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Question: The minimum value of \[{{2}^{\sin x}}+{{2}^{\cos x}}\]is: (a) \[{{2}^{1-\dfrac{1}{\sqrt{2}}}}\] ...

The minimum value of 2sin⁡x+2cos⁡x{{2}^{\sin x}}+{{2}^{\cos x}}is:
(a) 21−12{{2}^{1-\dfrac{1}{\sqrt{2}}}}
(b) 21+12{{2}^{1+\dfrac{1}{\sqrt{2}}}}
(c) 22{{2}^{\sqrt{2}}}
(d) 22

Explanation

Solution

Hint: Use Arithmetic mean\left( AM \right)$$$$\ge Geometric mean(GM)\left( GM \right)between 2sin⁡x{{2}^{\sin x}}and 2cos⁡x{{2}^{\cos x}}.

Here we have to find the minimum value of 2sin⁡x+2cos⁡x{{2}^{\sin x}}+{{2}^{\cos x}}.
We know that
Arithmetic mean ≥\ge Geometric mean
Or, AM≥GM....(i)AM\ge GM....\left( i \right)
For any two values, say aaand bb,
AM=a+b2AM=\dfrac{a+b}{2}
And GM=abGM=\sqrt{ab}
Considering a=2sin⁡xa={{2}^{\sin x}}and b=2cos⁡xb={{2}^{\cos x}}
We get AM=2sin⁡x+2cos⁡x2AM=\dfrac{{{2}^{\sin x}}+{{2}^{\cos x}}}{2}
And GM=2sin⁡x.2cos⁡xGM=\sqrt{{{2}^{\sin x}}{{.2}^{\cos x}}}
Also, am.an=am+n{{a}^{m}}.{{a}^{n}}={{a}^{m+n}}
Therefore, GM=2sin⁡x+cos⁡xGM=\sqrt{{{2}^{\sin x+\cos x}}}
By putting value of AMAMand GMGMin equation (i)\left( i \right)
We get, 2sin⁡x+2cos⁡x2≥2sin⁡x+cos⁡x\dfrac{{{2}^{\sin x}}+{{2}^{\cos x}}}{2}\ge \sqrt{{{2}^{\sin x+\cos x}}}
By cross multiplying, we get
=2sin⁡x+2cos⁡x≥2(2sin⁡x+cos⁡x)12={{2}^{\sin x}}+{{2}^{\cos x}}\ge 2{{\left( {{2}^{\sin x+\cos x}} \right)}^{\dfrac{1}{2}}}
We know that minimum value of
asin⁡x+bcos⁡x=−a2+b2a\sin x+b\cos x=-\sqrt{{{a}^{2}}+{{b}^{2}}}
Therefore, minimum value of
sin⁡x+cos⁡x=−12+12=−2\sin x+\cos x=-\sqrt{{{1}^{2}}+{{1}^{2}}}=-\sqrt{2}
Therefore, minimum value of
2sin⁡x+cos⁡x=2−2{{2}^{\sin x+\cos x}}={{2}^{-\sqrt{2}}}
Hence, 2sin⁡x+2cos⁡x≥21(2−2)12{{2}^{\sin x}}+{{2}^{\cos x}}\ge {{2}^{1}}{{\left( {{2}^{-\sqrt{2}}} \right)}^{\dfrac{1}{2}}}
=2sin⁡x+2cos⁡x≥21.2−12={{2}^{\sin x}}+{{2}^{\cos x}}\ge {{2}^{1}}{{.2}^{-\dfrac{1}{\sqrt{2}}}}
We know that am.an=am+n{{a}^{m}}.{{a}^{n}}={{a}^{m+n}}
Therefore, 2sin⁡x+2cos⁡x≥2(1−12){{2}^{\sin x}}+{{2}^{\cos x}}\ge {{2}^{\left( 1-\dfrac{1}{\sqrt{2}} \right)}}
Hence, 2sin⁡x+2cos⁡x{{2}^{\sin x}}+{{2}^{\cos x}}is always greater than or equal to 2(1−12){{2}^{\left( 1-\dfrac{1}{\sqrt{2}} \right)}}.
That means, the minimum value of 2sin⁡x+2cos⁡x{{2}^{\sin x}}+{{2}^{\cos x}}is 2(1−12){{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≥GMAM\ge GM for once and not always try to solve the question only through trigonometric
equations and functions.