Question

How do you verify the identity $ \cos \left( \pi -\theta \right)+\sin \left( \dfrac{\pi }{2}+\theta \right)=0 $ ?

Answer 1

Hint:
Hint: We will use the formula for the sine function of the sum of two angles. This formula is given as $ \sin \left( a+b \right)=\sin a\cos b+\cos a\sin b $ . We will also use the formula for the cosine function of the difference of two angles. This formula is given as $ \cos \left( a-b \right)=\cos a\cos b+\sin a\sin b $ . We will expand the right-hand side of the given identity using these two formulae.

Complete step by step answer:
We have to verify the identity $ \cos \left( \pi -\theta \right)+\sin \left( \dfrac{\pi }{2}+\theta \right)=0 $ . We know that there are formulae for the trigonometric functions of sum of angles and difference of angles. The formula for the sine function of sum of angles is given as $ \sin \left( a+b \right)=\sin a\cos b+\cos a\sin b $ . Substituting $ a=\dfrac{\pi }{2} $ and $ b=\theta $ in this formula, we get the following,
 $ \sin \left( \dfrac{\pi }{2}+\theta \right)=\sin \dfrac{\pi }{2}\cos \theta +\cos \dfrac{\pi }{2}\sin \theta $
We know that $ \sin \dfrac{\pi }{2}=1 $ and $ \cos \dfrac{\pi }{2}=0 $ . Substituting these values in the above equation, we get
 $ \begin{align}
  & \sin \left( \dfrac{\pi }{2}+\theta \right)=1\cdot \cos \theta +0\cdot \sin \theta \\
 & \therefore \sin \left( \dfrac{\pi }{2}+\theta \right)=\cos \theta \\
\end{align} $
Substituting this expression in place of the second term of the given identity, the left hand side becomes,
 $ LHS=\cos \left( \pi -\theta \right)+\cos \theta $
Now, we will look at the formula for the cosine function of the difference of two angles. This formula is given as $ \cos \left( a-b \right)=\cos a\cos b+\sin a\sin b $ . Substituting $ a=\pi $ and $ b=\theta $ in this formula, we get the following,
 $ \cos \left( \pi -\theta \right)=\cos \pi \cos \theta +\sin \pi \sin \theta $
Now, we know that $ \sin \pi =0 $ and $ \cos \pi =-1 $ . Substituting these values in the above equation, we get
 $ \begin{align}
  & \cos \left( \pi -\theta \right)=-1\cdot \cos \theta +0\cdot \sin \theta \\
 & \therefore \cos \left( \pi -\theta \right)=-\cos \theta \\
\end{align} $
Substituting this expression in place of the first term of the given identity, the left hand side becomes,
 $ \begin{align}
  & LHS=-\cos \theta +\cos \theta \\
 & \therefore LHS=0 \\
\end{align} $
Also, $ RHS=0 $ is given. Therefore, we have $ LHS=RHS $ . Thus, we have verified the given identity.

Note:
 It is important that we are familiar with the trigonometric functions and their identities. It is useful to know the values of the trigonometric functions for standard angles. While working with an equation having trigonometric functions, we should always take the help of the identities and relations to simplify the given equation.