evaluate sin 72^.


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Question Number 40640 by upadhyayrakhi20@gmail.com last updated on 25/Jul/18

$e v a l u a t e$ $\sin 72^{.}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jul/18

	$s i n 72 = c o s 18 = \sqrt{1 - {s i n}^{2} 18}$ $a = 18^{0}$ $5 a = 90^{o}$ $2 a = 90^{o} - 3 a$ $s i n 2 a = s i n (90^{o} - 3 a)$ $2 s i n a c o s a = c o s 3 a$ $2 s i n a c o s a = 4 {c o s}^{3} a - 3 c o s a$ $2 s i n a = 4 {c o s}^{2} a - 3$ $s i n a = t$ $2 t = 4 (1 - t^{2}) - 3$ $2 t + 4 t^{2} - 1 = 0$ $4 t^{2} + 2 t - 1 = 0$ $t = \frac{- 2 \pm \sqrt{4 + 16}}{8} = \frac{- 2 \pm 2 \sqrt{5}}{8} = \frac{- 1 \pm \sqrt{5}}{4}$ $s i n 18^{o} > 0 s o t = \frac{- 1 - \sqrt{5}}{4} n o t f e a s i b l e$ $t = \frac{\sqrt{5} - 1}{4} \to s i n 18^{o}$ $t^{2} = \frac{5 - 2 \sqrt{5} + 1}{16} = \frac{6 - 2 \sqrt{5}}{16}$ $1 - t^{2} = \frac{16 - 6 + 2 \sqrt{5}}{16} = \frac{10 + 2 \sqrt{5}}{16}$ $c o s 18^{o} = \sqrt{1 - t^{2}} = \frac{\sqrt{10 + 2 \sqrt{5}}}{4}$
	Commented by Tawa1 last updated on 25/Jul/18

	$great . .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 25/Jul/18

	$t h a n k y o u s i r . . .$
	Commented by upadhyayrakhi20@gmail.com last updated on 25/Jul/18

	$thnx$
	Commented by tanmay.chaudhury50@gmail.com last updated on 25/Jul/18

	$i t s o k . . .$
	Answered by MJS last updated on 25/Jul/18

	$\sin 72 ° = \frac{\sqrt{10 + 2 \sqrt{5}}}{4}$
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