....nice calculus.. in AB^Δ C : sin^2 (A)+sin^2 (B)+sin^2 (C)=2 prove that: AB^Δ C is right triangle Good luck.


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Question Number 124501 by mnjuly1970 last updated on 03/Dec/20

$. . . . n i c e c a l c u l u s . .$ $i n A \overset{Δ}{B C} : {s i n}^{2} (A) + {s i n}^{2} (B) + {s i n}^{2} (C) = 2$ $p r o v e t h a t : A \overset{Δ}{B C} i s r i g h t t r i a n g l e$ $G o o d l u c k .$
	Answered by mindispower last updated on 03/Dec/20

	$s i n (c) = s i n (a + b)$ $\Leftrightarrow {s i n}^{2} (A) + {s i n}^{2} (B) + {(s i n (A + B))}^{2}) = 2$ ${s i n}^{2} (A + B) = {s i n}^{2} (A) {c o s}^{2} (B) + {c o s}^{2} (A) {s i n}^{2} (B)$ $+ \frac{1}{2} s i n (2 A) s i n (2 B)$ ${s i n}^{2} (A) - 1 + {s i n}^{2} (B) - 1 + \frac{s i n (2 A) s i n (2 B)}{2} +$ ${s i n}^{2} (A) {c o s}^{2} (B) + {s i n}^{2} (B) {c o s}^{2} (A) = 0$ $\Leftrightarrow - {c o s}^{2} (A) + {c o s}^{2} (A) {s i n}^{2} (B) - {c o s}^{2} (B) + {c o s}^{2} (B) {s i n}^{2} (A)$ $+ \frac{s i n (2 A) s i n (2 B)}{2} = 0$ $\Leftrightarrow - 2 {c o s}^{2} (A) {c o s}^{2} (B) + 2 s i n (A) s i n (B) c o s (A) c o s (B) = 0$ $\Leftrightarrow 2 c o s (A) c o s (B) (s i n (A) s i n (B) - c o s (A) c o s (B)) = 0$ $\Rightarrow c o s (A) c o s (B) c o s (A + B) = 0$ $\Rightarrow A = \frac{π}{2}, B = \frac{π}{2}, o r A + B = \frac{π}{2} \Rightarrow C = \frac{π}{2}$
	Commented by mnjuly1970 last updated on 03/Dec/20

	$t h a n k y o u s o m u c h$ $m r m i n d s p o w e r$ $n i c e a s a l w a y s . . .$
	Answered by Dwaipayan Shikari last updated on 03/Dec/20

	$2 {s i n}^{2} A + 2 {s i n}^{2} B + 2 {s i n}^{2} C = 4$ $1 - c o s 2 A + 1 - c o s 2 B + 1 - s i n 2 C = 4$ $c o s 2 A + c o s 2 B + c o s 2 C = - 1$ $2 c o s (A + B) c o s (A - B) + c o s 2 C = - 1$ $2 c o s (A + B) c o s (A - B) + 2 {c o s}^{2} C = 0$ $c o s (A + B) c o s (A - B) + {c o s}^{2} C = 0$ $c o s C c o s (A - B) = {c o s}^{2} C$ $c o s C (c o s C - c o s (A - B)) = 0$ $c o s C = 0 C = \frac{π}{2} o r c o s C = c o s (A - B) \Rightarrow C = A - B \Rightarrow C + B = A$
	Commented by mnjuly1970 last updated on 03/Dec/20

	$t h a n k y o u s o m u c h$ $e x t r a o r d i n a r y . . .$
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