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) $$2$$

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Vedantu Content Team · Accepted Answer

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 meanOr, $$AM\ge GM....\left( i \right)$$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.

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) \[2\]

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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) \[2\]