Q.328. Application of De Moivere's Theorem.

Question 328.
Using the relationship e^(ix) = cos x + i sin x, express cos5x in term of cosx.
Hence show that x = cos(π/10) is a root of the equation 16x^4 - 20x^2  + 5 = 0.

Answer 328.
cos x + i sin x = e^(ix)
=> (cosx + isinx)^5 = e^(5ix)
=> cos^5 x + 5C1 cos^4 x * (isinx) + 5C2 cos^3 x (isinx)^2 + 5C3 cos^2 x (isinx)^3 + 5C4 cosx (isinx)^4 + 5C5 (isinx)^5 = cos5x + isin5x

Comparing real parts on both sides,
cos5x
= cos^5 x + 5C2 cos^3 x (isinx)^2 + 5C4 cosx (isinx)^4
= cos^5 x - 10cos^3 x sin^2 x + 5cosx sin^4 x
= cos^5 x - 10cos^3 x (1 - cos^2 x) + 5cosx (1 - cos^2 x)^2
= cos^5 x - 10cos^3 x + 10cos^5 x + 5cosx - 10cos^3 x + 5cos^5 x
= 16cos^5 x - 20cos^3 x + 5cosx

Plugging cos(π/10) in 16x^4 - 20x^2 + 5, we get
16cos^4 (π/10) - 20cos^2 (π/10) + 5
= [16cos^5 (π/10) - 20cos^3 (π/10) + 5cos(π/10)] / cos(π/10)
= cos [5(π/10)] / cos(π/10)
= cos(π/2) / cos(π/10)
= 0. [because cos(π/2) = 0].

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