# How do you simplify cos^-1(sqrt2/2)?

Aug 3, 2017

${\cos}^{-} 1 \left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}$

#### Explanation:

This value is actually a standard value in disguise. Surds are often removed from the denominator of a fraction by multiplying top and bottom by the surd in question. We can do exactly the same thing to put the surd back in the denominator:

$\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{2}{2 \sqrt{2}} = \frac{1}{\sqrt{2}}$

Maybe you recognise the value now.

What value of theta gives this value? It is one that should be memorised if you have exams on trigonometry.

${\cos}^{-} 1 \left(\frac{\sqrt{2}}{2}\right) = {\cos}^{-} 1 \left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$

Writing this another way, we have:

$\cos \left(\theta\right) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$

$\theta = \frac{\pi}{4}$

Aug 3, 2017

pi/4; (7pi)/4

#### Explanation:

Trig table and unit circle give -->
$\cos x = \frac{\sqrt{2}}{2}$ --> arc $x = \pm \frac{\pi}{4}$,
or, using co-terminal arcs:
$x = \frac{\pi}{4}$ and $x = \frac{7 \pi}{4}$