TS EAMCET · Maths · Trigonometric Ratios & Identities
If \(\cos \alpha+\cos \beta+\cos \gamma=0=\sin \alpha+\sin \beta+\sin \gamma\), then \(\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma=\)
- A \(\cos (\alpha+\beta)+\cos (\beta+\gamma)+\cos (\gamma+\alpha)\)
- B \(\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma\)
- C \(\sin ^2 \alpha+\sin ^2 \beta+\sin ^2 \gamma\)
- D \(\cos (2 \alpha-\beta-\gamma)+\cos (2 \beta-\gamma-\alpha)+\cos (2 \gamma-\alpha-\beta)\)
Answer & Solution
Correct Answer
(A) \(\cos (\alpha+\beta)+\cos (\beta+\gamma)+\cos (\gamma+\alpha)\)
Step-by-step Solution
Detailed explanation
Let \(a=e^{i\alpha}, b=e^{i\beta}, c=e^{i\gamma}\). \(a+b+c = (\cos \alpha+\cos \beta+\cos \gamma) + i(\sin \alpha+\sin \beta+\sin \gamma) = 0+i \cdot 0 = 0\). \(\bar{a}+\bar{b}+\bar{c} = 0 \Rightarrow 1/a+1/b+1/c=0 \Rightarrow ab+bc+ca=0\).…
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