If cosα+cosβ+cosγ=0 = sinα+sinβ+sinγ, prove that (i) cos3α+cos3β+cosγ = 3cos(a+β+γ) (ii) sin3α+sin3β+sinγ = 3sin(α+β+γ) (iii) cos2α+cos2β+cos2γ = 0 (iv) sin2α+sin2β+sin2γ = 0 (Hints: Take a=cis α, b=cis β, c=cis γ, a+b+c=0 ⇒ a^3 +b^3 +c^3 =3abc (1/a)+(1/b)+(1/c)=0 ⇒ a^2 +b^2 +c^2 =0) (v) cos^2 α+cos^2 β+cosγ = sin^2 α+sin^2 β+sin^2 γ = (3/2).


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Question Number 93836 by oustmuchiya@gmail.com last updated on 15/May/20

$I f c o s α + c o s β + c o s γ = 0 = s i n α + s i n β + s i n γ, p r o v e t h a t$ $(i) c o s 3 α + c o s 3 β + c o s γ = 3 c o s (a + β + γ)$ $(ii) s i n 3 α + s i n 3 β + s i n γ = 3 s i n (α + β + γ)$ $(iii) c o s 2 α + c o s 2 β + c o s 2 γ = 0$ $(iv) s i n 2 α + s i n 2 β + s i n 2 γ = 0$ $(Hints : T a k e a = cis α, b = cis β, c = cis γ, a + b + c = 0 \Rightarrow a^{3} + b^{3} + c^{3} = 3 abc$ $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \Rightarrow a^{2} + b^{2} + c^{2} = 0)$ $(v) {c o s}^{2} α + {c o s}^{2} β + c o s γ = {s i n}^{2} α + {s i n}^{2} β + {s i n}^{2} γ = \frac{3}{2} .$
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