Find the value of the trigonometric function $\,\sec(v-u)\,$ given that
$\sin u=−\frac{20}{29}\;\;,\;\; \cos v =−\frac{3}{5}$
(Both u and v are in Quadrant III.)
Thanks!
Find the value of the trigonometric function $\,\sec(v-u)\,$ given that
$\sin u=−\frac{20}{29}\;\;,\;\; \cos v =−\frac{3}{5}$
(Both u and v are in Quadrant III.)
Thanks!
$\sec(v-u)=\frac{1}{\cos(v-u)}=\frac{1}{\cos v\cos u+\sin v\sin u}=$
$\frac{1}{-\frac{3}{5}\cos u-\frac{20}{29}\sin v}$
But since $\,u,v\,$ in the third quadrant then
$\cos u=-\sqrt{1-\sin^2u}=-\sqrt{1-\frac{20^2}{29^2}}=-\frac{21}{29}$
$\sin v=-\sqrt{1-\cos^2v} \,...\,etc.$
End the exercise now.
So we want $\dfrac{1}{\cos(v-u)}$. We will be nearly finished once we know $\cos(v-u)$. By a standard formula, $\cos(v-u)=\cos v\cos u+\sin v\sin u.$
To find $\cos u$, note that we are in the third quadrant so $\cos u$ is negative. We have $\cos^2 u+\sin^2 u=1$. So $\cos^2u=1-\frac{400}{841}=\frac{441}{841}$ and therefore $\cos u=-\frac{21}{29}$. Finding $\cos v$ is similar, but the arithmetic is simpler.
Use the ID:
$\cos(v-u)=\cos(v)\cos(v)+\sin(u)\sin(u)$ $\frac{1}{\cos(v-u)}=\frac{1}{\cos(v)\cos(v)+\sin(u)\sin(u)}$ $\sec(v-u)=\frac{1}{\cos(v)\cos(v)+\sin(u)\sin(u)}$
Now, use the Pythagorean thm and quadrant number to find $\cos(u)$ and $\sin(v)$:
$\sin(u) = \frac y r$ $x = -\sqrt{r^2-y^2}$ $\cos(u) = \frac{-\sqrt{r^2-y^2}}{r}$ $\cos(u) = \frac{-\sqrt{29^2-20^2}}{29}$
$\sin(v)$ is similar, and so omitted.
Now plug in to the secant formula.