The maximum value of sin left( x + frac{pi}{6 | vedclass

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Question

(A) Let $f(x) = \sin \left( x + \frac{\pi}{6} \right) + \cos \left( x + \frac{\pi}{6} \right)$.We can rewrite this expression as $f(x) = \sqrt{2} \lef

Vedclass · Accepted Answer

$x = \frac{\pi}{12}$

Vedclass · Answer

(A) Let $f(x) = \sin \left( x + \frac{\pi}{6} \right) + \cos \left( x + \frac{\pi}{6} \right)$.We can rewrite this expression as $f(x) = \sqrt{2} \left[ \frac{1}{\sqrt{2}} \sin \left( x + \frac{\pi}{6} \right) + \frac{1}{\sqrt{2}} \cos \left( x + \frac{\pi}{6} \right) \right]$.Using the identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,we get $f(x) = \sqrt{2} \sin \left( x + \frac{\pi}{6} + \frac{\pi}{4} \right) = \sqrt{2} \sin \left( x + \frac{5\pi}{12} \right)$.The maximum value of the sine function is $1$,which occurs when the argument is $\frac{\pi}{2}$.Setting $x + \frac{5\pi}{12} = \frac{\pi}{2}$,we get $x = \frac{\pi}{2} - \frac{5\pi}{12} = \frac{6\pi - 5\pi}{12} = \frac{\pi}{12}$.Since $\frac{\pi}{12}$ lies within the interval $\left( 0, \frac{\pi}{2} \right)$,the maximum value is attained at $x = \frac{\pi}{12}$.

The maximum value of $\sin \left( x + \frac{\pi}{6} \right) + \cos \left( x + \frac{\pi}{6} \right)$ in the interval $\left( 0, \frac{\pi}{2} \right)$ is attained at

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