# How do you evaluate sin (arccos.5 + arcsin .6)?

Aug 15, 2015

Evaluate sin [arccos 0.5 + arcsin 0.6]

Ans: 0,99

#### Explanation:

Using calculator:
cos x = 0.5 --> arc x = 60 deg
sin x = 0.6 --> arc x = 36.87 deg
sin (60 + 36.87) = sin 96.87 = 0.99

Aug 15, 2015

$\sin \left(\arccos 0.5 + \arcsin 0.6\right) = 0.3 + 0.8 \sqrt{0.75}$

#### Explanation:

If we want to do this without a calculator or trig tables, use the following:

$\arccos 0.5$ is some $\alpha$ in $\left[0 , \pi\right]$ with $\cos \alpha = 0.5$.
We will need $\sin \alpha$ so we note that with $\alpha$ in $\left[0 , \pi\right]$, we have $\sin \alpha$ is positive.

Therefore $\sin \alpha = \sqrt{1 - {\cos}^{2} \alpha} = \sqrt{0.75}$

By similar reasoning, $\arcsin 0.6$ is some $\beta \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$ with $\sin \beta = 0.6$ and $\cos \beta = \sqrt{1 - 0.36} = \sqrt{0.64} = 0.8$

We have been asked to find $\sin \left(\alpha + \beta\right)$.

Use

$\sin \left(\alpha + \beta\right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$ and the values above to get:

$\sin \left(\alpha + \beta\right) = \left(\sqrt{0.75}\right) \left(0.8\right) + \left(0.5\right) \left(0.6\right)$

$= 0.3 + 0.8 \sqrt{0.75}$