I'm having a little trouble with this homework problem.
If tan(a) = -3/2 and sin(a) > 0, find the following:
sin(a)
cos(a)
sec(a)
csc(a)
cot(a)
I've already gotten that cot(a) = -2/3 but I have no idea how to do any of the others.
Hello,
First of all, translate the information you're given :
$\displaystyle \tan(a)=-\frac 32=\frac{\sin(a)}{\cos(a)}$
Since -3/2 < 0 and sin(a) > 0, we have that cos(a) < 0.
Next step, use the most common formula : $\displaystyle \sin^2(x)+\cos^2(x)=1$
$\displaystyle \implies \cos(a)=-\sqrt{1-\sin^2(a)}$ (it could have been + sqrt, but since it is negative...)
So substitute this formula of cos(a) into tan(a) and get sin(a) !
Then get cos(a) etc..
tan(a) < 0 and sin(a) > 0 ... "a" is in quad II
sketch a reference triangle in quad II
opposite side, y = 3
adjacent side x = -2
hypotenuse $\displaystyle r = \sqrt{13}$
now use your definitions for the six trig ratios.
sin(a) = y/r
cos(a) = x/r
tan(a) = y/x
cot(a) = x/y
sec(a) = r/x
csc(a) = r/y
$\displaystyle \tan A$ is negative and $\displaystyle \sin A$ is positive,therefore $\displaystyle A$ is in 2nd quadrant
$\displaystyle
\sin A=\frac{\tan A}{\sqrt{1+\tan^2 A}}
$ ;$\displaystyle
\cos A=-\frac{1}{\sqrt{1+\tan^2 A}}
$
Consequently,you can find $\displaystyle \sec A$ and $\displaystyle \csc A$ by taking reciprocals