Question

If $\sin \theta +\cos \theta =\sqrt{2}\cos \theta $, then find the general value of $\theta $?

Answer 1

Hint: We start solving the problem by dividing both sides of the given trigonometric equation with $\sqrt{2}$. We then make use of the results $\sin \dfrac{\pi }{4}=\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$ and $\cos A\cos B+\sin A\sin B=\cos \left( A-B \right)$ to proceed through the problem. We then make use of the fact that if $\cos \theta =\cos \alpha $, then the general solution is $\theta =2n\pi \pm \alpha $. We then make the necessary calculations to get the required answer.

Complete step by step answer:
According to the problem, we are given that $\sin \theta +\cos \theta =\sqrt{2}\cos \theta $ and we need to find the general value of $\theta $.
Now, we have $\sin \theta +\cos \theta =\sqrt{2}\cos \theta $ ---(1).
Let us divide both sides of equation (1) with $\sqrt{2}$.
$\Rightarrow \dfrac{\sin \theta +\cos \theta }{\sqrt{2}}=\dfrac{\sqrt{2}\cos \theta }{\sqrt{2}}$.
$\Rightarrow \dfrac{1}{\sqrt{2}}\sin \theta +\dfrac{1}{\sqrt{2}}\cos \theta =\cos \theta $ ---(2).
We know that $\sin \dfrac{\pi }{4}=\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$. Let us use this in equation (2).
$\Rightarrow \sin \dfrac{\pi }{4}\sin \theta +\cos \dfrac{\pi }{4}\cos \theta =\cos \theta $.
$\Rightarrow \cos \dfrac{\pi }{4}\cos \theta +\sin \dfrac{\pi }{4}\sin \theta =\cos \theta $ ---(3).
We know that $\cos A\cos B+\sin A\sin B=\cos \left( A-B \right)$. Let us use this result in equation (3).
$\Rightarrow \cos \left( \dfrac{\pi }{4}-\theta \right)=\cos \theta $ ---(4).
We know that if $\cos \theta =\cos \alpha $, then the general solution is defined as $\theta =2n\pi \pm \alpha $, $n\in Z$.
So, we have $\theta =2n\pi \pm \left( \dfrac{\pi }{4}-\theta \right)$.
$\Rightarrow \theta =2n\pi +\left( \dfrac{\pi }{4}-\theta \right)$, $\theta =2n\pi -\left( \dfrac{\pi }{4}-\theta \right)$.
$\Rightarrow 2\theta =2n\pi +\dfrac{\pi }{4}$, $\theta =2n\pi -\dfrac{\pi }{4}+\theta $(which is a contradiction).
$\Rightarrow \theta =n\pi +\dfrac{\pi }{8}$, $n\in Z$.

$\therefore $ We have found the general solution for the given trigonometric equation $\sin \theta +\cos \theta =\sqrt{2}\cos \theta $ as $n\pi +\dfrac{\pi }{8}$, $n\in Z$.

Note: We should perform each step carefully in order to avoid confusion and calculation mistakes. We should not confuse between principal and general value while solving the problems involving trigonometric equations. We can also solve this problem as shown below:
We have given $\sin \theta +\cos \theta =\sqrt{2}\cos \theta $ ---(5).
Let us divide both sides of equation (5) with $\sqrt{2}$.
$\Rightarrow \dfrac{\sin \theta +\cos \theta }{\sqrt{2}}=\dfrac{\sqrt{2}\cos \theta }{\sqrt{2}}$.
$\Rightarrow \dfrac{1}{\sqrt{2}}\sin \theta +\dfrac{1}{\sqrt{2}}\cos \theta =\cos \theta $ ---(6).
We know that $\sin \dfrac{\pi }{4}=\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$. Let us use this in equation (6).
$\Rightarrow \cos \dfrac{\pi }{4}\sin \theta +\sin \dfrac{\pi }{4}\cos \theta =\cos \theta $.
$\Rightarrow \sin \theta \cos \dfrac{\pi }{4}+\cos \theta \sin \dfrac{\pi }{4}=\cos \theta $ ---(7).
We know that $\sin A\cos B+\cos A\sin B=\sin \left( A+B \right)$. Let us use this result in equation (7).
\[\Rightarrow \sin \left( \theta +\dfrac{\pi }{4} \right)=\cos \theta \] ---(8).
We know that $\cos \theta =\sin \left( \dfrac{\pi }{2}-\theta \right)$. Let us use this result in equation (8).
\[\Rightarrow \sin \left( \theta +\dfrac{\pi }{4} \right)=\sin \left( \dfrac{\pi }{2}-\theta \right)\].
We know that if $\sin \theta =\sin \alpha $, then the general solution is defined as $\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha $, $n\in Z$.
\[\Rightarrow \theta +\dfrac{\pi }{4}=n\pi +{{\left( -1 \right)}^{n}}\left( \dfrac{\pi }{2}-\theta \right)\].
Let us assume n is even.
\[\Rightarrow \theta +\dfrac{\pi }{4}=n\pi +\left( \dfrac{\pi }{2}-\theta \right)\].
\[\Rightarrow 2\theta =n\pi +\dfrac{\pi }{4}\].
\[\Rightarrow \theta =\dfrac{n\pi }{2}+\dfrac{\pi }{8}\], $n\in Z$.
Let us assume n is odd.
\[\Rightarrow \theta +\dfrac{\pi }{4}=n\pi -\left( \dfrac{\pi }{2}-\theta \right)\].
\[\Rightarrow \theta +\dfrac{\pi }{4}=n\pi -\dfrac{\pi }{2}+\theta \]
\[\Rightarrow \dfrac{\pi }{4}=n\pi -\dfrac{\pi }{2}\], which is a contradiction.