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# Prove the identity, $\sin (A + \pi ) = - \sin A$

Last updated date: 23rd Feb 2024
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Hint: Imagine a unit circle and assume an angle, θ. Mark the corresponding coordinates $(\cos \theta ,\sin \theta )$ on the circle. The angle in the question is π more than the unknown angle. In the circle the coordinates corresponding to $(\pi + \theta )$ is $( - \cos \theta , - \sin \theta )$ . Thus, we see the sine function for θ has changed to negative sine function upon addition of π.

Let us visualize a circle with unit radius and center at origin. Then we mark the angle A on the positive $(x,y)$ axis. The corresponding point on the circle “M” is $(\cos A,\sin A)$ . When the angle is changed to $(\pi + A)$ , the point “M’” has the coordinates $( - \cos A, - \sin A)$ as they are located in the III quadrant, i.e., negative x and y axis.
And hence we see, $\sin (A + \pi ) = - \sin A$ .
We know that, $\sin (a + b) = \sin a.\cos b + \sin b.\cos a$
Putting $a = \pi \& b = A$ in the above identity,
$\sin (\pi + A) = \sin \pi .\cos A + \sin A.\cos \pi \\ Since \sin \pi = 0 and \cos \pi = -1 we have$ \sin (A + \pi ) = - \sin A \$