2
$\begingroup$

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 $?

  • 0
    @chris Tha$n$ks! 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.