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Question: If \(\sin \theta -\cos \theta =\sqrt{2}\sin \theta \) , then prove that \(\sin \theta +\cos \theta =...

If sin⁡θ−cos⁡θ=2sin⁡θ\sin \theta -\cos \theta =\sqrt{2}\sin \theta , then prove that sin⁡θ+cos⁡θ=−2cos⁡θ\sin \theta +\cos \theta =-\sqrt{2}\cos \theta ?

Explanation

Solution

Here we have been given an equation having trigonometric function and we have to prove another equation with trigonometric function. Firstly we will take the common term in the first equation together and then by rationalizing the denominator solve it. Finally we will rearrange our terms such that we get the value in the second equation from and get our desired answer.

Complete step by step solution:
We have been given that:
sin⁡θ−cos⁡θ=2sin⁡θ\sin \theta -\cos \theta =\sqrt{2}\sin \theta ….(1)\left( 1 \right)
Prove that:
sin⁡θ+cos⁡θ=−2cos⁡θ\sin \theta +\cos \theta =-\sqrt{2}\cos \theta ….(2)\left( 2 \right)
We will take equation (1) and simplify it to get our equation (2) as follows:
sin⁡θ−cos⁡θ=2sin⁡θ\sin \theta -\cos \theta =\sqrt{2}\sin \theta
Take sine terms on left side and rest term on other side as follows:
⇒sin⁡θ−2sin⁡θ=cos⁡θ\Rightarrow \sin \theta -\sqrt{2}\sin \theta =\cos \theta
Take sine common in left hand side as follows:
⇒sin⁡θ(1−2)=cos⁡θ\Rightarrow \sin \theta \left( 1-\sqrt{2} \right)=\cos \theta
Now rationalize the bracket value by multiplying and dividing by 1+21+\sqrt{2}on left side as follows:
⇒sin⁡θ(1−2×1+21+2)=cos⁡θ\Rightarrow \sin \theta \left( 1-\sqrt{2}\times \dfrac{1+\sqrt{2}}{1+\sqrt{2}} \right)=\cos \theta
Using algebraic identity (a−b)(a+b)=a2−b2\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}} where a=1,b=2a=1,b=\sqrt{2} we get,
⇒sin⁡θ(12−(2)21+2)=cos⁡θ\Rightarrow \sin \theta \left( \dfrac{{{1}^{2}}-{{\left( \sqrt{2} \right)}^{2}}}{1+\sqrt{2}} \right)=\cos \theta
⇒sin⁡θ(1−21+2)=cos⁡θ\Rightarrow \sin \theta \left( \dfrac{1-2}{1+\sqrt{2}} \right)=\cos \theta
Simplify and take the denominator value on the right hand side as follows:
⇒sin⁡θ(−11+2)=cos⁡θ\Rightarrow \sin \theta \left( \dfrac{-1}{1+\sqrt{2}} \right)=\cos \theta
⇒−sin⁡θ=cos⁡θ(1+2)\Rightarrow -\sin \theta =\cos \theta \left( 1+\sqrt{2} \right)
Simplifying further we get,
⇒−sin⁡θ=cos⁡θ+2cos⁡θ\Rightarrow -\sin \theta =\cos \theta +\sqrt{2}\cos \theta
⇒−sin⁡θ−cos⁡θ=2cos⁡θ\Rightarrow -\sin \theta -\cos \theta =\sqrt{2}\cos \theta
Multiplying the whole equation by −1-1 we get,
⇒sin⁡θ+cos⁡θ=−2cos⁡θ\Rightarrow \sin \theta +\cos \theta =-\sqrt{2}\cos \theta
So we got the equation (2).
Hence Proved

Note:
In this type of question we just need to rearrange the terms in such a way that we get the value we have to prove by the value already given. We can’t take value from the given equation and substitute it in the second equation as we have the unknown variable which can’t be simplified further. As we have the square root values rationalizing the denominator is used to simplify it as we want the square root value multiplied with cosine term and not with sine term.