JEE Main
Exam JEE Main
Subject Mathematics
Chapter Trigonometric Equations JEE Main · Mathematics · Trigonometric Equations
If ^3 θ+ ^3((2 π)/(3)+θ)+ ^3((4 π)/(3)+θ)=α 3 θ , then α is equal to
If
cos 3 θ + cos 3 ( 2 π 3 + θ ) + cos 3 ( 4 π 3 + θ ) = α cos 3 θ \cos ^3 \theta+\cos ^3\left(\frac{2 \pi}{3}+\theta\right)+\cos ^3\left(\frac{4 \pi}{3}+\theta\right)=\alpha \cos 3 \theta cos 3 θ + cos 3 ( 3 2 π + θ ) + cos 3 ( 3 4 π + θ ) = α cos 3 θ , then
α \alpha α is equal to
Options A 1 4 \frac{1}{4} 4 1 B 3 4 \frac{3}{4} 4 3 C 5 4 \frac{5}{4} 4 5 D 7 4 \frac{7}{4} 4 7 Correct answer B. 3 4 \frac{3}{4} 4 3
Step-by-step solution
We have,
cos 3 θ + cos 3 ( 2 π 3 + θ ) + cos 3 ( 4 π 3 + θ ) = a cos 3 θ ⇒ cos 3 θ + 3 cos θ 4 + cos 3 ( 2 π 3 + θ ) + 3 cos ( 2 π 3 + θ ) 4 + cos 3 ( 4 π 3 + θ ) + 3 cos ( 4 π 3 + θ ) 4 = a cos 3 θ ⇒ cos 3 θ + 3 cos θ + cos ( 2 π + 3 θ ) + 3 cos ( 2 π 3 + θ ) + cos ( 4 π + 3 θ ) + 3 cos ( 4 π 3 + θ ) = 4 a cos 3 θ ⇒ cos 3 θ + 3 cos θ + cos 3 θ + cos 3 θ + 3 [ cos ( 2 π 3 + θ ) ⇒ 3 cos 3 θ + 3 cos θ + 3 [ 2 cos ( π + θ ) cos π 3 ] = 4 a cos 3 θ ⇒ 3 cos 3 θ + 3 cos θ + 3 × 2 × ( − cos θ ) × 1 2 = 4 a cos 3 θ ⇒ 3 cos 3 θ + 3 cos θ − 3 cos θ = 4 a cos 3 θ ⇒ 3 cos 3 θ = 4 a cos 3 θ \Rightarrowa = 3 4 \begin{gathered}
\cos ^3 \theta+\cos ^3\left(\frac{2 \pi}{3}+\theta\right)+\cos ^3\left(\frac{4 \pi}{3}+\theta\right)=a \cos 3 \theta\\
\Rightarrow \frac{\cos 3 \theta+3 \cos \theta}{4}+\frac{\cos 3\left(\frac{2 \pi}{3}+\theta\right)+3 \cos \left(\frac{2 \pi}{3}+\theta\right)}{4}\\
+\frac{\cos 3\left(\frac{4 \pi}{3}+\theta\right)+3 \cos \left(\frac{4 \pi}{3}+\theta\right)}{4}=a \cos 3 \theta\\
\Rightarrow \cos 3 \theta+3 \cos \theta+\cos (2 \pi+3 \theta)+3 \cos \left(\frac{2 \pi}{3}+\theta\right)\\
+\cos (4 \pi+3 \theta)+3 \cos \left(\frac{4 \pi}{3}+\theta\right)=4 a \cos 3 \theta\\
\Rightarrow \cos 3 \theta+3 \cos \theta+\cos3 \theta+\cos3 \theta+3\left[\cos \left(\frac{2 \pi}{3}+\theta\right)\right.\\
\Rightarrow 3 \cos 3 \theta+3 \cos \theta+3\left[2 \cos (\pi+\theta) \cos \frac{\pi}{3}\right]
\end{gathered} \begin{gathered}
=4 a \cos 3 \theta\\
\Rightarrow 3 \cos 3 \theta+3 \cos \theta+3 \times2 \times(-\cos \theta) \times \frac{1}{2}\\
=4 a \cos 3 \theta\\
\Rightarrow 3 \cos 3 \theta+3 \cos \theta-3 \cos \theta=4 a \cos 3 \theta\\
\Rightarrow \quad 3 \cos 3 \theta=4 a \cos 3 \theta \Rightarrowa=\frac{3}{4}
\end{gathered} cos 3 θ + cos 3 ( 3 2 π + θ ) + cos 3 ( 3 4 π + θ ) = a cos 3 θ ⇒ 4 cos 3 θ + 3 cos θ + 4 cos 3 ( 3 2 π + θ ) + 3 cos ( 3 2 π + θ ) + 4 cos 3 ( 3 4 π + θ ) + 3 cos ( 3 4 π + θ ) = a cos 3 θ ⇒ cos 3 θ + 3 cos θ + cos ( 2 π + 3 θ ) + 3 cos ( 3 2 π + θ ) + cos ( 4 π + 3 θ ) + 3 cos ( 3 4 π + θ ) = 4 a cos 3 θ ⇒ cos 3 θ + 3 cos θ + cos 3 θ + cos 3 θ + 3 [ cos ( 3 2 π + θ ) ⇒ 3 cos 3 θ + 3 cos θ + 3 [ 2 cos ( π + θ ) cos 3 π ] = 4 a cos 3 θ ⇒ 3 cos 3 θ + 3 cos θ + 3 × 2 × ( − cos θ ) × 2 1 = 4 a cos 3 θ ⇒ 3 cos 3 θ + 3 cos θ − 3 cos θ = 4 a cos 3 θ ⇒ 3 cos 3 θ = 4 a cos 3 θ \Rightarrowa = 4 3
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