Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am having trouble proving the following identity (where $m,n \in \mathbb{R}$ are arbitrary): $$\sin(mx)\sin(nx) = \frac{1}{2}[\cos(m -n )x - \cos(m + n)x] \quad (1)$$ By expanding the RHS, I can prove the following: $$ \begin{align*} &= \frac{1}{2}[x(\cos(m)\cos(-n) - \sin(m)\sin(-n)) - x(\cos(m)\cos(n) - \sin(m)\sin(n))] \\ &= \frac{1}{2}[x\cos(m)(\cos(-n) - \cos(n)) + x\sin(m)(\sin(n) - \sin(-n))] \\ &= \frac{1}{2}[x\cos(m)\cdot0 + x\sin(m)\cdot2\sin(n)] \\ &= x\sin(m)\sin(n) \end{align*} $$ but I do not see how this equals the LHS in $(1)$.

I eventually relented and checked the answer key, but to my dismay it gave a proof depending on the "identity": $$\cos(m - n)x = \cos(mx - nx) \quad (2)$$ Now if I pick $m = n = 0$ and $x = 100$, the LHS of (2) is $$\cos(0 + 0)\cdot100 =1\cdot100 = 100$$ while the RHS of (2) is $$\cos(0\cdot 100 + 0\cdot100) = \cos(0) = 1$$ so, I don't think the proof provided is valid.

How should I proceed?

share|improve this question

2 Answers 2

up vote 3 down vote accepted

It seems you've made a mistake: The RHS should be written as $\frac{1}{2}\left( \cos[(m-n)x]-\cos[(m+n)x] \right)$ (i.e. the $x$'s belong inside the cosines.

share|improve this answer
    
That's indeed the rhs the OP started from. The mistake is the sudden arrival of $x$ in the second line. –  1015 Feb 21 '13 at 19:00
    
@julien No, I do not interpret the $x$ to be inside the brackets. This is exactly why I believe brackets should not be optional on trig functions (or linear transformations). –  providence Feb 21 '13 at 19:07
    
@providence Then the formula you are trying to prove is wrong. I agree with you, there should be brackets to make things clear. The formula is correct if you take the rhs to be $(\cos((m-n)x)-\cos((m+n)x))/2$. –  1015 Feb 21 '13 at 19:09

Using

$$\cos(a\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$$

we have:

$$\frac{1}{2}\left[\cos(m-n)x-\cos(m+n)x\right]=\frac{1}{2}\left(\cos mx\cos nx+\sin mx\sin nx-\cos mx\cos nx+\sin mx\sin nx\right)=\sin mx\sin nx$$

share|improve this answer
    
So I'm now quite confused as to where the $x$ is. To me, $\cos(m-n)x$ reads as equal to $x\cos(m-n)$. Should I be reading it $\cos((m-n)x)$? –  providence Feb 21 '13 at 19:04
    
The second option is, imo, the most natural and widespread one, as any more or less normal mathematician (yes, the two of us both!) would never write something so confusing: we'd always write $\,x\cos m\,$ to denote the product of the identity with the cosine. –  DonAntonio Feb 21 '13 at 19:06

Your Answer

 
discard

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.