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Question: If \[\cos \theta = \left( {\dfrac{1}{2}} \right)\left( {a + \dfrac{1}{a}} \right)\]then the value of...

If cos⁡θ=(12)(a+1a)\cos \theta = \left( {\dfrac{1}{2}} \right)\left( {a + \dfrac{1}{a}} \right)then the value of cos⁡3θ\cos 3\theta is
A.(18)(a3+1a3)\left( {\dfrac{1}{8}} \right)\left( {{a^3} + \dfrac{1}{{{a^3}}}} \right)
B.(32)(a+1a)\left( {\dfrac{3}{2}} \right)\left( {a + \dfrac{1}{a}} \right)
C.(12)(a3+1a3)\left( {\dfrac{1}{2}} \right)\left( {{a^3} + \dfrac{1}{{{a^3}}}} \right)
D.(13)(a3+1a3)\left( {\dfrac{1}{3}} \right)\left( {{a^3} + \dfrac{1}{{{a^3}}}} \right)

Explanation

Solution

The question is related to the trigonometric topic, the value of trigonometric ratio is known. By using that we are determining the value of cos⁡3θ\cos 3\theta , by considering the formula of cubic power of the cosine trigonometric function. Since the value is in the form of algebraic expression we use the algebraic identities and on simplification we obtain the solution.

Complete answer:
On considering the given question cos⁡θ=(12)(a+1a)\cos \theta = \left( {\dfrac{1}{2}} \right)\left( {a + \dfrac{1}{a}} \right) -------(1)
As we know the formula cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta
In RHS part of the above formula, take cos⁡θ\cos \theta as a common we get
⇒cos⁡3θ=cos⁡θ(4cos⁡2θ−3)\Rightarrow \cos 3\theta = \cos \theta \left( {4{{\cos }^2}\theta - 3} \right)------(2)
Squaring on both sides of the equation (1) we get
⇒(cos⁡θ)2=((12)(a+1a))2\Rightarrow {\left( {\cos \theta } \right)^2} = {\left( {\left( {\dfrac{1}{2}} \right)\left( {a + \dfrac{1}{a}} \right)} \right)^2}
On simplification we get
⇒cos⁡2θ=((12)2(a+1a)2)\Rightarrow {\cos ^2}\theta = \left( {{{\left( {\dfrac{1}{2}} \right)}^2}{{\left( {a + \dfrac{1}{a}} \right)}^2}} \right)
On applying the algebraic identity
⇒cos⁡2θ=14(a2+1a2+2)\Rightarrow {\cos ^2}\theta = \dfrac{1}{4}\left( {{a^2} + \dfrac{1}{{{a^2}}} + 2} \right)-------(3)
Substitute equation (1) and equation (3) in the equation (2) we get
⇒cos⁡3θ=12(a+1a)(4(14(a2+1a2+2))−3)\Rightarrow \cos 3\theta = \dfrac{1}{2}\left( {a + \dfrac{1}{a}} \right)\left( {4\left( {\dfrac{1}{4}\left( {{a^2} + \dfrac{1}{{{a^2}}} + 2} \right)} \right) - 3} \right)
On simplifying we get
⇒cos⁡3θ=12(a+1a)(a2+1a2+2−3)\Rightarrow \cos 3\theta = \dfrac{1}{2}\left( {a + \dfrac{1}{a}} \right)\left( {{a^2} + \dfrac{1}{{{a^2}}} + 2 - 3} \right)
⇒cos⁡3θ=12(a+1a)(a2+1a2−1)\Rightarrow \cos 3\theta = \dfrac{1}{2}\left( {a + \dfrac{1}{a}} \right)\left( {{a^2} + \dfrac{1}{{{a^2}}} - 1} \right)
As we know that the algebraic identity (a3+b3)=(a+b)(a2−ab+b2)\left( {{a^3} + {b^3}} \right) = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right), the above term is similar to the algebraic identity, so we get
⇒cos⁡3θ=12(a3+1a3)\Rightarrow \cos 3\theta = \dfrac{1}{2}\left( {{a^3} + \dfrac{1}{{{a^3}}}} \right)
Therefore option (3) is the correct answer.

Note:
The algebraic expressions have a standard formula, the square, cubic of a binomial expression are mostly used to solve the problems. So here we have to remember the formulas like (a+b)2=a2+2ab+b2{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}, (a−b)2=a2−2ab+b2{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}, (a+b)3=a3+b3+3ab(a+b){\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right) and so on. Likewise in trigonometry we have the formula related to square and cube.