# Thread: Feel pretty dumb, basic trig question

1. ## Feel pretty dumb, basic trig question

If $\displaystyle \sin\theta = \frac{1}{3}$ and $\displaystyle \tan\theta < 0$ then what is $\displaystyle \cos\theta$

Like I said I feel completely dumb here. I know this is so basic I'm just having a mental block with it.

a) $\displaystyle \frac{1}{3}$

b) $\displaystyle \frac{-3}{\sqrt{10}}$

c) $\displaystyle \frac{-1}{2\sqrt{2}}$

d) $\displaystyle \frac{-2\sqrt{2}}{3}$

e) $\displaystyle \frac{1}{\sqrt{10}}$

Because of the nature of the answers I know I'd have to use the unit triangle somewhere. Because tan is negative and sin is positive the answer is in quadrent 2. In other words sin1/3 = 180 - cos.

What I don't know though is how we've got the unit triangle to give sin 1/3, my stupid question is how do I get sin 1/3 on the unit triangle?

2. ## Re :

Originally Posted by Peleus
If $\displaystyle \sin\theta = \frac{1}{3}$ and $\displaystyle \tan\theta < 0$ then what is $\displaystyle \cos\theta$

Like I said I feel completely dumb here. I know this is so basic I'm just having a mental block with it.

a) $\displaystyle \frac{1}{3}$

b) $\displaystyle \frac{-3}{\sqrt{10}}$

c) $\displaystyle \frac{-1}{2\sqrt{2}}$

d) $\displaystyle \frac{-2\sqrt{2}}{3}$

e) $\displaystyle \frac{1}{\sqrt{10}}$

Because of the nature of the answers I know I'd have to use the unit triangle somewhere. Because tan is negative and sin is positive the answer is in quadrent 2. In other words sin1/3 = 180 - cos.

What I don't know though is how we've got the unit triangle to give sin 1/3, my stupid question is how do I get sin 1/3 on the unit triangle?

From the information given , we know that $\displaystyle \theta$ is in the second quadrant . Draw a triangle and find $\displaystyle cos\theta$

3. Hello, Peleus!

If $\displaystyle \sin\theta = \tfrac{1}{3}$ and $\displaystyle \tan\theta < 0$, find $\displaystyle \cos\theta.$

. . $\displaystyle (a)\;\frac{1}{3}\qquad (b)\;-\frac{3}{\sqrt{10}} \qquad (c)\;-\frac{1}{2\sqrt{2}} \qquad (d)\;-\frac{2\sqrt{2}}{3} \qquad (e)\;\frac{1}{\sqrt{10}}$

Because $\displaystyle \tan\theta < 0\text{ and }\sin\theta > 0$, the answer is in quadrant 2. . Good!
I don't use the unit circle . . .

We have: .$\displaystyle \sin\theta \:=\:\frac{1}{3} \:=\:\frac{opp}{hyp}$

So $\displaystyle \theta$ is in a right triangle with: .$\displaystyle opp = 1,\;hyp = 3$

Pythagorus says: .$\displaystyle opp^2 + adj^2 \:=\:hyp^2$

So we have: .$\displaystyle 1^2 + adj^2 \:=\:3^2 \quad\Rightarrow\quad adj^2 \:=\:8 \quad\Rightarrow\quad adj \:=\:\pm2\sqrt{2}$

Then: .$\displaystyle \cos\theta \:=\:\frac{adj}{hyp} \:=\:\pm\frac{2\sqrt{2}}{3}$

In quadrant 2, cosine is negative: .$\displaystyle {\color{blue}\cos\theta\:=\:-\frac{2\sqrt{2}}{3}}\quad\hdots$ answer (d)

4. It's times like this that the answer smacks you in the face and tells you how silly you are.

Thank you very much and quite obvious now. I guess I just have to not get focused on looking at one method to solve it (unit circle) and look at the big picture.

Thanks again.