Question
Mathematics Question on Trigonometric Equations
If cosα+2cosβ+3cosγ=0,sinα+2sinβ+3sinγ=0 and α+β+γ=π, then sin3α+8sin3β+27sin3γ =
A
-18
B
0
C
3
D
9
Answer
0
Explanation
Solution
The correct answer is B:0
Let z1=cosα+isinα=eiα
z2=cosβ+isinβ=eiβ
and z3=cosγ+isinγ=eiγ
Then, z1+2z2+3z3=(cosα+2cosβ+3cosγ)+i(sinα+2sinβ+3sinγ)=0
if a+b+c=0, then, a3+b3+c3=3abc
(z1)3+(2z2)3+(3z3)3=3(z1)(2z2)(3z3)
⇒e3iα+e3iβ+e3iγ=18ei(α+β+γ)
⇒(cos3α+8cos2β+27cos3γ)+i(sin3α+8sin2β+27sinα)=18iα
⇒(cos3α+8cos2β+27cos3γ)+i(sin3α+8sin2β+27sinγ)=18(cosπ+isinπ)
∴sin3α+8sin3β+27sinγ=0(∴cosπ=−1,sinπ=0)