evaluate-sin-72-

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 \dots

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 \dots

Answered by MJS last updated on 25/Jul/18

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

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