Prove that-i sin-sin-sin-sin -4 sin-2 sin-Y-2 siny-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) sinα +sinβ +sinγ sin(α +β +γ )=4sin ( α +β/2 )sin ( β +γ/2 )sin ( γ +α/2 ) We have, LHS=sinα +sinβ +sin γ sin(α +β +γ ) =(sin α +sinβ )+[sinγ sin(α +β + ...

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Byju's Answer

Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

Prove that:i ...

Question

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

(ii) $c o s (A + B + C) + c o s (A - B + C) + c o s (A + B - C) + c o s (- A + B + C) = 4 c o s A c o s B c o s C$

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Solution

(i) $s i n α + s i n β + s i n γ - s i n (α + β + γ) = 4 s i n (\frac{α + β}{2}) s i n (\frac{β + γ}{2}) s i n (\frac{γ + α}{2})$
We have,
$L H S = s i n α + s i n β + s i n γ - s i n (α + β + γ) = (s i n α + s i n β) + [s i n γ - s i n (α + β + γ)] = 2 s i n (\frac{α + β}{2}) c o s (\frac{α + β}{2}) + 2 s i n (\frac{γ - (α + β + γ)}{2}) c o s (\frac{γ + (α + β + γ)}{2}) = 2 s i n (\frac{α + β}{2}) c o s (\frac{α - β}{2}) + 2 s i n (\frac{- α - β}{2}) c o s (\frac{α + β + 2 γ}{2}) = 2 s i n (\frac{α + β}{2}) c o s (\frac{α - β}{2}) - 2 s i n (\frac{α + β}{2}) c o s (\frac{α + β + 2 γ}{2}) = 2 s i n (\frac{α + β}{2}) [c o s (\frac{α - β}{2}) - c o s (\frac{α + β + 2 γ}{2})] = 2 s i n (\frac{α + β}{2}) ⎡ ⎢ ⎣ - 2 s i n \frac{[\frac{α - β}{2} +] s i n [\frac{α + β + 2 γ}{2}]}{2} ⎤ ⎥ ⎦ = 2 s i n (\frac{α + β}{2}) [- 2 s i n [\frac{1 α + 2 γ}{2 \times 2}] s i n [\frac{- 2 β - 2 γ}{2 \times 2}]] = - 4 (\frac{α + β}{2}) [s i n (\frac{α + γ}{2}) s i n [\frac{- (β + γ)}{2}]] = 4 s i n (\frac{α + β}{2}) s i n (\frac{α + γ}{2}) s i n (\frac{β + γ}{2}) = 4 s i n (\frac{α + β}{2}) s i n (\frac{β + γ}{2}) s i n (\frac{α + γ}{2}) = R H S ∴ s i n α + s i n β + s i n γ - s i n (α + β + γ) = 4 s i n (\frac{α + β}{2}) s i n (\frac{β + γ}{2}) s i n (\frac{γ + α}{2})$

(ii) $c o s (A + B + C) + c o s (A - B + C) + c o s (A + B - C) + c o s (- A + B + C) = 4 c o s A c o s B c o s C$
We have,
$L H S = c o s (A + B + C) + c o s (A - B + C) + c o s (A + B - C) + c o s (- A + B + C) = 2 c o s {\frac{A + B + C + A - B + C}{2}} c o s {\frac{A + B + C + A + B - C}{2}} + 2 ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} c o s {\frac{A + B - C - A - B + C}{2}} c o s {\frac{A + B - C + A - B - C}{2}} \end{matrix} ⎫ ⎪ ⎬ ⎪ ⎭ = 2 c o s {\frac{2 A + 2 C}{2}} c o s {\frac{2 B}{2}} + 2 c o s {\frac{2 B}{2}} c o s {\frac{2 A - 2 C}{2}} = 2 c o s (A + C) c o s (B) + 2 c o s (B) c o s (A - C) = 2 c o s (B) [c o s (A + C) + c o s (A - C)] = 2 c o s (B) [2 c o s (\frac{A + C + A - C}{2}) c o s (\frac{A + C - A + C}{2})] = 2 c o s (B) [2 c o s A c o s C] = 4 c o s A c o s B c o s C$

Suggest Corrections

Prove that: (i) sinα+sinβ+sinγ−sin(α+β+γ)=4sin(α+β2)sin(β+γ2)sin(γ+α2) (ii) cos(A+B+C)+cos(A−B+C)+cos(A+B−C)+cos(−A+B+C)=4cosA cosB cosC

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

(ii) $c o s (A + B + C) + c o s (A - B + C) + c o s (A + B - C) + c o s (- A + B + C) = 4 c o s A c o s B c o s C$