3
$\begingroup$

Given two supplementary angles (for instance, 30 degrees and 150 degrees), why is $\sin(30^\circ) = \sin(150^\circ)$?

Where can I find a proof for this? Or the derivative of such proofs?

$\endgroup$
1
  • 1
    $\begingroup$ Draw a circle, supplementary angles, and indicate the sine values at those angles. You should immediately see why this is true. $\endgroup$
    – davidlowryduda
    Commented Dec 11, 2014 at 20:52

4 Answers 4

9
$\begingroup$

UNIT CIRCLE

I like looking at unit circles and seeing how the y value of the angle is the same no matter which side of the y axis the angle goes to (80 and 100, 45 and 135, pairs have the same y value), enter image description here

but...

PROOF $$\sin(\pi-\theta)=\sin\pi \cos\theta - \cos\pi \sin\theta$$ $$\sin(\pi-\theta)=0\times \cos\theta - (-1) \sin\theta$$ $$\sin(\pi-\theta)=\sin\theta$$ GRAPH

Finally, you can think of the sine function graph and start from $\theta=0$ and $\theta=180^\circ$ and move towards each other. You can see how the values mirror each other along the way. enter image description here

$\endgroup$
2
$\begingroup$

It depends a lot on how you define the sine. Let's stick to right triangles, which allow to define the sine for acute angles: if $\alpha=\widehat{BAC}$ is the angle of a right triangle $ABC$ (the angle in $B$ is right), then $$ \sin\alpha=\frac{BC}{AB}. $$ Now consider this figure:

enter image description here

The triangle $BHO$ is right at $H$ and the angle at $O$ is $\alpha$ by an important theorem in elementary geometry. If $r$ is the radius of the circle and $a$ is the length of $BC$, then $$ a=2r\sin\alpha. $$ This is valid for an acute triangle, because the incenter will be inside the triangle. Consider now a point $A'$ on the arc $BC$ not containing $A$. The angle $\widehat{BA'C}$ will be supplementary to $\alpha$, but the triangle $BCO$ plays exactly the same role for $A'BC$ as for $ABC$. If we want to give a meaning to $\sin(\pi-\alpha)$ ($\pi$ denotes the measure of the straight angle, $180°$, if you prefer), you are forced to define $$ \sin(\pi-\alpha)=\sin\alpha $$ so that the same statement as before holds: if $\widehat{BA'C}=\alpha'$, that is, $\alpha'=\pi-\alpha$, then $$ a=2r\sin\alpha'. $$

If you define the sine by means of the unit circle, then this image should explain the fact:

enter image description here

The rays corresponding to supplementary angles intersect the unit circles in points having the same $y$-coordinate, so the two angles have the same sine (and opposite cosines).

$\endgroup$
1
$\begingroup$

From $0$ to $180$, sine goes up and down symmetrically around 90. (Note: this is not rigorous.)

$\endgroup$
1
$\begingroup$

You can use the difference formula:

$$\sin 150^{\circ} = \sin(180^{\circ} - 30^{\circ}) = \sin 180^{\circ} \cos 130^{\circ} - \cos 180^{\circ} \sin 30^{\circ} \\ = 0 \cdot \cos 130^{\circ} - (-1) \cdot \sin 30^{\circ} = \sin 30^{\circ}.$$

Or, in general:

$$\sin x = \sin(\pi - (\pi - x)) = \sin \pi \cos (\pi - x) - \cos \pi \sin (\pi - x) = \\ 0 \cdot \cos (\pi - x) - (-1) \cdot \sin (\pi - x) = \sin(\pi - x).$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .