Mi question is: How to prove $$\sin(180^\circ-\theta)=\sin\theta$$ ?
Here, sine is defined for any angle such as 'alpha'
This is the question mi college teacher asked me to derive it but i could not. plz help me. Iam 12 grade. thank you....
$\endgroup$ 54 Answers
$\begingroup$By symmetry with respect to the vertical axis: reflect the line $OA$ forming the $\theta$ angle. The triangle $OAA'$ is isosceles, hence the $y$ axis is orthogonal to the base $AA'$ and the altitude is the sine. The reflected angle is $\pi-\theta$.
By a similar argument, $\cos(\theta)=\cos(-\theta)$.
$\endgroup$ $\begingroup$Use:
$$ \sin(x-y) = \sin x\cos y-\cos x \sin y $$
and
$\cos 180^\circ =-1$
$\endgroup$ $\begingroup$Note that
$$\sin(A - B) = \sin(A)\cos(B) - \cos(A)\sin(B).$$
Then
$$\sin(180° - \theta) = \sin(180°)\cos(\theta) - \cos(180°)\sin(\theta).$$
Since $$\sin(180°) = 0\ \ \textrm{ and }\ \cos(180°) = -1,$$
we get
$$\sin(180° - \theta) = 0 - (-1)\sin(\theta) = \sin(\theta).$$
We are going to need these two identities
$\sin\Big( \frac{\pi}{2} - x \Big) = \cos(x)$
$\cos\Big( \frac{\pi}{2} - x \Big) = \sin(x)$
Now here is the setup:
$\sin(\pi - \theta) = \sin\Big( \frac{\pi}{2} - \Big[ \theta - \frac{\pi}{2} \Big] \Big) = \cos\Big( \theta - \frac{\pi}{2} \Big)$
From here we have to use the fact that cosine is an even function, $\cos(x) = \cos(-x)$:
$\cos\Big( \theta - \frac{\pi}{2} \Big) = \cos\Big( \frac{\pi}{2} - \theta \Big) = \sin(\theta)$
$\endgroup$
