2^{sin theta} + 2^{cos theta} is greater than | vedclass

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

Question

(D) Using the Arithmetic Mean-Geometric Mean inequality $(AM \ge GM)$:$\frac{2^{\sin 	heta} + 2^{\cos 	heta}}{2} \ge \sqrt{2^{\sin 	heta} \cdot 2^{

Vedclass · Accepted Answer

$2^{(1 - \frac{1}{\sqrt{2}})}$

Vedclass · Answer

(D) Using the Arithmetic Mean-Geometric Mean inequality $(AM \ge GM)$:$\frac{2^{\sin 	heta} + 2^{\cos 	heta}}{2} \ge \sqrt{2^{\sin 	heta} \cdot 2^{\cos 	heta}}$$2^{\sin 	heta} + 2^{\cos 	heta} \ge 2 \cdot 2^{\frac{\sin 	heta + \cos 	heta}{2}}$We know that $\sin 	heta + \cos 	heta = \sqrt{2} \sin(	heta + \frac{\pi}{4})$.The minimum value of $\sin(	heta + \frac{\pi}{4})$ is $-1$,so the minimum value of $\sin 	heta + \cos 	heta$ is $-\sqrt{2}$.Thus,$2^{\sin 	heta} + 2^{\cos 	heta} \ge 2 \cdot 2^{\frac{-\sqrt{2}}{2}} = 2^1 \cdot 2^{-\frac{1}{\sqrt{2}}} = 2^{(1 - \frac{1}{\sqrt{2}})}$.

$2^{\sin \theta} + 2^{\cos \theta}$ is greater than

Similar Questions

If the sum of $n$ terms of a $G.P.$ is $255$,the $n^{th}$ term is $128$,and the common ratio is $2$,then the first term will be:

If ${a_1}, {a_2}, {a_3}, \dots, {a_n}$ are in $H.P.$,then ${a_1}{a_2} + {a_2}{a_3} + \dots + {a_{n-1}}{a_n}$ is equal to:

The sum of the squares of the first $n$ natural numbers exceeds their sum by $330$. Then,$n = $

If the sum of the series $2 + 5 + 8 + 11 + \dots$ is $60100$, then the number of terms is:

The sum of the first $20$ terms of the series $1 + \frac{3}{2} + \frac{7}{4} + \frac{15}{8} + \frac{31}{16} + \dots$ is?

Vedclass Test Series

Exam Paper Generator

Online Exam Module