KCET · Maths · Trigonometric Equations
If \(\cos \alpha+2 \cos \beta+3 \cos \gamma=0\),
\(\sin \alpha+2 \sin \beta+3 \sin \gamma=0\) and \(\alpha+\beta+\gamma=\pi\), then \(\sin 3 \alpha+8 \sin 3 \beta+27 \sin 3 \gamma\) is equal to
- A \(-18\)
- B 0
- C 3
- D 9
Answer & Solution
Correct Answer
(B) 0
Step-by-step Solution
Detailed explanation
Let \(a=\cos \alpha+i \sin \alpha\)
\(\quad b=\cos \beta+i \sin \beta\)
and \(\mathrm{c}=\cos \gamma+i \sin \gamma\)
Then, \(a+2 b+3 c=(\cos \alpha+2 \cos \beta+3 \cos \gamma)\)
\(\quad+i(\sin \alpha+2 \sin \beta+3 \sin \gamma)=0\)
\(\Rightarrow \quad a^{3}+8 b^{3}+27 c^{3}=18 a b c\)
\(\left(\because\right.\) if \(\left.x+y+z=0 \Rightarrow x^{3}+y^{3}+z^{3}=3 x y z\right)\)
\(\Rightarrow \quad \cos 3 \alpha+8 \cos 3 \beta+27 \cos 3 \gamma\)
and \(\quad \sin 3 \alpha+8 \sin 3 \beta+27 \sin 3 \gamma\)
\(\quad=18 \sin (\alpha+\beta+\gamma)\)
\(=18 \sin \pi=0\)
\(\quad b=\cos \beta+i \sin \beta\)
and \(\mathrm{c}=\cos \gamma+i \sin \gamma\)
Then, \(a+2 b+3 c=(\cos \alpha+2 \cos \beta+3 \cos \gamma)\)
\(\quad+i(\sin \alpha+2 \sin \beta+3 \sin \gamma)=0\)
\(\Rightarrow \quad a^{3}+8 b^{3}+27 c^{3}=18 a b c\)
\(\left(\because\right.\) if \(\left.x+y+z=0 \Rightarrow x^{3}+y^{3}+z^{3}=3 x y z\right)\)
\(\Rightarrow \quad \cos 3 \alpha+8 \cos 3 \beta+27 \cos 3 \gamma\)
and \(\quad \sin 3 \alpha+8 \sin 3 \beta+27 \sin 3 \gamma\)
\(\quad=18 \sin (\alpha+\beta+\gamma)\)
\(=18 \sin \pi=0\)
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