AP EAMCET · Maths · Trigonometric Equations
If \(A+B+C=270^{\circ}\), then \(\cos 2 A+\cos 2 B+\cos 2 C\) is equal to :
- A \(4 \sin A \sin B \sin C\)
- B \(4 \cos A \cos B \cos C\)
- C \(1-4 \sin A \sin B \sin C\)
- D \(1-4 \cos A \cos B \cos C\)
Answer & Solution
Correct Answer
(C) \(1-4 \sin A \sin B \sin C\)
Step-by-step Solution
Detailed explanation
\(\cos 2 A+\cos 2 B+\cos 2 C\) \(=2 \cos (A+B) \cos (A-B)+1-2 \sin ^2 C\) \(=2 \cos \left(\frac{3 \pi}{2}-C\right) \cos (A-B)+1-2 \sin ^2 C\) \(\left[\because A+B+C=270^{\circ} \Rightarrow B+C=\frac{3 \pi}{2}-C\right\rfloor\) \(=1-2 \sin C[\cos (A-B)+\sin C]\)…
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