# How do you evaluate sin(cos^-1(1/2)) without a calculator?

Mar 5, 2018

$\sin \left({\cos}^{- 1} \left(\frac{1}{2}\right)\right) = \frac{\sqrt{3}}{2}$

#### Explanation:

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

$\rightarrow \sin x = \sqrt{1 - {\cos}^{2} x} = \sqrt{1 - {\left(\frac{1}{2}\right)}^{2}} = \frac{\sqrt{3}}{2}$

$\rightarrow x = {\sin}^{- 1} \left(\frac{\sqrt{3}}{2}\right) = {\cos}^{- 1} \left(\frac{1}{2}\right)$

Now, $\sin \left({\cos}^{- 1} \left(\frac{1}{2}\right)\right) = \sin \left({\sin}^{- 1} \left(\frac{\sqrt{3}}{2}\right)\right) = \frac{\sqrt{3}}{2}$

Mar 5, 2018

sin cos ^-1 (1/2)) = sqrt 3 / 2

#### Explanation:

To find value of $\sin \left({\cos}^{-} 1 \left(\frac{1}{2}\right)\right)$

Let theta = cos^-1 (1/2)

$\cos \theta = \left(\frac{1}{2}\right)$ We know, from the above table, $\cos 60 = \frac{1}{2}$

Hence theta = 60^@

Replacing ${\cos}^{-} 1 \left(\frac{1}{2}\right)$ with $\theta = {60}^{\circ}$,

The sum becomes, $\implies \sin \theta = \sin 60 = \frac{\sqrt{3}}{2}$ (As per table above)