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Use of Trigonometric Table

Prove: sin 5...

Question

Prove:
$sin 5 θ = 16 {sin}^{5} θ - 20 {sin}^{3} θ + 5 sin θ$

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Solution

$s i n 5 A = s i n (3 A + 2 A)$
$= s i n 3 A c o s 2 A + c o s 3 A s i n 2 A$
$= (3 s i n A - 4 s i n ³ A) (1 - 2 s i n ² A) + c o s (2 A + A) s i n 2 A$
$= 3 s i n A - 10 s i n ³ A + 8 s i n^{5} A + [c o s 2 A c o s A - s i n 2 A s i n A] s i n 2 A$
$= 3 s i n A - 10 s i n ³ A + 8 s i n^{5} A + [(1 - 2 s i n ² A) c o s A - 2 s i n ² A c o s A] 2 s i n A c o s A$
$= 3 s i n A - 10 s i n ³ A + 8 s i n^{5} A + [c o s A - 4 s i n ² A c o s A] 2 s i n A c o s A$
$= 3 s i n A - 10 s i n ³ A + 8 s i n^{5} A + 2 s i n A c o s ² A - 8 s i n ³ A c o s ² A$
$= 3 s i n A - 10 s i n ³ A + 8 s i n^{5} A + 2 s i n A (1 - s i n ² A) - 8 s i n ³ A (1 - s i n ² A)$
$= 3 s i n A - 10 s i n ³ A + 8 s i n^{5} A + 2 s i n A - 2 s i n ³ A - 8 s i n ³ A + 8 s i n^{5} A$
$= 5 s i n A - 20 s i n ³ A + 16 s i n^{5} A$ [Proved]

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Prove that following identities:

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

\frac{s i n 7 θ - s i n 5 θ}{c o s 7 θ + c o s 5 θ} = t a n 6 θ

\frac{(sin 5 θ)}{(sin θ)} = 16 {cos}^{4} θ - 12 {cos}^{2} θ + 1

Q. Prove the following trigonometric identities.

If 3 sin θ + 5 cos θ = 5, prove that 5 sin θ − 3 cos θ = ± 3.

3 sin θ + 5 cos θ = 5

5 sin θ - 3 cos θ = \pm 3

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Use of Trigonometric Table

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