Evaluate the following trigonometric equation.
$\sin \theta .{\cos ^3}\theta - \cos \theta .{\sin ^3}\theta $ is equal to?
$
  {\text{A}}{\text{. }}\dfrac{{ - 1}}{4}\sin \theta \\
  {\text{B}}{\text{. }}\dfrac{{\sin 4\theta }}{4} \\
  {\text{C}}{\text{. }}\dfrac{{\cos 4\theta }}{4} \\
  {\text{D}}{\text{. }}\dfrac{{\cos 4\theta }}{3} \\
$

Question

Evaluate the following trigonometric equation.
$\sin \theta .{\cos ^3}\theta - \cos \theta .{\sin ^3}\theta $ is equal to?
$
  {\text{A}}{\text{. }}\dfrac{{ - 1}}{4}\sin \theta \\
  {\text{B}}{\text{. }}\dfrac{{\sin 4\theta }}{4} \\
  {\text{C}}{\text{. }}\dfrac{{\cos 4\theta }}{4} \\
  {\text{D}}{\text{. }}\dfrac{{\cos 4\theta }}{3} \\
$

Vedantu Content Team · Accepted Answer

Hint: For solving this complex equation first you have to take common whichever can be taken and then proceed using trigonometric results and shorten the equation as much as you can.

Complete step-by-step answer:
From given
$\sin \theta .{\cos ^3}\theta - \cos \theta .{\sin ^3}\theta $
Take $\sin \theta .\cos \theta $ common then we get
$\sin \theta .\cos \theta \left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right)$
$\left( {\because \cos 2\theta = {{\cos }^2}\theta - {{\sin }^2}\theta } \right)$ (on multiplying and dividing by 2)
$\dfrac{{2\sin \theta .\cos \theta }}{2}\left( {\cos 2\theta } \right)$
($\because \sin 2\theta = 2\sin \theta .\cos \theta $)
$\dfrac{{\sin 2\theta .\cos 2\theta }}{2}$ (on multiplying and dividing 2 we get)
$\dfrac{{2.\sin 2\theta .\cos 2\theta }}{{2.2}}$
$\left( {\because \sin 4\theta = 2\sin 2\theta .\cos 2\theta } \right)$
=$\dfrac{{\sin 4\theta }}{4}$
Hence option B is the correct option.

Note: Whenever you get this type of question the key concept of solving is you have to shorten the complex equation using trigonometric results like $\left( {\cos 2\theta = {{\cos }^2}\theta - {{\sin }^2}\theta } \right)$and use basic mathematics to proceed further.

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