If cos theta = frac{1}{2}left( a + frac{1}{a} | vedclass

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(C) We know the trigonometric identity: $\cos 3	heta = 4\cos^3 	heta - 3\cos 	heta$.Given $\cos 	heta = \frac{1}{2}\left( a + \frac{1}{a} ight)$

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$\frac{1}{2}\left( a^3 + \frac{1}{a^3} \right)$

Vedclass · Answer

(C) We know the trigonometric identity: $\cos 3	heta = 4\cos^3 	heta - 3\cos 	heta$.Given $\cos 	heta = \frac{1}{2}\left( a + \frac{1}{a} ight)$.Substituting the value of $\cos 	heta$ in the identity:$\cos 3	heta = 4 \left[ \frac{1}{2}\left( a + \frac{1}{a} ight) ight]^3 - 3 \left[ \frac{1}{2}\left( a + \frac{1}{a} ight) ight]$$\cos 3	heta = 4 \cdot \frac{1}{8} \left( a + \frac{1}{a} ight)^3 - \frac{3}{2} \left( a + \frac{1}{a} ight)$$\cos 3	heta = \frac{1}{2} \left( a + \frac{1}{a} ight)^3 - \frac{3}{2} \left( a + \frac{1}{a} ight)$$\cos 3	heta = \frac{1}{2} \left( a + \frac{1}{a} ight) \left[ \left( a + \frac{1}{a} ight)^2 - 3 ight]$$\cos 3	heta = \frac{1}{2} \left( a + \frac{1}{a} ight) \left[ a^2 + 2 + \frac{1}{a^2} - 3 ight]$$\cos 3	heta = \frac{1}{2} \left( a + \frac{1}{a} ight) \left[ a^2 - 1 + \frac{1}{a^2} ight]$$\cos 3	heta = \frac{1}{2} \left( a^3 - a + \frac{1}{a} + a - \frac{1}{a} + \frac{1}{a^3} ight)$$\cos 3	heta = \frac{1}{2} \left( a^3 + \frac{1}{a^3} ight)$.

If $\cos \theta = \frac{1}{2}\left( a + \frac{1}{a} \right),$ then the value of $\cos 3\theta$ is

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