Courses
Courses for Kids
Free study material
Free LIVE classes
More LIVE
Join Vedantu’s FREE Mastercalss

# The minimum value of ${{2}^{\sin x}}+{{2}^{\cos x}}$is:(a) ${{2}^{1-\dfrac{1}{\sqrt{2}}}}$(b) ${{2}^{1+\dfrac{1}{\sqrt{2}}}}$(c) ${{2}^{\sqrt{2}}}$(d) $2$ Verified
365.1k+ views
Hint: Use Arithmetic mean$\left( AM \right)$$\ge$Geometric mean$\left( GM \right)$between ${{2}^{\sin x}}$and ${{2}^{\cos x}}$.

Here we have to find the minimum value of ${{2}^{\sin x}}+{{2}^{\cos x}}$.
We know that
Arithmetic mean $\ge$ Geometric mean
Or, $AM\ge GM....\left( i \right)$
For any two values, say $a$and $b$,
$AM=\dfrac{a+b}{2}$
And $GM=\sqrt{ab}$
Considering $a={{2}^{\sin x}}$and $b={{2}^{\cos x}}$
We get $AM=\dfrac{{{2}^{\sin x}}+{{2}^{\cos x}}}{2}$
And $GM=\sqrt{{{2}^{\sin x}}{{.2}^{\cos x}}}$
Also, ${{a}^{m}}.{{a}^{n}}={{a}^{m+n}}$
Therefore, $GM=\sqrt{{{2}^{\sin x+\cos x}}}$
By putting value of $AM$and $GM$in equation $\left( i \right)$
We get, $\dfrac{{{2}^{\sin x}}+{{2}^{\cos x}}}{2}\ge \sqrt{{{2}^{\sin x+\cos x}}}$
By cross multiplying, we get
$={{2}^{\sin x}}+{{2}^{\cos x}}\ge 2{{\left( {{2}^{\sin x+\cos x}} \right)}^{\dfrac{1}{2}}}$
We know that minimum value of
$a\sin x+b\cos x=-\sqrt{{{a}^{2}}+{{b}^{2}}}$
Therefore, minimum value of
$\sin x+\cos x=-\sqrt{{{1}^{2}}+{{1}^{2}}}=-\sqrt{2}$
Therefore, minimum value of
${{2}^{\sin x+\cos x}}={{2}^{-\sqrt{2}}}$
Hence, ${{2}^{\sin x}}+{{2}^{\cos x}}\ge {{2}^{1}}{{\left( {{2}^{-\sqrt{2}}} \right)}^{\dfrac{1}{2}}}$
$={{2}^{\sin x}}+{{2}^{\cos x}}\ge {{2}^{1}}{{.2}^{-\dfrac{1}{\sqrt{2}}}}$
We know that ${{a}^{m}}.{{a}^{n}}={{a}^{m+n}}$
Therefore, ${{2}^{\sin x}}+{{2}^{\cos x}}\ge {{2}^{\left( 1-\dfrac{1}{\sqrt{2}} \right)}}$
Hence, ${{2}^{\sin x}}+{{2}^{\cos x}}$is always greater than or equal to ${{2}^{\left( 1-\dfrac{1}{\sqrt{2}} \right)}}$.
That means, the minimum value of ${{2}^{\sin x}}+{{2}^{\cos x}}$is ${{2}^{\left( 1-\dfrac{1}{\sqrt{2}} \right)}}$.
Therefore, option (a) is correct.

Note: In questions involving maxima and minima in trigonometry, students must try to use the approach
of $AM\ge GM$ for once and not always try to solve the question only through trigonometric
equations and functions.
Last updated date: 30th Sep 2023
Total views: 365.1k
Views today: 8.65k