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Let f_{k}\left ( x \right )= \frac{1}{k}\left ( \sin ^{k}x+\cos ^{k}x \right )where x\epsilon R\: \: and\: \: k\geqslant 1      Then f_{4}(x)-f_{6}(x)  equals :

  • Option 1)

    \frac{1}{4}

  • Option 2)

    \frac{1}{12}

  • Option 3)

    \frac{1}{6}

  • Option 4)

    \frac{1}{3}

 

Answers (3)

best_answer

As we have learnt in

 

Double Angle Formula -

Double angle formula

- wherein

These are formulae for double angles.

 

 

\frac{1}{4}(sin^4x+cos^4x)-\frac{1}{6}(sin^6x+cos^6x)\\=\frac{1}{4}(1-2sin^2x)-\frac{1}{6}(1-3sin^2xcos^2x)=\frac{1}{12}

 


Option 1)

\frac{1}{4}

Option 2)

\frac{1}{12}

Option 3)

\frac{1}{6}

Option 4)

\frac{1}{3}

Posted by

Himanshu

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JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Use this Double Sin Formula

 

Posted by

Vakul

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Option 2

Posted by

ANKIT KUMAR

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