# How did they get this result?

1) $-\left(\frac{1}{2}\right)^2 +1 = \cos^2x$

$\frac{\sqrt{3}}{2} = \cos x$

How did we get $\frac{\sqrt{3}}{2}$ from $-\left(\frac{1}{2}\right)^2 +1$?

2) $-\left(\frac{\sqrt{2}}{2}\right)^2 +1 = \cos^2x$

$\frac{\sqrt{2}}{2} = \cos x$

How did we get $\frac{\sqrt{2}}{2}$ from $-\left(\frac{\sqrt{2}}{2}\right)^2 +1$?

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For 1, note that $-(1/2)^2+1=3/4$ and so the square root of it is $\sqrt{3}/2$. – Alex Becker May 18 '12 at 18:44
@DrStrangeLove I fixed the formatting using \LaTeX; is everything how you originally intended it to look? – chris May 18 '12 at 18:50
@chris Thanks! It's OK. – DrStrangeLove May 18 '12 at 18:54

You have:

1. $$\cos^2 x = -(\frac{1}{2})^2 + 1$$

$$\cos^2 x = -\frac{1}{4} + 1=\frac{3}{4}$$

Which is:

$$\sqrt \cos^2 x = \sqrt \frac{3}{4}$$

$$\cos x = \pm \frac{\sqrt 3}{2}$$

Same works for the second one.

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Here's another way:

$$\cos^2(x) = 1 - (\frac{1}{2})^2$$ and

$$\cos^2(x) = 1 - \sin^2(x)$$ so immediately we have

$$\sin(x) = \pm\frac{1}{2}$$ then, since $\sin(x) = \frac{opp}{hyp}$ we have from the reference triangle, $$\cos(x) = \frac{adj}{hyp} = \pm\frac{\sqrt{3}}{2}$$ and also $$\tan(x) = \frac{opp}{adj} =\pm \frac{1}{\sqrt{3}}$$

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$$\sqrt{-\left(\frac12\right)^2 +1} = \sqrt{-\frac14 +1} = \sqrt{\frac34} = \frac{\sqrt{3}}{2}$$

$$\sqrt{-\left(\frac{\sqrt{2}}{2}\right)^2 +1} = \sqrt{-\frac24 +1} = \sqrt{\frac24} = \frac{\sqrt{2}}{2}$$

though you should also consider the negative square roots.

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