Question

Prove the following equation: $\sqrt{2}\sin \left( \dfrac{\pi }{4}-\theta \right)=\cos \theta -\sin \theta $

Answer 1

Hint: We need to prove the LHS and RHS to be equal. To do this, we will first take the LHS and solve it by the formula $\sin \left( x-y \right)=\sin x\cos y-\cos x\sin y$ and also require standard values like $\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$ and $\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$ . By solving the whole LHS, we will try to bring it in the form of RHS. Thus, the required condition will be proved.

Complete step-by-step solution:
We will take the LHS and solve it and try to bring it in the form of RHS.
Now, we know that for any two angles ‘x’ and ‘y’ $\sin \left( x-y \right)$ is given by the formula:
$\sin \left( x-y \right)=\sin x\cos y-\cos x\sin y$
Now, the LHS is given as:
$\sqrt{2}\sin \left( \dfrac{\pi }{4}-\theta \right)$
Now, we will solve the $\sin \left( \dfrac{\pi }{4}-\theta \right)$ part of the LHS by the above mentioned formula.
Here, we have $x=\dfrac{\pi }{4}$ and $y=\theta $
Thus, the value of $\sin \left( \dfrac{\pi }{4}-\theta \right)$ will be calculated as:
$\sin \left( \dfrac{\pi }{4}-\theta \right)=\sin \dfrac{\pi }{4}\cos \theta -\cos \dfrac{\pi }{4}\sin \theta $ ……(i)
We know that $\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$ and $\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$
Thus, putting in these values in equation (i) we get:
$\begin{align}
  & \sin \left( \dfrac{\pi }{4}-\theta \right)=\sin \dfrac{\pi }{4}\cos \theta -\cos \dfrac{\pi }{4}\sin \theta \\
 & \Rightarrow \sin \left( \dfrac{\pi }{4}-\theta \right)=\dfrac{1}{\sqrt{2}}\cos \theta -\dfrac{1}{\sqrt{2}}\sin \theta \\
\end{align}$
$\Rightarrow \sin \left( \dfrac{\pi }{4}-\theta \right)=\dfrac{1}{\sqrt{2}}\left( \cos \theta -\sin \theta \right)$ ……(ii)
Now, if we multiply the value we obtained for $\sin \left( \dfrac{\pi }{4}-\theta \right)$ by $\sqrt{2}$ , we will get the value of LHS.
Thus, multiplying equation (ii) by $\sqrt{2}$ we get our LHS as:
$\begin{align}
  & LHS=\sqrt{2}\sin \left( \dfrac{\pi }{4}-\theta \right) \\
 & \Rightarrow LHS=\sqrt{2}\left( \dfrac{1}{\sqrt{2}}\left( \cos \theta -\sin \theta \right) \right) \\
 & \Rightarrow LHS=1\left( \cos \theta -\sin \theta \right) \\
 & \Rightarrow LHS=\cos \theta -\sin \theta \\
\end{align}$
 We have our RHS as:
$RHS=\cos \theta -\sin \theta $
This is the same as LHS we obtained.
Thus, $LHS=RHS$
Hence, proved.

Note: Take care while taking the angles ‘x’ and ‘y’. They should not be combined when the formula is applied because $\sin \left( x-y \right)$ is not equal to $\sin \left( y-x \right)$ . The relation between them is given as $\sin \left( x-y \right)=-\sin \left( y-x \right)$ . Therefore, care should be taken while applying the formula.