# Trigonometric proof query

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?

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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.

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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:

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

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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