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: In this problem we need to show that $\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)=\sqrt{2}\sin x$. When we observe the given expression, we can say that the given expression is in the form of $\cos \left( A+B \right)-\cos \left( A-B \right)$. In trigonometry we have the formulas for $\cos \left( A+B \right)$ and $\cos \left( A-B \right)$as $\cos \left( A+B \right)$ is $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ and $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$. Now we will use these two formulas and simplify the obtained equation and write the solution of the equation.

Formula use:
1. $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$.
2. $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$.
3.$\sin \left( \dfrac{3\pi }{4} \right)=-\sin \left( \dfrac{\pi }{4} \right)$.
4.$\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$.

Complete Step by Step Solution:
Given that, $\cos \left( \dfrac{3\pi }{4}+X \right)-\cos \left( \dfrac{3\pi }{4}-X \right)=\sqrt{2}\sin x$.
Considering the LHS part of the above equation, then we will get
$LHS=\cos \left( \dfrac{3\pi }{4}+x \right)-\cos \left( \dfrac{3\pi }{4}-x \right)$
We can observe that the above LHS part is in the form of $\cos \left( A+B \right)-\cos \left( A-B \right)$.
In trigonometry we have the formula for $\cos \left( A+B \right)$ as $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$.
Now we will evaluate $\cos \left( \dfrac{3\pi }{4}+x \right)$ by using the formula of $\cos \left( A+B \right)$, then
$\cos \left( \dfrac{3\pi }{4}+x \right)=\cos \left( \dfrac{3\pi }{4} \right)\cos x-\sin \left( \dfrac{3\pi }{4} \right)\sin x$.
In trigonometry we also have the formula for $\cos \left( A-B \right)$ as $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$
Now we will evaluate $\cos \left( \dfrac{3\pi }{4}-x \right)$ by using the formula of $\cos \left( A-B \right)$, then
$\cos \left( \dfrac{3\pi }{4}-x \right)=\cos \left( \dfrac{3\pi }{4} \right)\cos x+\sin \left( \dfrac{3\pi }{4} \right)\sin x$.
Now we substitute the two values in the given expression, then we will get
$LHS=\cos \left( \dfrac{3\pi }{4} \right)\cos x-\sin \left( \dfrac{3\pi }{4} \right)\sin x-\left( \cos \left( \dfrac{3\pi }{4} \right)\cos x+\sin \left( \dfrac{3\pi }{4} \right)\sin x \right)$.
Now we will multiply the $-$ to the $\cos \left( \dfrac{3\pi }{4}-x \right)$, then we will get
$LHS=\cos \left( \dfrac{3\pi }{4} \right)\cos x-\sin \left( \dfrac{3\pi }{4} \right)\sin x-\cos \left( \dfrac{3\pi }{4} \right)\cos x-\sin \left( \dfrac{3\pi }{4} \right)\sin x$.
Now we will cancel the opposite sign terms, then we will have
$\begin{align}
  & LHS=-\sin \left( \dfrac{3\pi }{4} \right)\sin x-\sin \left( \dfrac{3\pi }{4} \right)\sin x \\
 & \Rightarrow LHS=-2\sin \left( \dfrac{3\pi }{4} \right)\sin x \\
\end{align}$
We have the value $\sin \left( \dfrac{3\pi }{4} \right)=-\sin \left( \dfrac{\pi }{4} \right)$ . now we will substitute it in the above expression, then,
$\Rightarrow LHS=-2\sin x\left( -\sin \left( \dfrac{3\pi }{4} \right) \right)$.
Now we will simplify the above expression, then we will get
$\Rightarrow LHS=2\sin \left( \dfrac{\pi }{4} \right)\sin x$.
We know that$\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$. Now we will use it for above expression, then
$\Rightarrow LHS=2\left( \dfrac{1}{\sqrt{2}} \right)\sin x$.
Now we will simplify the above expression, then we will have
$\begin{align}
  & \Rightarrow LHS=\sqrt{2}\sin x \\
 & \therefore LHS=RHS \\
\end{align}$.

Note:
For this problem we also use the simple method to get the result. In advanced trigonometry we have the formula $\cos \left( A+B \right)-\cos \left( A-B \right)=-2\sin A\sin B$. We will use this formula directly and substitute all the values we have in the above formula and simplify it to get the result.