First slide

Trigonometric equations

Question

$The min. value of 2^{\sin x} + 2^{\cos x} =$

Difficult

A

1
B

$2^{1 - \frac{1}{\sqrt{2}}}$
C

$2^{1 / 3}$
D

0

Solution

We have
$\begin{array}{l} \frac{2^{\sin x} + 2^{\cos x}}{2} \geq \sqrt{2^{\sin x} \cdot 2^{\cos x}} (∵ A M \geq G M) \\ \geq 2^{\frac{\sin x + \cos x}{2}} \\ \geq 2^{\frac{- \sqrt{2}}{2}} \\ \geq 2^{- \frac{1}{\sqrt{2}}} \\ \Rightarrow 2^{\sin x} + 2^{\cos x} \geq 2 . 2^{\frac{- 1}{\sqrt{2}}} \\ \geq 2^{1 - \frac{1}{\sqrt{2}}} \end{array}$

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