Thread: Trig problem help

1. Trig problem help

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.

2. Hello,
Originally Posted by john11235
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.
First of all, translate the information you're given :

$\tan(a)=-\frac 32=\frac{\sin(a)}{\cos(a)}$

Since -3/2 < 0 and sin(a) > 0, we have that cos(a) < 0.

Next step, use the most common formula : $\sin^2(x)+\cos^2(x)=1$

$\implies \cos(a)=-\sqrt{1-\sin^2(a)}$ (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..

3. Originally Posted by Moo
Hello,

First of all, translate the information you're given :

$\tan(a)=-\frac 32=\frac{\sin(a)}{\cos(a)}$

Since -3/2 < 0 and sin(a) > 0, we have that cos(a) < 0.

Next step, use the most common formula : $\sin^2(x)+\cos^2(x)=1$

$\implies \cos(a)=-\sqrt{1-\sin^2(a)}$ (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..
So put the $-sqrt(1-((sin^2)(a))$ into the $sin(a)/cos(a)$ equation?

4. Originally Posted by john11235
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.
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 $r = \sqrt{13}$

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

5. $\tan A$ is negative and $\sin A$ is positive,therefore $A$ is in 2nd quadrant
$
\sin A=\frac{\tan A}{\sqrt{1+\tan^2 A}}
$
; $
\cos A=-\frac{1}{\sqrt{1+\tan^2 A}}
$

Consequently,you can find $\sec A$ and $\csc A$ by taking reciprocals