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

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EAMCET Engineering 2008 Solved Paper

Section: Mathematics

Question : 107

Total: 160

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