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2\sin ^{2}B+ 4\cos (A+B) \sin A \sin B+ \cos 2 (A+B)

  • Option 1)

    \sin 2A

  • Option 2)

    \cos 2B

  • Option 3)

    \cos 2A

  • Option 4)

    \sin 2B

 

Answers (1)

best_answer

As we learnt in 

Double Angle Formula -

Double angle formula

- wherein

These are formulae for double angles.

 

 2\sin ^{2}B+ 4\cos \left ( A+B \right )\sin A\sin B+\cos 2\left ( A+B \right )

= 1 - \cos 2B+4\cos \left ( A+B \right )\sin A\sin B+2\cos ^{2}\left ( A+B \right )-1

= -\cos 2B+2\cos ^{2}\left ( A+B \right )+4\cos \left ( A+B \right )\sin A\sin B

= - \cos 2B+2\cos \left ( A+B \right )\left [ \cos \left ( A+B \right ) +2\sin A\sin B\right ]

= - \cos 2B+2\cos \left ( A+B \right )\left [ \cos \left ( A+B \right ) +\cos \left ( A-B \right ) - \cos \left ( A+B \right ) ]

= - \cos 2B+ \cos 2A+\cos 2B

= \cos 2A


Option 1)

\sin 2A

This option is incorrect

Option 2)

\cos 2B

This option is incorrect

Option 3)

\cos 2A

This option is correct

Option 4)

\sin 2B

This option is incorrect

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Plabita

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