If sin x = 2/3 and cos y = -2/7, find the possible values of cos(x+y). Anyone please?
I tried cos(x+y) = cos x cos y - sin x sin y
= -2/7cos x - 2/3 sin y
now i'm stuck
Since you know that sin(x) is positive, that means x could be in the first and second quadrants, which means cos(x) could be positive or negative.
Since you know that cos(y) is negative, that means y could be in the second or third quadrants, which means sin(x) could be positive or negative.
So yes, since there are four possibilities, you will need to substitute all four values to get four possible results
Hello, jessm001!
You were on your way.
We are expected to find those missing values.
$\displaystyle \text{If }\sin x = \tfrac{2}{3}\text{ and }\cos y = \text{-}\tfrac{2}{7}\text{, find the possible values of }\cos(x+y). $
$\displaystyle \sin x \:=\:\frac{2}{3} \:=\:\frac{opp}{hyp} \qquad x\text{ is in Quadrant 1 or 2.}$
. . $\displaystyle adj \:=\:\pm\sqrt{5} \quad\Rightarrow\quad \cos x \:=\:\pm\frac{\sqrt{5}}{3}$
$\displaystyle \cos y \:=\:\frac{\text{-}2}{7} \:=\:\frac{adj}{hyp}\qquad y\text{ is in Quadrant 2 or 3.}$
. . $\displaystyle opp \:=\:\pm\sqrt{45} \:=\:\pm3\sqrt{5} \quad\Rightarrow\quad \sin y \:=\:\pm\frac{3\sqrt{5}}{7}$
$\displaystyle \cos(x + y) \;=\;\cos x\cos y - \sin x\sin y$
. . . . . . . . $\displaystyle =\;\left(\pm\frac{\sqrt{5}}{3}\right)\left(-\frac{2}{7}\right) - \left(\frac{2}{3}\right)\left(\pm\frac{3\sqrt{5}}{ 7}\right)$
There are four possible values: . $\displaystyle \frac{8\sqrt{5}}{21},\;-\frac{8\sqrt{5}}{21},\;\frac{4\sqrt{5}}{21},\;-\frac{4\sqrt{5}}{21}$