If \alpha+\beta+\gamma=2 \theta, then \cos \theta+\cos (\theta-\alpha) +\cos (\theta-\beta)+\cos (\theta-\gamma) is equal to

Correct Answer is : 4cosα2⋅cosβ2⋅cosγ2

4 \sin \frac{\alpha}{2} \cdot \cos \frac{\beta}{2} \cdot \sin \frac{\gamma}{2}

4 \cos \frac{\alpha}{2} \cdot \cos \frac{\beta}{2} \cdot \cos \frac{\gamma}{2}

4 \sin \frac{\alpha}{2} \cdot \sin \frac{\beta}{2} \cdot \sin \frac{\gamma}{2}

4 \sin \alpha \cdot \sin \beta \cdot \sin \gamma

EAMCET Engineering 2008 Solved Paper

Section: Mathematics

Question : 107 of 160

Marks: +1, -0

If α+β+γ=2θ, then cos‌θ+cos(θ−α) +cos(θ−β)+cos(θ−γ) is equal to

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