Question

How do you verify the identity $\sin \left( {\dfrac{x}{2}} \right)\cos \left( {\dfrac{x}{2}} \right) = \dfrac{{\sin x}}{2}$?

Answer 1

Hint: First we will evaluate the right-hand of equation and then further the left-hand side of the equation. We will use the relation $\sin 2A = 2\sin A\cos A$. Then we will try to factorise and simplify the terms so that the left-hand side matches the right-hand side. Then finally evaluate the solution which lies within the given interval.

Complete step-by-step solution:
First we will start solving this question by mentioning the double angle formula $\sin 2A = 2\sin A\cos A$
Now, if we substitute $A = \dfrac{x}{2}$ we will get.
$
  \sin 2A = 2\sin A\cos A \\
  \dfrac{{\sin x}}{2}\,\,\,\, = \dfrac{1}{2}\sin \left( {2\left( {\dfrac{x}{2}} \right)} \right) \\
 $
Now if we simplify the equation further, we will get,
$
   = 2\left( {\dfrac{1}{2}} \right)\sin \left( {\dfrac{x}{2}} \right)\cos \left( {\dfrac{x}{2}} \right) \\
   = \sin \dfrac{x}{2}\cos \dfrac{x}{2} \\
 $
Hence, proved that $\sin \left( {\dfrac{x}{2}} \right)\cos \left( {\dfrac{x}{2}} \right)$ is equal to the expression $\dfrac{{\sin x}}{2}$.

Additional information: The trigonometric double angle formulas give a relationship between the basic trigonometric functions applied to twice an angle in terms of trigonometric functions of the angle itself. Always use double angle formulas to find the exact values. You can also use double angle formulas to verify the identities.

Note: While choosing the side to solve, always choose the side where you can directly apply the trigonometric identities. Also, remember the trigonometric identities ${\sin ^2}x + {\cos ^2}x = 1$ and $\sin 2A = 2\sin A\cos A$. While applying the double angle identities, first choose the identity according to the terms you have then choose the terms from the expression involving which you are using the double angle identities. While modifying any identity make sure that when you back trace the identity, you get the same original identity. Also, remember that the range of $\cos $ function is from $ - 1\,$ to $ + 1$ and the range of $\sin $ function is also from $ - 1\,$ to $ + 1$.