The value of cos(60^{circ} - A) cdot cos A cd | vedclass

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(A) Using the formula $\cos(x - y) \cos(x + y) = \cos^2 x - \sin^2 y$, we have:$\cos(60^{\circ} - A) \cos(60^{\circ} + A) = \cos^2 60^{\circ} - \sin^2

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$\frac{1}{4} \cos 3A$

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(A) Using the formula $\cos(x - y) \cos(x + y) = \cos^2 x - \sin^2 y$, we have:$\cos(60^{\circ} - A) \cos(60^{\circ} + A) = \cos^2 60^{\circ} - \sin^2 A$Substitute $\cos 60^{\circ} = \frac{1}{2}$:$= (\frac{1}{2})^2 - \sin^2 A = \frac{1}{4} - \sin^2 A$Now multiply by $\cos A$:$= \cos A (\frac{1}{4} - \sin^2 A) = \frac{1}{4} \cos A - \cos A \sin^2 A$Since $\sin^2 A = 1 - \cos^2 A$:$= \frac{1}{4} \cos A - \cos A (1 - \cos^2 A) = \frac{1}{4} \cos A - \cos A + \cos^3 A = \cos^3 A - \frac{3}{4} \cos A$Factor out $\frac{1}{4}$:$= \frac{1}{4} (4 \cos^3 A - 3 \cos A)$Using the identity $\cos 3A = 4 \cos^3 A - 3 \cos A$, we get:$= \frac{1}{4} \cos 3A$

The value of $\cos(60^{\circ} - A) \cdot \cos A \cdot \cos(60^{\circ} + A)$ is

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