Given sin(u)=3/5, sin(v)=4/5, u and v between 0 and pi/2,

[goldenfuture5]

Given sin(u)=3/5, sin(v)=4/5, u and v lie between 0 and pi/2... Find cos(u-v)

Confused, need answer and some work. THanks!

 

[soroban]

Hello, goldenfuture5!

$\displaystyle \text{Given: }\:\sin(u)=\tfrac{3}{5},\;\sin(v)=\tfrac{4}{5},\:u\text{ and }v\text{ lie between 0 and }\tfrac{\pi}{2}$

. . $\displaystyle \text{Find }\:\cos(u-v)$

$\displaystyle \text{We're expected to know that: }\;\cos(u-v) \;=\;\cos(u)\cos(v) + \sin(u)\sin(v)\;\;[1]$

$\displaystyle \text{We know the values of }\sin(u)\text{ and }\sin(v).$
. . $\displaystyle \text{We must find the values of }\cos(u)\text{ and }\cos(v).$


$\displaystyle \text{Since }\,\sin(u) \:=\:\tfrac{3}{5} \:=\:\tfrac{opp}{hyp}$
. . $\displaystyle u\text{ is an angle in Quadrant 1 with: }opp = 3,\:hyp = 5.$

$\displaystyle \text{Using Pythagorus, we find that: }\:adj = 4$
. . $\displaystyle \text{Hence: }\;\cos(u) \:=\:\tfrac{adj}{hyp} \quad\Rightarrow\quad \cos(u) \:=\:\tfrac{4}{5}$

$\displaystyle \text{Similarly: }\:\sin(v) \:=\:\tfrac{4}{5} \quad\Rightarrow\quad \cos(v) \:=\:\tfrac{3}{5}$


$\displaystyle \text{Substitute into [1]:}\;\cos(u-v) \;=\;\left(\tfrac{4}{5}\right)\left(\tfrac{3}{5}\right) + \left(\tfrac{3}{5}\right)\left(\tfrac{4}{5}\right) \;=\;\tfrac{12}{25} + \tfrac{12}{25} \;=\;\frac{24}{25}$