Question

X,y apartin (0;pi/2)
Sin x=3/5
Cos y=5/13
Cos (x-y)=?

Answer 1

$sin^2x+cos^2x=1 \\ cos^2x=1-sin^2x=1- \frac{9}{25}= \frac{16}{25} \\ cosx= \frac{4}{5}$

$sin^2y+cos^2y=1 \\ sin^y=1-cos^2y=1- \frac{25}{169} = \frac{144}{169} \\ siny= \frac{12}{13}$

$cos(x-y)=cosx*cosy+sinx*siny= \frac{4}{5} * \frac{5}{13} + \frac{3}{5}* \frac{12}{13} = \\ = \frac{20}{65}+ \frac{36}{65}= \frac{56}{65}$


Formule folosite:

$sin^2a+cos^2a=1 \\ cos(a-b)=cosa*cosb+sina*sinb$