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tainted
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Homework Statement
Prove the following formula
[itex]
\int_{-\pi}^{\pi} \sin(mx)\cos(nx)\,dx = 0\\
(m, n = \pm 1, \pm 2, \pm 3, ...)
[/itex]
Homework Equations
[itex]
\sin(A)\cos(B) = \frac{1}{2}[\sin(A-B)+\sin(A+B)]
[/itex]
The Attempt at a Solution
[itex]
\int_{-\pi}^{\pi} \sin(mx)\cos(nx)\,dx\\
\int_{-\pi}^{\pi} \sin(mx-nx)+\sin(mx+nx)\,dx\\
\int_{-\pi}^{\pi} \sin((m-n)x)+\sin((m+n)x)\,dx
[/itex]
Edit
[itex]
\int_{-\pi}^{\pi} \sin((m-n)x)\,dx + \int_{-\pi}^{\pi} \sin((m+n)x)\,dx
[/itex]
the function, sine of any constant times x, is an odd function, therefore when it is integraded from -a to a, its value is 0.
Therefore, we get
[itex]
\int_{-\pi}^{\pi} \sin((m-n)x)\,dx + \int_{-\pi}^{\pi} \sin((m+n)x)\,dx = 0 + 0 = 0
[/itex]
Would that be sufficient in the proof?
Thanks guys!
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