... nice calculus ... in AB^Δ C prove : sin^2 (B)+sin^2 (C)−sin^2 (A)=2cos(A)sin(B)sin(C)


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Question Number 125389 by mnjuly1970 last updated on 10/Dec/20

$. . . n i c e c a l c u l u s . . .$ $i n A \overset{Δ}{B C} p r o v e :$ ${s i n}^{2} (B) + {s i n}^{2} (C) - {s i n}^{2} (A) = 2 c o s (A) s i n (B) s i n (C)$
	Answered by mnjuly1970 last updated on 10/Dec/20

	$l . h . s :: {s i n}^{2} (B) + {s i n}^{2} (C) - {s i n}^{2} (A) = \frac{b^{2} + c^{2} - a^{2}}{4 R^{2}} = \frac{2 b c c o s (A)}{4 R^{2}}$ $= \frac{2 (2 R s i n (B) . 2 R s i n (C) c o s (A)}{4 R^{2}}$ $= 2 s i n (B) s i n (C) c o s (A) ✓$
	Answered by Dwaipayan Shikari last updated on 10/Dec/20

	$2 {s i n}^{2} (B) + 2 {s i n}^{2} (C) - 2 {s i n}^{2} (A) = 2 y$ $1 - c o s 2 B + 1 - c o s 2 C - 1 + c o s 2 A = 2 y$ $- 2 c o s (B + C) c o s (B - C) + c o s 2 A = 2 y - 1$ $2 c o s (A) c o s (B - C) + 2 {c o s}^{2} A = 2 y$ $y = c o s A (c o s A + c o s (B - C)) = 2 c o s A c o s (\frac{A + B - C}{2}) c o s (\frac{A - B + C}{2})$ $= 2 c o s A c o s (\frac{π - 2 C}{2}) c o s (\frac{π - 2 B}{2}) = 2 c o s A s i n B s i n C$
	Commented by mnjuly1970 last updated on 10/Dec/20

	$t h a n k y o u s o m u c h$ $h a r d - w o r k i n g a n d p o w e r f u l$ $. . . . g r a t e f u l . .$
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