Prove that-i sin-sin-sin-sin -4 sin- 2 sin- 2 sin- 2ii cos A-B-C-cos A-B-C-cos A-B-C-cos -A-B-C-4 cos A cos B cos C

(i)Consider #160;LHS: #160;sin #160; #945; #160;+ #160;sin #160; #946; #160;+ #160;sin #160; #947; #160; #160;sin #160;( #945; ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Compound Angles

Prove that:i ...

Question

Prove that:
(i) $\sin α + \sin β + \sin γ - \sin (α + β + γ) = 4 \sin (\frac{α + β}{2}) \sin (\frac{β + γ}{2}) \sin (\frac{γ + α}{2})$

(ii) cos (A + B + C) + cos (A − B + C) + cos (A + B − C) + cos (− A + B + C) = 4 cos A cos B cos C

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Solution

$(i) Consider LHS : \sin α + \sin β + \sin γ - \sin (α + β + γ) = 2 si n (\frac{α + β}{2}) \cos (\frac{α - β}{2}) + 2 \cos (\frac{γ + α + β + γ}{2}) \sin (\frac{γ - α - β - γ}{2}) = 2 \sin (\frac{α + β}{2}) \cos (\frac{α - β}{2}) + 2 \cos (\frac{2 γ + α + β}{2}) \sin (\frac{- α - β}{2}) = 2 \sin (\frac{α + β}{2}) \cos (\frac{α - β}{2}) + 2 \cos (\frac{2 γ + α + β}{2}) \sin [- (\frac{α + β}{2})] = 2 \sin (\frac{α + β}{2}) [\cos (\frac{α - β}{2}) - \cos (\frac{2 γ + α + β}{2})] = 2 \sin (\frac{α + β}{2}) [- 2 \sin (\frac{α - β + 2 γ + α + β}{4}) \sin (\frac{α - β - 2 γ - α - β}{4})] = 2 \sin (\frac{α + β}{2}) [- 2 \sin (\frac{α + γ}{2}) \sin (\frac{- β - γ}{2})] = 2 \sin (\frac{α + β}{2}) [2 \sin (\frac{α + γ}{2}) s in (\frac{β + γ}{2})] = 4 \sin (\frac{α + β}{2}) \sin (\frac{α + γ}{2}) \sin (\frac{β + γ}{2}) = RHS Hence, LHS = RHS .$

$(ii) Consider LHS : \cos (A + B + C) + \cos (A - B + C) + \cos (A + B - C) + \cos (- A + B + C) = 2 \cos (\frac{A + B + C + A - B + C}{2}) \cos (\frac{A + B + C - A + B - C}{2}) + 2 \cos (\frac{A + B - C - A + B + C}{2}) \cos (\frac{A + B - C + A - B - C}{2}) = 2 \cos (A + C) \cos B + 2 \cos B \cos (A - C) = 2 \cos B [\cos (A + C) + \cos (A - C)] = 2 \cos B [2 \cos (\frac{A + C + A - C}{2}) \cos (\frac{A + C - A + C}{2})] = 2 \cos B [2 \cos A \cos C] = 4 \cos A \cos B \cos C = RHS Hence, LHS = RHS .$

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Similar questions

Q. Prove that:
(i)

\frac{\cos (A + B + C) + \cos (- A + B + C) + \cos (A - B + C) + \cos (A + B - C)}{\sin (A + B + C) + \sin (- A + B + C) + \sin (A - B + C) - \sin (A + B - C)} = \cot C

(ii) sin (B − C) cos (A − D) + sin (C − A) cos (B − D) + sin (A − B) cos (C − D) = 0

Prove that:
(i) $\frac{c o s (A + B + C) + c o s (- A + B + C) + c o s (A - B + C) + c o s (A + B - C)}{s i n c o s (A + B + C) + s i n c o s (- A + B + C) + s i n c o s (A - B + C) - s i n (c o s (A + B - C))}$
(ii) $s i n (B - C) c o s (A - D) + s i n (C - A) c o s (B - D) + s i n (A - B) c o s (C - D) = 0$