AP EAMCET · Maths · Trigonometric Ratios & Identities
Let \(\alpha, \beta\) be two real number such that \(\pi < (\alpha-\beta) < 3 \pi\). If \(\sin \alpha+\sin \beta=\frac{-21}{65}\) and \(\cos \alpha+\cos \beta=\frac{-2}{65}\), then \(\cos \left(\frac{\beta-\alpha}{2}\right)=\)
- A \(\frac{-\sqrt{89}}{26 \sqrt{5}}\)
- B \(\frac{-\sqrt{8}}{26 \sqrt{5}}\)
- C \(\frac{-\sqrt{91}}{26 \sqrt{5}}\)
- D \(\frac{-\sqrt{72}}{26 \sqrt{5}}\)
Answer & Solution
Correct Answer
(A) \(\frac{-\sqrt{89}}{26 \sqrt{5}}\)
Step-by-step Solution
Detailed explanation
\((\sin \alpha+\sin \beta)^2=\left(\frac{-21}{65}\right)^2\) \(\sin ^2 \alpha+\sin ^2 \beta+2 \sin \alpha \sin \beta=\frac{441}{(65)^2}\) ...(i) \((\cos \alpha+\cos \beta)^2=\left(\frac{-2}{65}\right)^2\) \(\cos ^2 \alpha+\cos ^2 \beta+2 \cos \alpha \cos \beta=\frac{4}{(65)^2}\)…
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