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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $f(x) = \sin \left( x + \frac{\pi}{6} \right) + \cos \left( x + \frac{\pi}{6} \right)$.We can rewrite this as $f(x) = \sqrt{2} \left( \frac{1}

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $f(x) = \sin \left( x + \frac{\pi}{6} ight) + \cos \left( x + \frac{\pi}{6} ight)$.We can rewrite this as $f(x) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \left( x + \frac{\pi}{6} ight) + \frac{1}{\sqrt{2}} \cos \left( x + \frac{\pi}{6} ight) ight)$.Using the identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$,we get $f(x) = \sqrt{2} \sin \left( x + \frac{\pi}{6} + \frac{\pi}{4} ight) = \sqrt{2} \sin \left( x + \frac{5\pi}{12} ight)$.The maximum value of $\sin 	heta$ is $1$,which occurs when $	heta = \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 in the interval $\left( 0, \frac{\pi}{2} ight)$,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

Similar Questions

If the minimum value of $\cos(\sinh(\log x) + \cosh(\log x))$ is $k$,then $\cosh(k+1) =$

Maximum value of $f(x) = \sin x + \cos x$ is

If $A+B+C=\pi$ and $\cos B=\cos A \cos C$,then $\tan A \tan C=$

Let $x, y, z$ be real numbers such that $x \geq y \geq z \geq \frac{\pi}{12}$. If $x+y+z = \frac{\pi}{2}$,then the minimum value of $\cos x \cdot \sin y \cdot \cos z$ is

If $\alpha \in \left( 0, \frac{\pi}{2} \right)$,then $\sqrt{x^2 + x} + \frac{\tan^2 \alpha}{\sqrt{x^2 + x}}$ is always greater than or equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module