Question

Prove that, $\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-\sqrt{2}\sin x$.

Answer 1

Hint: We will first start by using the identity that $\cos C-\cos D=2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{D-C}{2} \right)$. Then we will use the fact that $\sin \left( \pi -\theta \right)=\sin \theta $ to find the value of $\sin \left( \dfrac{3\pi }{4} \right)$. Then we will use this value to prove the left hand side of the equation to be equal to the right side.

Complete step-by-step answer:
Now, we have been given to prove that, $\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-\sqrt{2}\sin x$.
Now, we will take the left hand side of the equation and prove it to be equal to the right hand side. So, in left hand side we have,
$\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)$
Now, we know that the trigonometric identity that,
$\cos C-\cos D=2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{D-C}{2} \right)$
Now, we have the left hand side as,
$\begin{align}
  & \Rightarrow \cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=2\sin \left( \dfrac{6\pi }{8} \right)\sin \left( \dfrac{-2x}{2} \right) \\
 & =2\sin \left( \dfrac{3\pi }{4} \right)\sin \left( -x \right) \\
\end{align}$
Now, we know the trigonometric identity that,
$\begin{align}
  & \sin \left( -x \right)=-\sin \left( x \right) \\
 & \sin \left( \pi -\theta \right)=\sin \left( \theta \right) \\
\end{align}$
So, using this we have,
$\begin{align}
  & \Rightarrow \cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=2\sin \left( \pi -\dfrac{\pi }{4} \right)\left( -\sin x \right) \\
 & =-2\sin \left( \dfrac{\pi }{4} \right)\sin \left( x \right) \\
\end{align}$
Now, we know the fact that $\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$.So, we have,
$\begin{align}
  & \Rightarrow \cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-2\times \dfrac{1}{\sqrt{2}}\sin x \\
 & =-\sqrt{2}\sin x \\
\end{align}$
Hence, we have LHS = RHS and therefore, it is proved that $\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=-\sqrt{2}\sin x$.

Note: It is important to note that to find the value of $\sin \left( \dfrac{3\pi }{4} \right)$ we have re-written it as $\sin \left( \dfrac{4\pi -\pi }{4} \right)\ or\ \sin \left( \pi -\dfrac{\pi }{4} \right)$ and then used the fact that $\sin \left( \pi -\theta \right)=\sin \theta $ to find its value. Also, we have used the property of sine that $\sin \left( -x \right)=-\sin x$ to finalize the proof.