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Question: If we have a trigonometric expression as \(\sin \theta +\cos \theta =m\), then value of \(\sin \thet...

If we have a trigonometric expression as sin⁡θ+cos⁡θ=m\sin \theta +\cos \theta =m, then value of sin⁡θ−cos⁡θ\sin \theta -\cos \theta is
A. 2+m2\sqrt{2+{{m}^{2}}}
B. mm
C. 2+m2\sqrt{2+\dfrac{m}{2}}
D. 2−m2\sqrt{2-{{m}^{2}}}

Explanation

Solution

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

Complete step-by-step solution:
We first take the square of sin⁡θ+cos⁡θ=m\sin \theta +\cos \theta =m. We have the identity sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1.
(sin⁡θ+cos⁡θ)2=m2 ⇒sin⁡2θ+cos⁡2θ+2sin⁡θcos⁡θ=m2 \begin{aligned} & {{\left( \sin \theta +\cos \theta \right)}^{2}}={{m}^{2}} \\\ & \Rightarrow {{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta ={{m}^{2}} \\\ \end{aligned}
Simplifying we get
sin⁡2θ+cos⁡2θ+2sin⁡θcos⁡θ=m2 ⇒1+2sin⁡θcos⁡θ=m2 ⇒2sin⁡θcos⁡θ=m2−1 \begin{aligned} & {{\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{aligned}
We now multiply with −1-1 to get
2sin⁡θcos⁡θ=m2−1 ⇒−2sin⁡θcos⁡θ=1−m2 \begin{aligned} & 2\sin \theta \cos \theta ={{m}^{2}}-1 \\\ & \Rightarrow -2\sin \theta \cos \theta =1-{{m}^{2}} \\\ \end{aligned}
Now adding 1 on both sides we get 1−2sin⁡θcos⁡θ=1+1−m2=2−m21-2\sin \theta \cos \theta =1+1-{{m}^{2}}=2-{{m}^{2}}.
We use the identity sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1.
1−2sin⁡θcos⁡θ=2−m2 ⇒sin⁡2θ+cos⁡2θ−2sin⁡θcos⁡θ=2−m2 ⇒(sin⁡θ−cos⁡θ)2=2−m2 \begin{aligned} & 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{aligned}
We now take the square root to get
(sin⁡θ−cos⁡θ)2=2−m2 ⇒sin⁡θ−cos⁡θ=±2−m2 \begin{aligned} & {{\left( \sin \theta -\cos \theta \right)}^{2}}=2-{{m}^{2}} \\\ & \Rightarrow \sin \theta -\cos \theta =\pm \sqrt{2-{{m}^{2}}} \\\ \end{aligned}
The correct option is D.

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