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I had learnt that:

$\sin \theta = \frac {\text{perpendicular}} {\text{hypotenuse}}$

$\cos \theta = \frac {\text{base}} {\text{hypotenuse}}$

But in unit circles, we find the trigonometric values of obtuse angles. How is that even possible when there's no right angled triangle present?

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    $\begingroup$ You just define the sine of obtuse angles that way. There is no laws in mathematics. $\endgroup$ – Kenny Lau Jun 16 '17 at 4:13
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A lot of the time, mathematicians will devise something that works for a particular case, and then see if they can extend or generalise it to a bigger area. For example, you can use fairly basic proofs to show that if $n$ and $m$ are integers, then $a^n \times a^m = a^{n + m}$. So mathematicians ask - do they have to be integers? Can we extend the exponentiation rule so that it works for any $n$ and $m$? And if you just take it as given, then it works fine, and then when you devise some alternative ways of expressing exponentiation you can prove it explicitly and it doesn't really break anything.

So, similarly, you can start with $\sin \theta = \frac{\mbox{opposite}}{\mbox{hypotenuse}}$ for $0 \leq \theta \leq \pi$, then you derive a few neat properties of the function, then you draw a circle around your triangle and scale it down so the radius is equal to 1, and then you ask "but what if I used a different angle?" and you can see that you can assign values to the trigonometric functions based on the point on the circle, and then you just confirm that your nice properties still hold - things like $\sin^2 \theta + \cos^2 \theta = 1$ and so forth - and you just say "we will define the sine function to be like this" and it all works nicely.

Which is not to say that it always works nicely - after you make such an extension you still have to prove that it makes sense and doesn't break anything else, otherwise you wind up with things like $1 - 1 + 1 - 1 + \ldots = -\frac{1}{2}$.

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enter image description here

So you have the angle, then you just draw your triangle like that.

Then, you have $\displaystyle \sin(135º)=\frac{\sqrt{2}}{2}$, and $\displaystyle \cos(135º)=-\frac{\sqrt{2}}{2}$.

Hope I helped!

enter image description here

Another picture, for $225º$.

Here, $\displaystyle \cos(225º)=\sin(225º)=-\frac{\sqrt{2}}{2}$.

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I will 'define' the trigonometric ratios in this way.

Suppose that I am standing and holding a rod of length $1$ unit in front of me, with one end of the rod next to my foot on the ground. The rod has an inclination of $\theta$ to the ground. $\sin\theta$ is the distance between the other end of the rod and the ground. The sun is directly above me and casting a shadow of the rod on the ground. $\cos\theta$ is the length of the shadow. $\tan\theta$ is the quotient of the two lengths.

If $\theta$ is obtuse, the rod is behind me, with one end still next to my foot. The distance between the other end of the rod and the ground is still $\sin\theta$, and it is positive. The shadow of the rod is now behind me, and I take its length as negative. This is $\cos\theta$ and we have $\cos\theta<0$. $\tan\theta$ is the quotient of this two lengths, which is also negative.

I can extend this 'definition' to a general $\theta$. If I hold the rod vertically, $\theta=90^\circ$. $\sin90^\circ$ is the distance between the other end of the rod and the ground, which is now equal to the length of the rod. So $\sin90^\circ =1$. The length of the shadow is $0$ and so $\cos90^\circ =0$.

If we take $\theta=200^\circ$, the rod is below the ground (we have to imagine that). So $\sin200^\circ$ is negative....

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  • $\begingroup$ Can you add a diagram to make things clearer? $\endgroup$ – user342531 Jun 16 '17 at 6:27
  • $\begingroup$ @Abcd My 'definition' is essentially the unit circle definition of trigonometric ratio, but I avoid to use the abstract setting of the coordinate plane. I use this method to teach my students and they almost at once can tell why $\sin 150^\circ=\sin30^\circ$ or $\cos 310$ is positive or $\tan 270^\circ$ is undefined. I will take a long stick to my class to teach them. But I'm not good at drawing. It will be great if someone can draw a picture. But I'm not able to do that. Sorry for that. Besides, I am not native in English. So maybe I cannot express my idea very well. $\endgroup$ – CY Aries Jun 16 '17 at 6:36

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