What's the intuition behind the sin of an obtuse angle?
Mia Lopez 
The definition of sin of an angle is opposite / hypotenuse. Obviously, this definition is easy to use for acute angles within a right triangle, but it's hard to see how it carries over for obtuse angles.
I know there are trig identities and unit circle concepts that allow one to get to the sin of an obtuse angle, but I'm looking for a more intuitive explanation for what exactly the sin of an obtuse angle means.
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$\begingroup$"The definition of sin of and angle is opposite/hypotenuse" - well that only works for a right-angled triangle, and is the beginning of the definition of the sine function.
In order to extend the definition draw a unit circle with centre at the origin. Measure the angle counterclockwise from the positive $x$-axis and take a point $(x,y)$ on the circle. The sine of the angle (equivalent to opposite/hypotenuse in the first quadrant, with the hypotenuse made equal to $1$) is $y$ and the cosine of the angle is $x$. You can go round the circle more than once, so you can see that the functions are periodic.
This is why the functions are sometimes known as circular functions and underlies why they come in so surprisingly useful.
In case you are interested ...
In more advanced work the sine function is sometimes defined very differently, with the angle measured in radians ($2\pi$ radians $=360^{\circ}$). Then $$\sin x = \sum_{r=1}^\infty (-1)^{r-1}\frac{x^{2r-1}}{(2r-1)!}=\frac {e^{ix}-e^{-ix}}{2i}$$This can be applied in more general circumstances still, and the series is convenient because it converges rapidly and enables the sine function to be computed accurately for practical use. The first few terms are $$x-\frac {x^3}6+\frac {x^5}{120}-\frac {x^7}{5040}+\dots$$
$\endgroup$ $\begingroup$In a triangle $ABC$, we take the angle $A$ as example.
When $A$ is acute, the definition of $\sin(A)$ is the ratio of the hight $BH/AB$ where $BH$ is the height bewteen $B$ and the side $AC$ (we can also do with $CH$ which is the height between $C$ and side $AB$).
in this case, $\sin(A)=BH/AB$
when $A=\pi/2$, in the triangle $AB'C'$, we imagine the same way, $\sin(A)$ is the height from $B'$ to the side $AC'$, but in fact $AC'=0$ and $AB'=B'H'$, so $\sin(A)=AB'/B'H'=1$
when $A$ is opposite, we do the same thing in triangle $AB''C''$: $\sin(A)$ is the height from $B''$ to the side $AC''$ which is $BH''/AB''$. It is also the $\sin$ of angle $B''AH''$. Then $\sin(A)=\sin(180°-A)$.
$\endgroup$ $\begingroup$One way to think about it is to take the sine of the supplementary angle.
In this image, we can say that $\sin(\beta) = \sin(\alpha)$. This is because $\sin(x)$ satisfies certain properties so that $\sin(180^\circ - x) = \sin(x)$.
I believe this is a good geometric interpretation but from a pure "intuitive" standpoint you must think of $\sin(x)$ in terms of the unit circle.
$\endgroup$ 0 $\begingroup$Let's use the definition of the slope or tangent of the terminal line for each angle .... For the terminal line of an acute angle the slope is y/x.... For the terminal line of an obtuse angle the slope is also y/x.....(though negative in sign).... Now we multiply each slope by x to get us the sin values for the two angles (in the unit circle ) as y and -y which are equal in absolute values.....
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