Question

Prove that $\sin \left( \pi -x \right)=\sin \left( x \right)$

Answer 1

Hint: In the above question, we have been given a trigonometric identity to prove, which is written as $\sin \left( \pi -x \right)=\sin \left( x \right)$. To prove it, we need to consider the LHS of the given identity, which is equal to $\sin \left( \pi -x \right)$. We need to simplify the LHS by using the trigonometric identity which is given by $\sin \left( a-b \right)=\sin a\cos b-\cos a\sin b$. On substituting \[a=\pi \] and $b=x$ into the identity, we will be able to simplify the LHS as \[\sin \pi \cos x-\cos \pi \sin x\]. Then on substituting the values of $\sin \pi $ and $\cos \pi $ into the obtained expression, the LHS will be simplified and will come equal to the RHS and hence the given identity will be finally proved.

Complete step by step answer:
The trigonometric identity to be proved has been given in the above question as
$\Rightarrow \sin \left( \pi -x \right)=\sin \left( x \right)$
Let us first consider the LHS of the above equation.
$\Rightarrow LHS=\sin \left( \pi -x \right)$
We know the trigonometric identity given by $\sin \left( a-b \right)=\sin a\cos b-\cos a\sin b$. On substituting $a=\pi $ and $b=x$ into the identity, we can write the above expression as
$\Rightarrow LHS=\sin \pi \cos x-\cos \pi \sin x$
Now, we know that $\sin \pi =0$ and $\cos \pi =-1$. On substituting these in the above equation, we get
$\begin{align}
  & \Rightarrow LHS=\left( 0 \right)\cos x-\left( -1 \right)\sin x \\
 & \Rightarrow LHS=0+\sin x \\
 & \Rightarrow LHS=\sin x \\
\end{align}$
Now, considering the RHS of the given trigonometric identity, we have
$\Rightarrow RHS=\sin x$
From the above two equation, we can finally write
$\Rightarrow LHS=RHS$
Hence, the given trigonometric identity has been proved.

Note: For solving these types of questions, we must remember all of the important trigonometric identities. Also, we must be able to find out the values of the trigonometric functions at the multiples of $\dfrac{\pi }{2}$. For this, we need to have the knowledge of the four quadrants and the behavior of the trigonometric functions in the respective quadrants.