Question

If $\sin x+\cos x=\sqrt{2}\cos x$, then $\cos x-\sin x$ is equal to:
(a). $\sqrt{2}\cos x$
(b). $\sqrt{2}\sin x$
(c). $\sqrt{2}\left( \sin x+\cos x \right)$
(d). None of these

Answer 1

- Hint: First of all write $\sin x$ in terms of $\cos x$ from the given equation $\sin x+\cos x=\sqrt{2}\cos x$ and then substitute the value of $\sin x$ in the given expression $\cos x-\sin x$ and then simplifying this expression in terms of $\cos x$ and then compare the answer with the options given in the question.

Complete step-by-step solution -

The equation given in the question is:
$\sin x+\cos x=\sqrt{2}\cos x$
Rearranging the above equation and writing $\sin x$ in terms of $\cos x$ we get,
$\sin x=\cos x\left( \sqrt{2}-1 \right)$
Now, substituting this value of $\sin x$ in the expression given in the question $\cos x-\sin x$ we get$\cos x-\sin x$
$\begin{aligned} & \cos x-\cos x\left( \sqrt{2}-1 \right) \\\ & \Rightarrow 2\cos x-\sqrt{2}\cos x \\\ & \Rightarrow \sqrt{2}\cos x\left( \sqrt{2}-1 \right) \\\ \end{aligned}$
Rearranging this expression $\sin x+\cos x=\sqrt{2}\cos x$ we get,
$\begin{aligned} & \sin x+\cos x=\sqrt{2}\cos x \\\ & \Rightarrow \cos x\left( \sqrt{2}-1 \right)=\sin x \\\ \end{aligned}$
Now, substituting the above value of $\cos x\left( \sqrt{2}-1 \right)$ in $\sqrt{2}\cos x\left( \sqrt{2}-1 \right)$ we get,
$\sqrt{2}\sin x$
From the above calculations, we have found that the simplification of this expression $\cos x-\sin x$ is$\sqrt{2}\sin x$.
Now, we are going to compare this result with the options given in the question.
$\sqrt{2}\cos x$
The value of the above options is not equal to the answer that we have obtained from solving this expression $\cos x-\sin x$.
$\sqrt{2}\sin x$
The value of the above options is equal to the answer that we have obtained from solving this expression $\cos x-\sin x$.
$\sqrt{2}\left( \sin x+\cos x \right)$
Simplifying the above option by substituting the value of $\sin x+\cos x$ from $\sin x+\cos x=\sqrt{2}\cos x$ in the above expression we get,
$\begin{aligned} & \sqrt{2}\left( \sin x+\cos x \right) \\\ & =\sqrt{2}\left( \sqrt{2}\cos x \right) \\\ & =2\cos x \\\ \end{aligned}$
The solution of the above option is $2\cos x$ which is not matching with the result that we have got from solving this expression $\cos x-\sin x$.
From the above options, we have found that none of the values of option (b) is matched with $\sqrt{2}\sin x$.
Hence, the correct option is (b).

Note: If instead of writing $\sin x$ in terms of $\cos x$ from the given equation $\sin x+\cos x=\sqrt{2}\cos x$ we write $\cos x$ in terms of $\sin x$ from the given equation $\sin x+\cos x=\sqrt{2}\cos x$ then you will get.
$\begin{aligned} & \sin x+\cos x=\sqrt{2}\cos x \\\ & \Rightarrow \cos x\left( \sqrt{2}-1 \right)=\sin x \\\ & \Rightarrow \cos x=\dfrac{\sin x}{\left( \sqrt{2}-1 \right)} \\\ \end{aligned}$
Now, rationalizing the above expression we get,
$\begin{aligned} & \cos x=\dfrac{\sin x}{\sqrt{2}-1}\times \dfrac{\sqrt{2}+1}{\sqrt{2}+1} \\\ & \Rightarrow \cos x=\left( \sqrt{2}+1 \right)\sin x \\\ \end{aligned}$
Substituting this value of $\cos x$ in $\cos x-\sin x$ we get,
$\begin{aligned} & \left( \sqrt{2}+1 \right)\sin x-\sin x \\\ & =\sqrt{2}\sin x \\\ \end{aligned}$
Hence, the correct option is (b).