Question

For an acute angle, $\alpha$, $\sin \alpha +\cos \alpha$ takes the
greatest value when $\alpha$ is

• A. ${30}^0$
• B. ${45}^0$
• C. ${60}^0$
• D. ${90}^0$

Answer 1

Answer: (B)

${45}^0$

$(cos\alpha +sin\alpha )$
Multiplying and dividing by $\sqrt{2}$
Therefore
$\sqrt{2}(\frac{cos\alpha }{\sqrt{2}}+\frac{sin\alpha }{\sqrt{2}})$
$=\sqrt{2}(sin\frac{\pi }{4}cos\alpha +cos\frac{\pi }{4}sin\alpha )$ ...$(sin\frac{\pi }{4}=cos\frac{\pi }{4}=\frac{1}{\sqrt{2}})$
$=\sqrt{2}sin(\frac{\pi }{4}+alpha)$
Since $0<\theta <\frac{\pi }{2}$
$sin(\frac{\pi }{4}+\alpha )$ is maximum for $sin(\frac{\pi }{4}+\alpha )=1$
Therefore
$\frac{\pi }{4}+\alpha =\frac{\pi }{2}$
$\alpha =\frac{\pi }{4}={45}^0$