even now I'm unsure how I would prove this identity.

score 11 · Answer 1 · answered Jul 28 '14 at 03:50

11

$$2\sin(x)\cos(7x)+\sin(6x)=2\sin(x)\cos(7x)+\sin(7x-x)$$

$$=2\sin(x)\cos(7x)+\sin(7x)\cos(x)-\sin(x)\cos(7x)$$

$$=\sin(x)\cos(7x)+\sin(7x)\cos(x)=\sin(8x)$$

answered Jul 28 '14 at 03:50

user71352

13,038

1

Ah yes, I think if I was more experienced I could have seen that $6x = 7x - x$ – Mark Jul 28 '14 at 03:57
I found that tricky as well. – user71352 Jul 28 '14 at 04:00
I believe things can be largely simplified: see my answer below – lab bhattacharjee Jul 28 '14 at 15:42

score 4 · Answer 2 · answered Jul 28 '14 at 04:02

4

There is a group of identities called the product-to-sum identities (Wikipedia has a very useful list of such things); the one that's relevant here is as follows: $$\sin\alpha\cos\beta = \frac{\sin(\alpha+\beta) + \sin(\alpha-\beta)}{2}$$

Applying the above, we have:

$$\begin{align} 2\sin x \cos 7x + \sin 6x &= 2\big( \frac{\sin 8x + \sin (-6x)}{2}\big) + \sin 6x\\ &=\sin 8x -\sin6x + \sin 6x\\ &=\sin 8x \end{align}$$

answered Jul 28 '14 at 04:02

Théophile

24,627

1

Oh cool. I like this because it's more of a forward direction than the other answer which is like working backwards. – Mark Jul 28 '14 at 04:37
@Mark, $$3$rd of http://mathworld.wolfram.com/WernerFormulas.html – lab bhattacharjee Jul 28 '14 at 07:07

score 4 · Accepted Answer · answered Jul 28 '14 at 05:32

4

If you don't want to remember or look up any trig identities at all, you can turn everything into exponentials:

\begin{eqnarray} 2 \sin x \cos (7x) + \sin (6x) &=& 2\left(\frac{e^{ix}-e^{-ix}}{2i}\right)\left(\frac{e^{7ix}+e^{-7ix}}{2}\right) + \frac{e^{6ix}-e^{-6ix}}{2i}\\ &=&\frac{1}{2i}\left(e^{8ix}-e^{6ix}+e^{-6ix}-e^{-8ix} + e^{6ix}-e^{-6ix}\right)\\ &=&\frac{e^{8ix}-e^{-8ix}}{2i}\\ &=&\sin(8x) \end{eqnarray}

answered Jul 28 '14 at 05:32

Micah

38,108
15
85
133

This one is good because it doesnt involve playing with identities. No intuition required. – jgyou Jul 28 '14 at 06:26
This is my favorite one yet. Thanks! – Mark Jul 29 '14 at 10:59

score 2 · Answer 4 · answered Jul 28 '14 at 03:52

2

Another method. Take the derivative, which ends up being $8\cos(8x)$, and then integrate to get $\sin(8x)$

answered Jul 28 '14 at 03:52

Asimov

3,024

But is it easier to show the derivative is $8 \cos(8x)$? – Mark Jul 28 '14 at 03:53
Easier? Maybe, maybe not. I never liked using identities, so I usually do a calculus approach. This is just my way of doing it. – Asimov Jul 28 '14 at 03:54
The other answer is a bit more elegant, and is easier for those who dont know any calculus. – Asimov Jul 28 '14 at 03:55
If you do this, you also have to check that the original function matches $\sin(8x)$ at some point. Otherwise you could be off by a constant. – Micah Jul 28 '14 at 05:37
@Micah Good point, forgot to mention that. The $+c$ can be the hardest part to remember sometimes – Asimov Jul 28 '14 at 18:24

Simplify $2 \sin(x) \cos(7x) + \sin(6x)$

4 Answers4