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Trigonometric Ratio and Identites question

2021 · 27 Jul · Shift 1 · Q30
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  5. /2021 · 27 Jul · Shift 1 · Q30

Trigonometric Ratio and Identites question

2021 · 27 Jul · Shift 1 · Q30

JEE MainMathematicsTrigonometric Ratio and IdentitesMCQ+4 / −1
If sin⁡θ+cos⁡θ=12\sin \theta + \cos \theta = {1 \over 2}sinθ+cosθ=21​, then 16(sin(2 θ\thetaθ) + cos(4 θ\thetaθ) + sin(6 θ\thetaθ)) is equal to :
  1. A
    23
  2. B
    −-− 27
  3. C
    −-− 23
  4. D
    27
View written solutionFree

Correct answer: C

  1. Let s=sin⁡θ+cos⁡θ=12.s=\sin\theta+\cos\theta=\frac12.s=sinθ+cosθ=21​.

  2. Use the identity (sin⁡θ+cos⁡θ)2=sin⁡2θ+cos⁡2θ+2sin⁡θcos⁡θ=1+sin⁡2θ.(\sin\theta+\cos\theta)^2=\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=1+\sin 2\theta.(sinθ+cosθ)2=sin2θ+cos2θ+2sinθcosθ=1+sin2θ. Hence, (12)2=1+sin⁡2θ\left(\frac12\right)^2=1+\sin 2\theta(21​)2=1+sin2θ 14=1+sin⁡2θ\frac14=1+\sin 2\theta41​=1+sin2θ sin⁡2θ=−34.\sin 2\theta=-\frac34.sin2θ=−43​.

  3. Now find cos⁡4θ\cos 4\thetacos4θ using cos⁡4θ=1−2sin⁡22θ.\cos 4\theta=1-2\sin^2 2\theta.cos4θ=1−2sin22θ. Since sin⁡2θ=−34\sin 2\theta=-\frac34sin2θ=−43​, sin⁡22θ=916.\sin^2 2\theta=\frac{9}{16}.sin22θ=169​. Therefore, cos⁡4θ=1−2⋅916=1−1816=1−98=−18.\cos 4\theta=1-2\cdot \frac{9}{16}=1-\frac{18}{16}=1-\frac98=-\frac18.cos4θ=1−2⋅169​=1−1618​=1−89​=−81​.

  4. Next, find sin⁡6θ\sin 6\thetasin6θ. Write sin⁡6θ=sin⁡(2θ+4θ)=sin⁡2θcos⁡4θ+cos⁡2θsin⁡4θ.\sin 6\theta=\sin(2\theta+4\theta)=\sin 2\theta\cos 4\theta+\cos 2\theta\sin 4\theta.sin6θ=sin(2θ+4θ)=sin2θcos4θ+cos2θsin4θ. But this is lengthy. A better identity is: sin⁡3x=3sin⁡x−4sin⁡3x.\sin 3x=3\sin x-4\sin^3 x.sin3x=3sinx−4sin3x. Put x=2θx=2\thetax=2θ: sin⁡6θ=3sin⁡2θ−4sin⁡32θ.\sin 6\theta=3\sin 2\theta-4\sin^3 2\theta.sin6θ=3sin2θ−4sin32θ. With sin⁡2θ=−34\sin 2\theta=-\frac34sin2θ=−43​, sin⁡6θ=3(−34)−4(−34)3.\sin 6\theta=3\left(-\frac34\right)-4\left(-\frac34\right)^3.sin6θ=3(−43​)−4(−43​)3. Compute: 3(−34)=−94,3\left(-\frac34\right)=-\frac94,3(−43​)=−49​, (−34)3=−2764,\left(-\frac34\right)^3=-\frac{27}{64},(−43​)3=−6427​, −4(−2764)=+2716.-4\left(-\frac{27}{64}\right)=+\frac{27}{16}.−4(−6427​)=+1627​. So, sin⁡6θ=−94+2716=−3616+2716=−916.\sin 6\theta=-\frac94+\frac{27}{16}=-\frac{36}{16}+\frac{27}{16}=-\frac{9}{16}. sin6θ=−49​+1627​=−1636​+1627​=−169​.

  5. Now evaluate sin⁡2θ+cos⁡4θ+sin⁡6θ.\sin 2\theta+\cos 4\theta+\sin 6\theta.sin2θ+cos4θ+sin6θ. Substitute the values: =−34−18−916.=-\frac34-\frac18-\frac{9}{16}.=−43​−81​−169​. Take LCM 161616: =−1216−216−916=−2316.=-\frac{12}{16}-\frac{2}{16}-\frac{9}{16}=-\frac{23}{16}. =−1612​−162​−169​=−1623​.

  6. Therefore, 16(sin⁡2θ+cos⁡4θ+sin⁡6θ)=16(−2316)=−23.16(\sin 2\theta+\cos 4\theta+\sin 6\theta)=16\left(-\frac{23}{16}\right)=-23.16(sin2θ+cos4θ+sin6θ)=16(−1623​)=−23.

  7. Hence the correct option is C (−23).\boxed{\text{C }(-23)}.C (−23)​.

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