Question

If we have a trigonometric expression as $\sin \theta +\cos \theta =m$, then value of $\sin \theta -\cos \theta $ is
A. $\sqrt{2+{{m}^{2}}}$
B. $m$
C. $\sqrt{2+\dfrac{m}{2}}$
D. $\sqrt{2-{{m}^{2}}}$

Answer 1

Hint: We first take the square of the given expression $\sin \theta +\cos \theta =m$ and use the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. We find the value of $-2\sin \theta \cos \theta $. Then we add 1 and change it to the square of $\sin \theta -\cos \theta $. The square root gives the solution.

Complete step-by-step solution:
We first take the square of $\sin \theta +\cos \theta =m$. We have the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.
$\begin{align}
  & {{\left( \sin \theta +\cos \theta \right)}^{2}}={{m}^{2}} \\
 & \Rightarrow {{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta ={{m}^{2}} \\
\end{align}$
Simplifying we get
$\begin{align}
  & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta ={{m}^{2}} \\
 & \Rightarrow 1+2\sin \theta \cos \theta ={{m}^{2}} \\
 & \Rightarrow 2\sin \theta \cos \theta ={{m}^{2}}-1 \\
\end{align}$
We now multiply with $-1$ to get
$\begin{align}
  & 2\sin \theta \cos \theta ={{m}^{2}}-1 \\
 & \Rightarrow -2\sin \theta \cos \theta =1-{{m}^{2}} \\
\end{align}$
Now adding 1 on both sides we get $1-2\sin \theta \cos \theta =1+1-{{m}^{2}}=2-{{m}^{2}}$.
We use the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.
$\begin{align}
  & 1-2\sin \theta \cos \theta =2-{{m}^{2}} \\
 & \Rightarrow {{\sin }^{2}}\theta +{{\cos }^{2}}\theta -2\sin \theta \cos \theta =2-{{m}^{2}} \\
 & \Rightarrow {{\left( \sin \theta -\cos \theta \right)}^{2}}=2-{{m}^{2}} \\
\end{align}$
We now take the square root to get
$\begin{align}
  & {{\left( \sin \theta -\cos \theta \right)}^{2}}=2-{{m}^{2}} \\
 & \Rightarrow \sin \theta -\cos \theta =\pm \sqrt{2-{{m}^{2}}} \\
\end{align}$
The correct option is D.

Note: We cannot use the algebraic identity of ${{\left( \sin \theta +\cos \theta \right)}^{2}}={{\left( \sin \theta -\cos \theta \right)}^{2}}+4\sin \theta \cos \theta $ as the value of $\sin \theta \cos \theta $ is unknown. Therefore, we used the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ to first find the unknown value and then change to the form of ${{\left( \sin \theta -\cos \theta \right)}^{2}}$.