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Please, explain these computations:

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 $\sqrt{3}/2$.2012-05-18
  • 0
    @DrStrangeLove I fixed the formatting using \LaTeX; is everything how you originally intended it to look?2012-05-18
  • 0
    @chris Thanks! It's OK.2012-05-18

3 Answers 3

1

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

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

though you should also consider the negative square roots.

4

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 $$ \tan(x) = \frac{opp}{adj} =\pm \frac{1}{\sqrt{3}} $$ enter image description here

2

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.