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Question

Mathematics Question on Trigonometric Equations

If cosα+2cosβ+3cosγ=0,sinα+2sinβ+3sinγ=0\cos \alpha + 2 \cos \beta + 3 \cos \gamma = 0,\sin \alpha + 2 \sin \beta + 3 \sin \gamma = 0 and α+β+γ=π,\alpha + \beta + \gamma = \pi, then sin3α+8sin3β+27sin3γ \sin 3\alpha + 8 \sin 3\beta + 27 \sin 3\gamma =

A

-18

B

0

C

3

D

9

Answer

0

Explanation

Solution

The correct answer is B:0
Let z1=cosα+isinα=eiαz_1=cos\, \alpha \,+ \,i \,sin\, \alpha=e^{i\alpha}
z2=cosβ+isinβ=eiβz_2 = cos \,\beta \,+\, i \,sin \,\beta=e^{i\beta}
and z3=cosγ+isinγ=eiγz_3 = cos\, \gamma\, + \,i \,sin \,\gamma=e^{i\gamma}
Then, z1+2z2+3z3=(cosα+2cosβ+3cosγ)+i(sinα+2sinβ+3sinγ)=0z_1 + 2z_2 + 3z_3 = \left( cos\, \alpha \,+ \,2 \,cos \,\beta + 3 \,cos\, \gamma\right) \,+\, i \left(sin \,\alpha\,+ \,2\,sin \,\beta \,+ \,3 \,sin \,\gamma\right)= 0
if a+b+c=0a+b+c=0 , then, a3+b3+c3=3abca^3+b^3+c^3=3abc
(z1)3+(2z2)3+(3z3)3=3(z1)(2z2)(3z3)(z_1)^3+(2z_2)^3+(3z_3)^3=3(z_1)(2z_2)(3z_3)
e3iα+e3iβ+e3iγ=18ei(α+β+γ)e^{3i\alpha}+e^{3i\beta}+e^{3i\gamma}=18e^{i(\alpha+\beta+\gamma)}
(cos3α+8cos2β+27cos3γ)+i(sin3α+8sin2β+27sinα)=18iα(cos3\alpha+8cos2\beta+27cos3\gamma)+i(sin3\alpha+8sin2\beta+27sin\alpha)=18^{i\alpha}
(cos3α+8cos2β+27cos3γ)+i(sin3α+8sin2β+27sinγ)=18(cosπ+isinπ)(cos3\alpha+8cos2\beta+27cos3\gamma)+i(sin3\alpha+8sin2\beta+27sin\gamma)=18(cos\pi+isin\pi)
sin3α+8sin3β+27sinγ=0  (cosπ=1,sinπ=0)\therefore sin3\alpha+8sin3\beta+27sin\gamma=0\space (\therefore cos\pi=-1,sin\pi=0)
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