# Given: cos(x/4)=a and cos(y/3)+b, how do you find cos(x/4+y/3) in terms of a and b?

Jun 7, 2016

$a b - {\left(1 - {a}^{2}\right)}^{\frac{1}{2}} {\left(1 - {b}^{2}\right)}^{\frac{1}{2}}$

#### Explanation:

Use trig identity:
cos (a + b) = cos a.cos b - sin a.sin b.
$\cos \left(\frac{x}{4} + \frac{y}{3}\right) = \cos \left(\frac{x}{4}\right) . \cos \left(\frac{y}{3}\right) - \sin \left(\frac{x}{4}\right) . \sin \left(\frac{x}{3}\right) =$
Since $\cos \left(\frac{x}{4}\right) = a$ and $\cos \left(\frac{y}{3}\right) = b$, then -->
cos $\left(\frac{x}{4} + \frac{y}{3}\right) = a b - \sin \left(\frac{x}{4}\right) . \sin \left(\frac{y}{3}\right)$.
Find $\sin \left(\frac{x}{4}\right)$ and $\sin \left(\frac{y}{3}\right)$
${\sin}^{2} \left(\frac{x}{4}\right) = 1 - {a}^{2} \to \sin \left(\frac{x}{4}\right) = {\left(1 - {a}^{2}\right)}^{\frac{1}{2}}$
${\sin}^{2} \left(\frac{y}{3}\right) = 1 - {b}^{2} - \to \sin \left(\frac{y}{2}\right) = {\left(1 - {b}^{2}\right)}^{\frac{1}{2}}$
Finally,
$\cos \left(\frac{x}{4} + \frac{y}{3}\right) = a b - {\left(1 - {a}^{2}\right)}^{\frac{1}{2}} {\left(1 - {b}^{2}\right)}^{\frac{1}{2}} .$