# When finding $\sin\theta$ and $\cos\theta$ given $\tan\theta = \frac35$, shouldn't I get positive and negative answers?

Q: Find the exact value of $\cos\theta$ and $\sin\theta$ if $\tan\theta=\frac{3}{5}$

To solve it I got told to draw the triangle and label the sides and then find cos and sin. The answer from the answer sheet gives both cos and sin as positive solutions. However, I thought that, since tan is positive in both the 1st and 3rd quadrants, then you could draw the triangle in those two quadrants. So,

Why don't you get plus-minus solutions for $\sin\theta$ and $\cos\theta$ ?

• You should. ${}$ – arctic tern Aug 1 '16 at 10:21
• The answer sheet is wrong. (It happens.) – Blue Aug 1 '16 at 10:22
• If $\sin\theta,\cos\theta$ are solutions, so are $-\sin\theta,-\cos\theta=\sin(\theta+\pi),\cos(\theta+\pi)$. – Yves Daoust Aug 1 '16 at 10:50

You are correct in that with only the information given in this problem, you could be in either the first or third quadrant and therefore you have two possibilities:

• $\sin\theta$ and $\cos\theta$ are both positive.
• $\sin\theta$ and $\cos\theta$ are both negative.

There are two explanations for what is happening. Either there is (supposed to be) some understood assumption in class that you guys are only working in the first quadrant for now, or the answer sheet is wrong.

$$\dfrac{\sin\theta}{\cos\theta}=\dfrac35$$

$$\iff\dfrac{\sin\theta}3=\dfrac{\cos\theta}5=\pm\sqrt{\dfrac{\cos^2\theta+\sin^2\theta}{3^2+5^2}}=?$$

We have $\tan\theta=\frac{3}{5}$ so $$\sin { \theta =\frac { 3 }{ \sqrt { 34 } } ,\cos { \theta } =\frac { 5 }{ \sqrt { 34 } } }$$

• But why can't you also use the triangle with sides (-3 * -5* sqrt35)? – kjhg Aug 1 '16 at 10:25