Verified Solution • 2025–26 Session moderate

QJEEM21Ma06eb: JEE Main 2021 Mathematics - Class 11 JEEM21Ma06eb Solution: If , then 16(sin(2 ) + cos(4 ) + sin(6 )) is equal to :

If sin⁡θ+cos⁡θ=12\sin \theta + \cos \theta = {1 \over 2} , then 16(sin(2 θ\theta ) + cos(4 θ\theta ) + sin(6 θ\theta )) is equal to :

Multiple Choice Options

  • A 23
  • B −- 27
  • C −- 23
  • D 27

Step-by-Step Verified Solution

sin⁡θ+cos⁡θ=12\sin \theta + \cos \theta = {1 \over 2}

sin⁡2θ+cos⁡2θ+2sin⁡θcos⁡θ=14{\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta = {1 \over 4}

sin⁡2θ=−34\sin 2\theta = - {3 \over 4}

Now :

cos⁡4θ=1−2sin⁡22θ\cos 4\theta = 1 - 2{\sin ^2}2\theta

=1−2(−34)2= 1 - 2{\left( { - {3 \over 4}} \right)^2}

=1−2×916=−18= 1 - 2 \times {9 \over {16}} = - {1 \over 8}

sin⁡6θ=3sin⁡2θ−4sin⁡32θ\sin 6\theta = 3\sin 2\theta - 4{\sin ^3}2\theta

=(3−4sin⁡22θ).sin⁡2θ= (3 - 4{\sin ^2}2\theta ).\sin 2\theta

=[3−4(916)].(−34)= \left[ {3 - 4\left( {{9 \over {16}}} \right)} \right].\left( { - {3 \over 4}} \right)

⇒[34]×(−34)=−916\Rightarrow \left[ {{3 \over 4}} \right] \times \left( { - {3 \over 4}} \right) = - {9 \over {16}}

16[sin⁡2θ+cos⁡4θ+sin⁡6θ]16[\sin 2\theta + \cos 4\theta + \sin 6\theta ]

= 16(−34−18−916)=−2316\left( { - {3 \over 4} - {1 \over 8} - {9 \over {16}}} \right) = - 23

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