# can't seem to understand $\sin{\theta} = y$ on a unit circle

So I've been working very hard on my trigonometry on khan academy. However I'm constantly getting stumped by one type of question in particular.

There is some fundamental flaw in my understanding.

I know what the unit circle is completely and why $\sin{\theta} = y$, however, in the following question I completely DO NOT understand why the answer wasn't B and G. The answer apparently was only G!

Why Isn't the y co-ordinate of $B = \sin{\theta}$ ?

• Sorry for the weird question but "Gideon" sounds like an Israeli name yet you're Indian. Is it also an Indian name? – Shahar Aug 27 '14 at 16:04
• @Shahar Yes. Very much an Israeli name, my middle name is also Israel by the way. So both my first and my middle name have never gone down well in india. – gideon Aug 27 '14 at 16:05
• Oh yeah I just noticed that when I clicked on your profile. That's odd, lol. – Shahar Aug 27 '14 at 16:26
• @Shahar I'm christian but my father in his mad Biblical and Jewish fest named me so, along with taking me to the synagogue nearby every now and then. – gideon Aug 27 '14 at 17:04
• Oh, I see. Wow, that's really interesting. I have never heard of something like that ^_^. – Shahar Aug 27 '14 at 17:06

The line segment $OG$ is at an angle of $\theta$ from the positive $x$-axis, whilst the line segment $OB$ is at an angle of $\theta$ from the negative $x$-axis and hence an angle of $\pi + \theta$ from the positive $x$-axis, so that the $y$-coordinate of $B$ is $\sin(\pi + \theta) = -\sin(\theta)$.

• Ok, starting to make some sense. So you're saying the measure at the angle along OB is actually not $\theta$ but $\pi + \theta$ ? But how do I really know when this is the case? – gideon Aug 27 '14 at 16:10
• oh oh! Is it because the y co-ordinate is really negative not positive. I am however not seeing how this magically makes theta become theta+pi :/ – gideon Aug 27 '14 at 16:12
• For simplicity, let's use the convention that angles are always measured counterclockwise; let $P$ denote the point $(1,0)$, and let $Q$ denote the point $(-1,0)$, and recall that $P$ is the point on the unit circle such that for any point $R$ on the unit circle, $R = (\cos(\angle POR),\sin(\angle POR))$. Now, your drawing shows you that $\angle POG = \theta$ and that $\angle QOB = \theta$. In particular, the drawing does not show you that $\angle POB = \theta$; instead, because $\angle QOB = \theta$ and $\angle POQ = \pi$, $\angle POB = \angle POQ + \angle QOB = \pi + \theta$. – Branimir Ćaćić Aug 27 '14 at 16:27
• ok! I get it!!! I get it! So when we're talking from the positive axis I always have to count from there. (Or IOW the point P that you mentioned) – gideon Aug 28 '14 at 3:34

The value at $B$ is $-\sin(\theta),$ not $+\sin(\theta)$.

The value of the $y$ component on the unit circle centered at the origin, with the angle measured counterclockwise from the $+x$ axis, is $\sin(\theta)$. Since you're below the $x$ axis at this point, the value is negative.

But you did see part of the connection. Since the angle with respect to the negative $x$ axis is $\theta$, the magnitude of the $y$ component is indeed $\sin(\theta)$.

• Yes, but what suddenly makes it $-\sin{\theta}$ now? The diagram has the same theta and the double lines marked on both, leading me to believe that the measure of the angles are the same! – gideon Aug 27 '14 at 16:14
• The two $\theta$'s are there to make the congruency suggested by the diagram explicit: $\angle GOC$ is congruent to the angle made by the ray $\vec{OD}$ and the negative $x$ axis. – John Aug 27 '14 at 16:31