find minimum value of $2^{\sin^2(\theta)}+2^{\cos^2(\theta)}$

Question

find minimum value of $2^{\sin^2(\theta)}+2^{\cos^2(\theta)}$

I have found the minimum value using derivative method :

Let $f(\theta)=2^{\sin^2(\theta)}+2^{\cos^2(\theta)}$. Then calculate $f'(\theta)$ and $f''(\theta)$.

Is it possible to find minimum value by alternative process without using the concept of derivative?

Answer 1

HINT:

For real $a>0,$

$$(a^2)^{\sin^2\theta}+(a^2)^{\cos^2\theta}=a\left(a^{-\cos2\theta}+a^{\cos2\theta}\right)$$

Now $\dfrac{a^{-\cos2\theta}+a^{\cos2\theta}}2\ge\sqrt{a^{-\cos2\theta}\cdot a^{\cos2\theta}}=1$

Can you identify $a$ here?

Answer 2

For $a>0$

$$\dfrac{a^{\sin^2(\theta)}+a^{\cos^2(\theta)}}2\ge\sqrt{a^{\sin^2(\theta)}\cdot a^{\cos^2(\theta)}}=\sqrt{a^{\sin^2(\theta)+\cos^2(\theta)}}=\sqrt a$$

The equality occurs if $a^{\sin^2(\theta)}=a^{\cos^2(\theta)}$

Answer 3

If $a,b\ge 0$ then $$a+b\ge 2\sqrt{ab}$$ $$2^{\sin^2\theta}+2^{\cos^2\theta}\ge 2\sqrt{2^{\sin^2\theta+\cos^2\theta}}=2\sqrt{2}$$

Answer 4

Just use AM-GM,

$\frac{2^{\sin^2{\theta}}+2^{\cos^2{\theta}}}{2}$

$\geq \sqrt {2^{\sin^2{\theta}+\cos^2{\theta}}}$

$=\sqrt{2}$

$\implies 2^{\sin^2{\theta}}+2^{\cos^2{\theta}}\ge 2\sqrt{2}$

Here equality holds at $\theta=\frac{n\pi}{2}+\frac{\pi}{4}$