nice-calculus-in-AB-C-prove-sin-2-B-sin-2-C-sin-2-A-2cos-A-sin-B-sin-C-

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

\dots n i c e c a l c u l u s \dots

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

\dots . g r a t e f u l . .