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Trigonometry – II — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – II1 Mark
MCQ
The value of $\cos A \cos \left(60^{\circ}-A\right) \cos \left(60^{\circ}+A\right)$ is equal to
  • A
    $\frac{1}{2} \cos 3 \mathrm{~A}$
  • B
    $\cos 3 \mathrm{~A}$
  • ✓
    $\frac{1}{4} \cos 3 \mathrm{~A}$
  • D
    $4 \cos 3 \mathrm{~A}$
✓

Answer

Correct option: C.
$\frac{1}{4} \cos 3 \mathrm{~A}$
(c) $\frac{1}{4} \cos 3 \mathrm{~A}$
Hint:
$\begin{aligned}
& \cos A \cos \left(60^{\circ}-A\right) \cos \left(60^{\circ}+A\right) \\
& =(\cos A)\left(\cos 60^{\circ} \cos A+\sin 60^{\circ} \sin A\right) \\
& \quad \cdot\left(\cos 60^{\circ} \cos A-\sin 60^{\circ} \sin A\right) \\
& =(\cos A)\left(\frac{1}{2} \cos A+\frac{\sqrt{3}}{2} \sin A\right)
\end{aligned}$
$\left(\frac{1}{2} \cos \mathrm{A}-\frac{\sqrt{3}}{2} \sin \mathrm{A}\right)$
$\begin{aligned}
& =\frac{1}{4} \cos \mathrm{A}\left(\cos ^2 \mathrm{~A}-3 \sin ^2 \mathrm{~A}\right) \\
& =\frac{1}{4}\left[\cos ^3 \mathrm{~A}-3 \cos \mathrm{A}\left(1-\cos ^2 \mathrm{~A}\right)\right] \\
& =\frac{1}{4}\left(4 \cos ^3 \mathrm{~A}-3 \cos \mathrm{A}\right)=\frac{1}{4} \cos 3 \mathrm{~A}
\end{aligned}$

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