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Trigonometric Ratio and Identites question

2009 · Shift 0 · Q45
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  5. /2009 · Shift 0 · Q45

Trigonometric Ratio and Identites question

2009 · Shift 0 · Q45

JEE MainMathematicsTrigonometric Ratio and IdentitesMCQ+4 / −1
Let A and B denote the statements A: cos⁡α+cos⁡β+cos⁡γ=0\cos \alpha + \cos \beta + \cos \gamma = 0cosα+cosβ+cosγ=0 B: sin⁡α+sin⁡β+sin⁡γ=0\sin \alpha + \sin \beta + \sin \gamma = 0sinα+sinβ+sinγ=0 If cos⁡(β−γ)+cos⁡(γ−α)+cos⁡(α−β)=−32,\cos \left( {\beta - \gamma } \right) + \cos \left( {\gamma - \alpha } \right) + \cos \left( {\alpha - \beta } \right) = - {3 \over 2},cos(β−γ)+cos(γ−α)+cos(α−β)=−23​, then:
  1. A
    A is false and B is true
  2. B
    both A and B are true
  3. C
    both A and B are false
  4. D
    A is true and B is false
View written solutionFree

Correct answer: B

  1. Let x=cos⁡α+cos⁡β+cos⁡γ,y=sin⁡α+sin⁡β+sin⁡γ.x=\cos\alpha+\cos\beta+\cos\gamma,\qquad y=\sin\alpha+\sin\beta+\sin\gamma.x=cosα+cosβ+cosγ,y=sinα+sinβ+sinγ. Then x2+y2=(cos⁡α+cos⁡β+cos⁡γ)2+(sin⁡α+sin⁡β+sin⁡γ)2.x^2+y^2=(\cos\alpha+\cos\beta+\cos\gamma)^2+(\sin\alpha+\sin\beta+\sin\gamma)^2.x2+y2=(cosα+cosβ+cosγ)2+(sinα+sinβ+sinγ)2.

  2. Expand: \begin{align*} x^2+y^2 &= \sum (\cos^2\theta+\sin^2\theta) +2\big(\cos\alpha\cos\beta+\sin\alpha\sin\beta\big) \ &\quad +2\big(\cos\beta\cos\gamma+\sin\beta\sin\gamma\big) +2\big(\cos\gamma\cos\alpha+\sin\gamma\sin\alpha\big). \end{align*} Since cos⁡2θ+sin⁡2θ=1\cos^2\theta+\sin^2\theta=1cos2θ+sin2θ=1 and cos⁡ucos⁡v+sin⁡usin⁡v=cos⁡(u−v),\cos u\cos v+\sin u\sin v=\cos(u-v),cosucosv+sinusinv=cos(u−v), we get x2+y2=3+2[cos⁡(α−β)+cos⁡(β−γ)+cos⁡(γ−α)].x^2+y^2=3+2\big[\cos(\alpha-\beta)+\cos(\beta-\gamma)+\cos(\gamma-\alpha)\big].x2+y2=3+2[cos(α−β)+cos(β−γ)+cos(γ−α)].

  3. Given cos⁡(β−γ)+cos⁡(γ−α)+cos⁡(α−β)=−32,\cos(\beta-\gamma)+\cos(\gamma-\alpha)+\cos(\alpha-\beta)=-\frac32,cos(β−γ)+cos(γ−α)+cos(α−β)=−23​, so x2+y2=3+2(−32)=3−3=0.x^2+y^2=3+2\left(-\frac32\right)=3-3=0.x2+y2=3+2(−23​)=3−3=0.

  4. A sum of two squares is zero only when both squares are zero. Hence x=0andy=0.x=0\quad\text{and}\quad y=0.x=0andy=0. Therefore, cos⁡α+cos⁡β+cos⁡γ=0\cos\alpha+\cos\beta+\cos\gamma=0cosα+cosβ+cosγ=0 and sin⁡α+sin⁡β+sin⁡γ=0.\sin\alpha+\sin\beta+\sin\gamma=0.sinα+sinβ+sinγ=0.

  5. So statement AAA is true and statement BBB is also true.

  6. Therefore the correct option is B: both A and B are true.\boxed{\text{B: both A and B are true}}.B: both A and B are true​.

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