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Consider the following terms:

$cos5x$ and $sin^2x$

Are these terms equivalent to:

$5cosx$ and $(sinx)^2$

If not please explain. If so please confirm.

Thanks

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${\sin ^2}(x) = {(\sin x)^2}$ -- yes, but $\cos 5x \ne 5\cos x$. Writing ${\sin ^2}(x) = {(\sin x)^2}$ is just a convention adopted a long time ago. –  glebovg Oct 22 '12 at 5:53
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They aren't statements, they are terms, and, no, they are not equivalent (the ones with cosine), as you can tell by substituting in, say, $x=1$, and evaluating on a calculator. –  Gerry Myerson Oct 22 '12 at 5:54
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They are actually statements because $\cos 5x$ and ${\sin ^2}(x)$ is a statement declaring that the cosine function and the sine function have such and such properties, but we usually think of them as terms. –  glebovg Oct 22 '12 at 6:01
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3 Answers

up vote 0 down vote accepted

No.

The first $\cos 5x$ means $\cos(5x)$, while $\sin^2x$ means $(\sin x)^2$.

The parentheses are omitted in the first case as the argument $5x$ is understood. The square is placed after the $\sin$ but before the $x$ so as to differentiate $\sin(x^2)$ from $(\sin x)^2$.

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No! for the first one, you can take $x = \pi /5$, then $\cos(5x) = -1$ while $5\cos(x) = 0.9$. The second however are equivalent!

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The main problem is that there are traditional ways of writing arguments and powers in connection with trigonometric functions. Always adding parentheses would probably be helpful, especially for beginners.

Thus (with "$\ne$" here meaning "not equal in general") $$\sin^2x = (\sin(x))^2 \ne \sin(x^2)\text{ and }\ne\sin(\sin(x))$$ (though there are areas where the latter would be the preferred interpretation).

By the same writing traditions, $$\cos 5x = \cos(5x)\ne \cos(5)\cdot x$$ (and even more so $\ne5\cdot\cos(x)$ as checked by plugging in appropriate values for $x$). Beware however that one would always read $$\sin x\cos x=\sin(x)\cdot\cos(x)\ne \sin(x\cdot\cos(x)).$$

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