# How do you evaluate cos (arcsin (3/5) - arccos (1/2)) ?

Jun 8, 2015

$\cos \left[\arcsin \left(\frac{3}{5}\right) - \arccos \left(\frac{1}{2}\right)\right] =$

#### Explanation:

$\sin x = \frac{3}{5} = 0.6$--> x = 36.87 and x = 180 - 36.87 = 143.13 deg

$\cos y = \frac{1}{2}$ --> y = +- 60 deg

a. cos (x - y) = cos (143.13 - 60) = cos 83,13 $= 0.12$

b. cos (36.87 - 60) = cos (-23.23) = 0.92

Jun 8, 2015

An algebraic expression for this can be calculated using the properties of the right angled triangles with sides {3,4,5} and {1,sqrt(3),2} using the formula for $\cos \left(\alpha + \beta\right)$.

#### Explanation:

$\cos \left(\alpha + \beta\right) = \cos \left(\alpha\right) \cos \left(\beta\right) - \sin \left(\alpha\right) \sin \left(\beta\right)$

So

$\cos \left(\alpha - \beta\right) = \cos \left(\alpha\right) \cos \left(- \beta\right) - \sin \left(\alpha\right) \sin \left(- \beta\right)$

$= \cos \left(\alpha\right) \cos \left(\beta\right) + \sin \left(\alpha\right) \sin \left(\beta\right)$

If $\sin \left(\alpha\right) = \frac{3}{5}$ and $0 \le \alpha \le \frac{\pi}{2}$ then

$\cos \left(\alpha\right) = \frac{4}{5}$

If $\cos \left(\beta\right) = \frac{1}{2}$ and $0 \le \beta \le \frac{\pi}{2}$ then

$\sin \left(\beta\right) = \frac{\sqrt{3}}{2}$

Then

$\cos \left(\alpha - \beta\right)$

$= \cos \left(\alpha\right) \cos \left(\beta\right) + \sin \left(\alpha\right) \sin \left(\beta\right)$

$= \frac{4}{5} \cdot \frac{1}{2} + \frac{3}{5} \cdot \frac{\sqrt{3}}{2}$

$= \frac{4}{10} + \frac{3 \sqrt{3}}{10}$

$= \frac{4 + 3 \sqrt{3}}{10}$

$\cong 0.9196$