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

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?

• The answer is $2\sqrt{2}$ – COOLGUY Jul 21 '16 at 8:33
• Yes. Use the fact that $\sin^2(\theta)$ and $\cos^2(\theta)$ occupy all possible pairs of values between $[0,1]$ that sum to $1$. This makes the problem equivalent to minimizing $2^x + 2^{1-x}$ for $x \in [0,1]$. – Erick Wong Jul 21 '16 at 14:41

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?

• Is min value for $a=2$, $2\sqrt{2}$ – MKS Jul 21 '16 at 8:26
• @MdKutubuddinSardar, True – lab bhattacharjee Jul 21 '16 at 8:29
• Nice .......(+1) – Behrouz Maleki Jul 21 '16 at 8:30
• @LB, THANKS +1 for u – MKS Jul 21 '16 at 8:30

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

• I suppose using derivative to find max/min would be useless in this case. – StubbornAtom Jul 21 '16 at 8:34

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 $a^{\sin^2(\theta)}=a^{\cos^2(\theta)}$

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 $\theta=\frac{n\pi}{2}+\frac{\pi}{4}$