Question

The maximum value of $\sin \theta +\cos \theta$ in $[\theta ,\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.]$ is

• A. $\sqrt{2}$
• B. $2$
• C. $0$
• D. $-\sqrt{2}$

Answer 1

The correct option is A

$\sqrt{2}$

Find the maximum value of the expression $\sin \theta +\cos \theta$ be in interval $[\theta ,\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.]$

The maximum value of $a\cos x+b\sin x=\sqrt{a^2+b^2}$

Now, $\sin \theta +\cos \theta =\left(1\right)\cos \theta +\left(1\right)\sin \theta$

Therefore, the maximum value of $\sin \theta +\cos \theta$$=\sqrt{1^2+1^2}=\sqrt{2}$

Hence, the maximum value of $\sin \theta +\cos \theta$ in $[\theta ,\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.]$ is $\sqrt{2}$.