# How do you evaluate the expression cos(u-v) given sinu=3/5 with pi/2<u<p and cosv=-5/6 with pi<v<(3pi)/2?

Sep 27, 2017

see below

#### Explanation:

Use the composite argument formula $\cos \left(u - v\right) = \cos u \cos v - \sin u \sin v$.

But first we need to find $\cos u \mathmr{and} \sin v$. Since $u$ is in Quadrant II and sin u= 3/5=(opposite)/(hypote n use we can find the adjacent side by using pythagorean theorem.

That is,
${a}^{2} = {c}^{2} - {b}^{2}$

$a = \sqrt{{5}^{2} - {3}^{2}} = \sqrt{25 - 9} = \sqrt{16} = 4$ but since $u$ is in quadrant two then $a = - 4$. Hence, $\cos u = - \frac{4}{5}$.

Likewise, we need to find $\sin v$ and $v$ is in Quadrant III. So, to find the missing side from $\cos v = - \frac{5}{6} = \frac{a \mathrm{dj} a c e n t}{h y p o t e n u s e}$
we use pythagorean theorem to find the length of the opposite side

$o = \sqrt{{6}^{2} - {\left(- 5\right)}^{2}} = \sqrt{36 - 25} = \sqrt{11.}$ Since, $v$ is in Quadrant III then $o = - \sqrt{11}$ and $\sin v = - \frac{\sqrt{11}}{6}$

Therefore,

$\cos \left(u - v\right) = \cos u \cos v - \sin u \sin v$

$= \left(- \frac{4}{5}\right) \left(- \frac{5}{6}\right) - \left(\frac{3}{5}\right) \left(- \frac{\sqrt{11}}{6}\right)$

=((-2cancel4) /cancel5)(-cancel5/(3cancel6)-(cancel3/5)(-sqrt11/(2cancel6))

$= \frac{2}{3} + \frac{\sqrt{11}}{10}$

$\therefore = \frac{20 + 3 \sqrt{11}}{30}$