If \cos \alpha+\cos \beta+\cos \gamma=0= \sin \alpha+ \sin \beta+ \sin \gamma, then \sin 2 \alpha+ \sin 2 \beta+ \sin 2 \gamma=

Correct Answer is : cos(α+β)+cos(β+γ)+cos(γ+α)

\cos (\alpha+\beta)+\cos (\beta+\gamma)+\cos (\gamma+\alpha)

\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma

\sin ^2 \alpha+ \sin ^2 \beta+ \sin ^2 \gamma

\cos (2 \alpha-\beta-\gamma)+\cos (2 \beta-\gamma-\alpha)+\cos (2 \gamma-\alpha-\beta)

TG EAMCET 3 May 2025 Shift 1 Paper

Question : 10 of 160

Marks: +1, -0

If cos‌α+cos‌β+cos‌γ=0=sin‌α+sin‌β+sin‌γ, then sin‌2α+sin‌2β+sin‌2γ=

Go to Question:

Prev Question

Next Question