Question

Maximum value of $\sin x-\cos x$ is equal to

• A. $\sqrt{2}$
• B. $1$
• C. $0$
• D. None of these

Answer 1

The correct option is A $\sqrt{2}$

$y=\sin x-\cos x,\frac{dy}{dx}=\cos x+\sin x$

$\frac{d^{2}y}{d{x}^2}=-\sin x+\cos x=-(\sin x-\cos x)$

$=-y.$ (negative) hence a maxima.

Further for maxima $\frac{dy}{dx}=0$

$\Rightarrow \cos x+\sin x=0$

$\Rightarrow tanx=-1=tan\frac{3\pi }{4}$

$\therefore x=\frac{3\pi }{4}$

$\therefore$ Maximum value of (\sin x - \cos x)

$=\sin \frac{3\pi }{4}-\cos \frac{3\pi }{4}$

$=\left(\frac{1}{\sqrt{2}}\right)-(-\frac{1}{\sqrt{2}})=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$