Prove that
sin 5? = 5 sin ? -20sin^3 ? +16 sin^5 ?

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Solution

$S i n 5 x = S i n (2 x + 3 x) = S i n 2 x C o s 3 x + S i n 3 x C o s 2 x = 2 S i n x C o s x (4 C o s^{3} x - 3 C o s x) + (3 S i n x - 4 S i n^{3} x) (1 - 2 S i n^{2} x) = 2 S i n x (4 C o s^{4} x - 3 C o s^{2} x) + 3 S i n x - 4 S i n^{3} x - 6 S i n^{3} x + 8 S i n^{5} x = 2 S i n x (4 (1 - S i n^{2} x)^{2} - 3 (1 - S i n^{2} x)) + 3 S i n x - 4 S i n^{3} x - 6 S i n^{3} x + 8 S i n^{5} x = 2 S i n x (4 + 4 S i n^{4} x - 8 S i n^{2} x - 3 + 3 S i n^{2} x) + 3 S i n x - 4 S i n^{3} x - 6 S i n^{3} x + 8 S i n^{5} x = 2 S i n x (1 + 4 S i n^{4} x - 5 S i n^{2} x) + 3 S i n x - 4 S i n^{3} x - 6 S i n^{3} x + 8 S i n^{5} x = 2 S i n x + 8 S i n^{5} x - 10 S i n^{3} x + 3 S i n x - 4 S i n^{3} x - 6 S i n^{3} x + 8 S i n^{5} x = 16 S i n^{5} x - 20 S i n^{3} x + 5 S i n x$

$F o r m u l a u s e d S i n (A + B) = S i n A C o s B + C o s A S i n B S i n 2 x = 2 S i n x C o s x S i n 3 x = 3 S i n x - 4 S i n^{3} x C o s 3 x = 4 C o s^{3} x - 3 C o s x$

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Similar questions

\sin 5 x = 5 \sin x - 20 \sin^{3} x + 16 \sin^{5} x

\sin 5 θ = 5 \sin θ - 20 \sin^{3} θ + 16 \sin^{5} θ

Prove that following identities:

$s i n 5 θ = 5 s i n θ - 20 s i n^{3} θ + 16 s i n^{5} θ$

{sin}^{- 1} (\frac{4}{5}) + {sin}^{- 1} (\frac{5}{13}) + {sin}^{- 1} (\frac{16}{65}) = \frac{π}{2}

{sin}^{- 1} \frac{4}{5} + {sin}^{- 1} \frac{5}{13} + {sin}^{- 1} (\frac{16}{65}) = \frac{π}{2}

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