# What does $\sin(\theta) > 0$ mean here? If $\tan(\theta) = -\frac{8}{15}$, and $\sin(\theta) > 0$, then find $\cos(\theta)$.

The question is:

If $$\tan(\theta) = -\frac{8}{15}$$, and $$\sin(\theta) > 0$$, then find $$\cos(\theta)$$.

What I did was draw a triangle on the unit circle with sides $$8, 15$$ and therefore hypotenuse $$17$$ by Pythagoras theorem.

Then, to find $$\cos(\theta)$$, I did adjacent $$15$$ over hypotenuse $$17$$, which was my final answer: $$\frac{15}{17}$$.

However, the correct answer is $$-\frac{15}{17}$$. Can someone please explain the discrepancy here? I neglected the part were it says "and $$\sin(\theta) > 0$$", because I'm not sure what it means by that. I think that is why I got the signs wrong maybe.

You found the proper value of $\cos \theta$ for the reference angle for $\theta$, not for $\theta$ itself. The reference angle is the angle in the first quadrant that has the same trig values in absolute value. Your final step is to find the proper sign of your trig value, based on the quadrant your $\theta$ is in.

Your question tells you that the tangent is negative, which is true for quadrants II and IV. It also tells you that the sine is positive, which is true for quadrants I and II. The only quadrant that fits both requirements is the second quadrant, so that's where $\theta$ lies. In that quadrant, cosine is negative, so you must change your answer from $\frac{15}{17}$ to $-\frac{15}{17}$.

One way to remember the signs for each quadrant is the mnemonic All Students Take Calculus. This tells you which trig ratios are positive in each quadrant: all, sine, tangent, and cosine for quadrants I, II, III, and IV respectively.

There are other ways, such as trig identities, to solve your problem that give the correct sign automatically, but it is good to have several ways to solve your problem.

• How are you labelling your quadrants? If it's like this: onemathematicalcat.org/algebra_book/online_problems/graphics/… then there's something off with your post. For instance, tangent will be negative in Q II and IV, not Q I and III. – Deepak Aug 5 '14 at 12:28
• @Deepak: Yes, my labelling is the same as the one in your link. My link for "All Students Take Calculus" also explains that convention, though in the text rather than in the graphic. If you don't mind I'll steal your idea and add a link to a similar graphic in my answer. – Rory Daulton Aug 5 '14 at 12:31
• If your labelling is the same, then your post has mistakes (one example of which I pointed out). – Deepak Aug 5 '14 at 12:31
• @Deepak: Yes, you are correct and I was very wrong. How embarrassing! I have corrected my errors: many thanks for pointing them out! – Rory Daulton Aug 5 '14 at 12:36
• No worries, glad to help. – Deepak Aug 5 '14 at 12:40

One way to think about it simply (and that I find helpful) is to consider it in terms of quadrants of the Cartesian plane, which are labelled in a counterclockwise direction starting from the 1st quadrant (positive $x$ and $y$). All trig ratios are non-negative in the 1st quadrant. In the other quadrants, only one of the ratios is non-negative while the other two are negative. In the 2nd quadrant, sine is non-negative. In the third, tangent is non-negative and in the 4th quadrant, cosine is non-negative.

I remember this as "ASTC" (all, sine, tangent, cosine).

The given info allows you to localise the angle to the 2nd quadrant (negative tangent, positive sine). That means the cosine is also negative, so you take the negative value.

Tangent has period $\pi$ but $sin$ and $cos$ have period $2\pi$, so for the range of values of $sin$ and $cos$ there are two angles for which the tangent is $-\frac{8}{15}$. The difference between these two angles is $\pi$ so for one of those angles the sine is positive and for one its negative, your condition constrains the result to one angle, i.e. one value for cosine.

$\sin(\theta)>0$ means that the angle $\theta$ should lie in the first or second quadrant of the goniometric circle (in this case the second quadrant). Don't forget that there are $\textbf{two}$ angles for which $\tan(\theta)=-\frac{8}{15}$.