# How do you evaluate sin^-1(1/sqrt2) without a calculator?

Mar 7, 2018

${\sin}^{-} 1 \left(\frac{1}{\sqrt{2}}\right) = {45}^{0}$

#### Explanation:

Let ${\sin}^{-} 1 \left(\frac{1}{\sqrt{2}}\right) = \theta \therefore \sin \theta = \frac{1}{\sqrt{2}}$

In the right triangle sin theta = P/H :. P=1; H= sqrt2; P is

perpendiculur and $H$ is hypotenuse. In right triangle

${H}^{2} = {P}^{2} + {B}^{2} \therefore {B}^{2} = {H}^{2} - {P}^{2} = 2 - 1 \therefore B = 1$ . Since

$P = 1 \mathmr{and} B = 1$, it is isosceles triangle $\therefore \theta = {45}^{0}$

$\therefore {\sin}^{-} 1 \left(\frac{1}{\sqrt{2}}\right) = {45}^{0}$ [Ans]

Mar 7, 2018

$\frac{\pi}{4} + 2 k \pi$
$\frac{5 \pi}{4} + 2 k \pi$

#### Explanation:

$\sin x = - \frac{1}{\sqrt{2}} = - \frac{\sqrt{2}}{2}$. Find arcsin x
Trig Table and unit circle give 2 solutions -->
arc $x = - \frac{\pi}{4}$ , and
$x = \pi - \left(- \frac{\pi}{4}\right) = \pi + \frac{\pi}{4} = \frac{5 \pi}{4}$
Note. $x = \frac{7 \pi}{4}$ is co-terminal to $x = \left(- \frac{\pi}{4}\right)$
$x = \frac{\pi}{4} + 2 k \pi$
$x = \frac{5 \pi}{4} + 2 k \pi$