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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$\begingroup$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.
$\endgroup$ 3 $\begingroup$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:
$$\frac12 (1 - \cos (2x)) = \frac12 (1 - (1 - 2 \sin^2 x)) = \frac12 (2 \sin^2 x) = \sin^2 x.$$
$\endgroup$ 1 $\begingroup$The usual convention is that $\sin^2(X)=(\sin(X))^2$. So for your example $1/4$ is correct
$\endgroup$ $\begingroup$$\sin^2(30)=(\sin(30))^2$ so it is equal $(1/2)^2=1/4$
$\endgroup$ $\begingroup$$ \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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