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Let's take the sine of $30^\circ$ which is one-half. If you take $\sin^2(30^\circ)$, would that just be the sine of $900$? Or would it be equal to one-quarter, or would it be equal to something completely different?

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    $\begingroup$ $\sin^2 x =(\sin x)^2$ $\endgroup$
    – Pedja
    Commented Mar 22, 2012 at 20:40
  • $\begingroup$ Understanding what is sin(x) may make things clearer. One source is: mathworld.wolfram.com/Sine.html $\endgroup$
    – NoChance
    Commented Mar 22, 2012 at 22:45

5 Answers 5

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I think you are confused with the following notation:

$$\sin ^2(x)=\sin x \cdot \sin x \neq \sin(x^2) ~~\mbox{very often.}$$

So, $\sin^2(30^\circ)=\dfrac 1 4$.


And, $\sin 900^\circ$ is not untractable either.

$$\sin 900^\circ=\sin 5 \pi=0$$


I am being nitpicky now:

When you write $900$, I assume that you mean $900^\circ$. But, in Mathematics, it is a convention that $900$ means $900^c=900$ radians. For definition of a radian and other details, you may want to look up Wikipedia.

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    $\begingroup$ Actually, there are infinitely many solutions to $\sin^2(x) = \sin(x^2)$, as should be obvious if you plot both sides as functions of $x$. The smallest positive solution appears to be about 1.3644129631252872330 (in radians). $\endgroup$ Commented Mar 22, 2012 at 20:53
  • $\begingroup$ Thank you. I was not being careful.... $\endgroup$
    – user21436
    Commented Mar 22, 2012 at 20:55
  • $\begingroup$ @IlmariKaronen I have now edited to make it unambiguous. $\endgroup$
    – user21436
    Commented Mar 22, 2012 at 20:56
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As many people have pointed out by now, $\sin^2 x$ is simply a "nickname" for $(\sin x)^2$. Therefore, $\sin^2\ 30 = (\sin 30)^2 = (1/2)^2 = 1/4$.

As it happens, though, there is another useful thing we can say about $\sin^2 x$:

$$\sin^2 x = (\sin x)^2 = \frac12 (1 - \cos (2 x)).$$

We can see this using the double-angle formula for cosines:

$$\frac12 (1 - \cos (2x)) = \frac12 (1 - (1 - 2 \sin^2 x)) = \frac12 (2 \sin^2 x) = \sin^2 x.$$

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    $\begingroup$ +1, though I would have put $\sin^2(x)=1-\cos^2(x)$ before $\frac12 (1 - \cos (2 x))$ $\endgroup$
    – Henry
    Commented Mar 23, 2012 at 0:46
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The usual convention is that $\sin^2(X)=(\sin(X))^2$. So for your example $1/4$ is correct

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$\sin^2(30)=(\sin(30))^2$ so it is equal $(1/2)^2=1/4$

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$ \sin(30^{\circ}) = 1/2 $, thus $ \sin^{2}(30^{\circ}) = (\sin(30^{\circ}))^{2} = (1/2)^{2} = 1/4 $.

However, $$ \sin(900^{\circ}) = \sin(180^{\circ} \cdot 5) = \sin(180^{\circ}) = 0$$ because $ \sin(180^{\circ} \cdot k) = 0 $ for any integer $ k $.

So $$ \sin^{2}(30^{\circ}) \neq \sin(900^{\circ}) $$

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