2 - 3 complex no. and trig

Question 1

cos nθ in terms of cos and sin

Answer 1

let z = cosθ+ i sinθ
write z^n using demoivres
then z^n using binomal expansion
cosnθ = the real part of the binomial expansion
(sin is the imaginary

Question 2

cos θ and sin as imaginary and exponential

Answer 2

2cos θ= e^iθ + e^-iθ = z+1/z
2isinθ = e^iθ - e^-iθ = z-1/z

Question 3

if z = e^iθ then

Answer 3

z^n + 1/z^n = 2cos nθ
z^n - 1/z^n = 2isin nθ

Question 4

sin^nθ in terms of sin

Answer 4

let z = cosθ + i sinθ
(2isinθ)^n = (z-1/z)^n
2i^n sin^nθ = (z-1/z)^n
expand the RHS binomaly
group terms in the z^k -1/z^k form
rewrite using 2isinkθ
rearrange for sin^nθ
(same for cos bu no i and use z + 1/z)

Question 5

simplifying tirg series

Answer 5

write in exponential and imaginary form
simplify geometrically and take either imaginary or real part (im for sin and re for cos)