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We know that for odd function $f(-x) = -f(x)$ and for even function $f(-x) = f(x)$.

Therefore, $\cos^3(-x) = \cos(-x)\cos(-x)\cos(-x) = \cos{x}\cos{x}\cos{x} = \cos^3{x}$ (i.e. $\cos^3{x}$ must be even function). And similarly, since $\sin(-x) = - \sin{x}$, $\sin^3{x}$ must be odd function.

But in my text book they claimed that $\cos^3{x}$ is odd function while $\sin^3{x}$ is even function. Maybe I have done something wrong, but I am unable to understand how $\cos^3{x}$ is an odd function while $\sin^3{x}$ an even function. Please help. Thanks.

Ritu
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    Sounds like you may have spotted a mistake in your textbook. – Mike Apr 02 '15 at 07:49
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    You are right and the book is wrong. – zhw. Apr 02 '15 at 07:49
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    That doesn't mean anything. The definite integral of an odd function with bounds symmetric about $0$ is always $0$. In fact, you can rewrite that integral as $\int_{-\frac\pi2}^{\frac\pi2}\sin^3xdx$. And since that integrand is odd and that interval symmetric, it must integrate to 0. – Mike Apr 02 '15 at 08:06
  • And apparently the comment I was replying to was deleted... – Mike Apr 02 '15 at 08:07
  • @Mike - It was deleted because after posting it I realized I was wrong... – Ritu Apr 03 '15 at 02:15

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