Question

By using De Moivre's Theorem obtain tan 5θ in terms of tan θ and show that `1-10 tan^2(pi/10)+5tan^4(pi/10)=0`.

Answer 1

(cos⁵θ + isin⁵θ) = (cos θ + isin θ)⁵
= cos⁵θ + 5cos⁴θ isin θ + 10 cos³θ i² sin²θ + 10 cos²θ i³ sin³θ + 5 cos θ
i⁴ sin⁴θ + i⁵ sin⁵θ
= cos⁵θ + 5cos⁴θ isin θ - 10 cos³θ sin²θ – 10i cos²θ sin³θ + 5 cosθ sin⁴θ + i sin⁵θ
= (cos⁵θ- 10 cos³θ sin²θ + 5 cosθ sin⁴θ) + i (5cos⁴θ sin θ – 10 cos²θ
sin³θ + sin⁵θ)
Equating both sides we get,
∴ cos⁵θ = cos⁵θ- 10 cos³θ sin²θ + 5 cosθ sin⁴θ
∴ sin⁵θ = 5cos⁴θ sin θ – 10 cos²θ sin³θ + sin⁵θ
But tan⁵θ = `sin^2theta/cos^5theta`

=`(5cos^4theta sintheta-10cos^2thetasin^3theta+sintheta)/(cos^5theta-10cos^3thetasin^2theta+5costheta sin^4theta)`

Dividing by cos 5θ
tan 5θ =`(5tan theta-10tan^2theta+tan^5theta)/(1-10tan^2theta+5tan^4theta)`

for deduction, put θ = `pi/10`

`cot 5x pi/10=(1-10tan^2  pi/10+5tan^4  pi/10)/` 

`therefore 1-10tan^2(pi/10)+5tan^4(pi/10)=0.`

Hence proved.