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How would you find $\cos^2(x)$ in terms of $\sin(x)$? Please explain clearly, I am a high school student.

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Research a little more, please!. –  Pedro Tamaroff Mar 24 '12 at 2:51

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The key here is the Pythagorean identity, which states that $\sin^2x+\cos^2x=1$. You may have seen the Pythagorean identity in the form $a^2+b^2=c^2$ where $a,b$ are the lengths of the legs of a right triangle and $c$ is the length of the hypotenuse. The other form follows from this, as $$\sin^2x+\cos^2x=\frac{a^2}{c^2}+\frac{b^2}{c^2}=\frac{a^2+b^2}{c^2}=\frac{c^2}{c^2}=1.$$ From this, it is easy to see that $\cos^2x=1-\sin^2x$.

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Are you familiar with what is perhaps the most basic relationship between sine and cosine? For any $x$, $$\cos^2(x) + \sin^2(x) = 1.$$

How you establish this depends on how you define sine and cosine. One way is to use a right triangle whose hypothenuse has length $1$, and has an angle of size $x$. Then the opposite side has length $\sin(x)$, because $\sin(x)$ is equal to the length of the opposite side divided by the length of the hypothenuse, $$\sin(x) = \frac{\text{opposite}}{\text{hypothenuse}} = \frac{\text{opposite}}{1} = \text{opposite};$$ and in the same manner, the adjacent side has length $\cos(x)$, and so by the Pythagorean Theorem, we have that $$\begin{align*} (\text{opposite})^2 + (\text{adjacent})^2 &= (\text{hypothenuse})^2\\ \sin^2x + \cos^2x &= 1. \end{align*}$$

Once you have that $\cos^2(x) + \sin^2(x) = 1$, solving for $\cos^2(x)$ will give you the expression you want.

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