Prove that sin5 theta= 5sintheta -20sin³theta+16 sin⁵theta

Step-by-step explanation:

Express sin5θ in terms of sinθ and hence find the value of sin36

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By De-moivre's Theorem we know that

(cos θ + i sin θ)

5

= cos 5θ + i sin 5θ

L.H.S. on expansion by binomial theorem is

cos

5

θ+5cos

4

θ(isinθ)+10cos

3

θ(isinθ)

2

+10cos

2

θ(isinθ)

3

+5cosθ(isinθ)

4

+(isinθ)

5

Now equate real and imaginary parts and change

sin

2

θto1−cos

2

θandcos

2

θto1−sin

2

θ

depending on the answer.

∴ cos 5θ = cos θ (16 cos

4

θ - 20 cos

2

θ + 5)

sin = 5θ = sin θ (16 sin

4

θ - 20 sin

2

θ + 5)

Deduction : If θ = 36

∘

, then 5θ = 180

∘

∴ sin 5θ = 0

Also sin 36

∘

< sin 45

∘

or sin

2

36

∘

<

2

1

Now from (2), we get

0 = s(16 s

4

- 20 s

2

+ 5), s = sin 36

∘



= 0

∴s

2

=

32

20±

400−320

=

16

10−2

5

(∵s

2

<

2

1

)

∴s=

5

10−2

5

idk

step-by-step explanation:

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