# 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) \[2\]

Last updated date: 19th Mar 2023

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Answer

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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.

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.

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