Question

Find the result of the trigonometric identity which is given as cosA cos(60° – A) cos(60° + A) =
$
  {\text{A}}{\text{. cosA}} \\
  {\text{B}}{\text{. 0}} \\
  {\text{C}}{\text{. }}\dfrac{1}{4}{\text{sin3A}} \\
  {\text{D}}{\text{. }}\dfrac{1}{4}{\text{cos3A}} \\
$

Answer 1

Hint: To solve the given equation in the question we use a trigonometric formula, i.e.
2cosAcosB = cos (A+B) + cos (A−B).

Complete Step-by-Step solution:
Given Data – cosA cos(60°−A) cos(60°+A)
⟹$\dfrac{1}{2}$cosA × [2cos(60°+A)cos(60°−A)]
This is in the form of 2cosAcosB
⟹$\dfrac{1}{2}$cosA × [cos(60°+A+60°−A)+cos(60°+A−60°+A)]
⟹$\dfrac{1}{2}$cosA[cos120°+cos2A]
Cos(120°) = Cos( 90°+30°)= -sin30° =$ - \dfrac{1}{2}$
(According to the graph of cos function, it is negative in second quadrant and
cos(90°+θ)=-sinθ). Also sin30°=$\dfrac{1}{2}$, from the trigonometric table of sine function.

⟹$\dfrac{1}{2}$cosA [cos2A−$\dfrac{1}{2}$]
⟹$\dfrac{1}{2}$cosA[$\dfrac{{\left( {2{\text{cos2A - 1}}} \right)}}{2}$]
⟹$\dfrac{1}{4}$ (2cosAcos2A – cosA)
(2cosA cos2A = cos (2A+A) + cos(2A-A) = cos (3A) + cos (A))
Put this in the above equation we get
=$\dfrac{1}{4}$ (cos3A + cosA – cosA)
=$\dfrac{1}{4}$cos3A.
Hence, cosA cos(60°– A) cos(60° +A) = $\dfrac{1}{4}$cos3A.

Note: In order to solve questions of this type, the key is to reduce the given trigonometric equation into known trigonometric identities using appropriate formulas. Basic formulae of trigonometric functions expressed as sum or difference of two terms are applied. Then we solved the output to simplify it further.