Question

How do you verify the identity $\sin \left( {\dfrac{\pi }{6} + x} \right) + \sin \left( {\dfrac{\pi }{6} - x} \right) = \cos x$ ?

Answer 1

Hint : Here, we are given an equation in which the sum of sines of two angles is equal to cosine of an angle. The angles are $\dfrac{\pi }{6} + x$ , $\dfrac{\pi }{6} - x$ and $x$ . Note that the angles are expressed in radian which is unit of angle. What you can do here is to apply the trigonometric properties on the sines which are on the left-hand side and then figure out a way to simplify the left-hand side as much as possible. By simplifying what is meant, is that trying to prove left-hand side equals the right-hand side.

Complete step by step solution:
Let us first consider the left-hand side, in that too, first consider the first term which is
$\sin \left( {\dfrac{\pi }{6} + x} \right)$ . You can see that the angle inside the sine can be observed as the sum of two angles separately, that is $\dfrac{\pi }{6}$ and $x$ . So, $\sin \left( {\dfrac{\pi }{6} + x} \right)$ can be considered as $\sin \left( {A + B} \right)$ , where $A = \dfrac{\pi }{6}$ and $B = x$ .
The trigonometric property that gives you the expansion of sine of sum of two angles is
$\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$ . So, the first time can be written as $\sin \dfrac{\pi }{6}\cos x + \cos \dfrac{\pi }{6}\sin x$ .
The trigonometric property that gives the expansion of sine of difference between two angles is
$\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$ . So, the second term can be written as $\sin \dfrac{\pi }{6}\cos x - \cos \dfrac{\pi }{6}\sin x$ .
Let us add these two terms.
$\sin \dfrac{\pi }{6}\cos x + \cos \dfrac{\pi }{6}\sin x + \sin \dfrac{\pi }{6}\cos x - \cos \dfrac{\pi }{6}\sin x = 2\sin \dfrac{\pi }{6}\cos x$
$\dfrac{\pi }{6}radians = \dfrac{{180}}{6} = 30^\circ$ ,
we have $\sin 30^\circ = \dfrac{1}{2}$ , let us substitute the value of $\sin 30^\circ$ in the above obtained value.
$2\sin \dfrac{\pi }{6}\cos x = \left( 2 \right)\left( {\dfrac{1}{2}} \right)\cos x = \cos x$
As you can see, the left-hand side is equal to the right-hand side.
Hence proved
$\sin \left( {\dfrac{\pi }{6} + x} \right) + \sin \left( {\dfrac{\pi }{6} - x} \right) = \cos x$ , identity verified.

Note : In order to solve questions like this which includes verification of trigonometric identities, you need to bring all the material you have on trigonometry, that is trigonometric properties and identities. You need to memorize the property, formulae, expansion and values that are used in this question or are used in trigonometry.