Question

What is the value of $\sin \left( {{{60}^o}} \right)$?
$
  (a){\text{ 2sin3}}{{\text{0}}^0}\cos {30^0} \\
  (b){\text{ 2sin6}}{{\text{0}}^0}\cos {60^0} \\
  (c){\text{ 2sin6}}{{\text{0}}^0}\cos {30^0} \\
  (d){\text{ 2sin3}}{{\text{0}}^0}\cos {30^0} \\
 $

Answer 1

Hint – There can be two ways to solve these problems, the first one is using the approach of formula based. In this approach we will simply write $\sin \left( {{{60}^o}} \right)$ as $\sin \left( {2 \times {{30}^o}} \right)$ and then we will use formula $\sin \left( {2\theta } \right) = 2\sin \theta \cos \theta $. The second method we will discuss at the end of this solution.

Complete step-by-step answer:
Given trigonometric equation is
$\sin \left( {{{60}^o}} \right)$
This equation is also written as
$ \Rightarrow \sin \left( {2 \times {{30}^o}} \right)$
Now as we know that $\sin \left( {2\theta } \right) = 2\sin \theta \cos \theta $ so use this property in above equation we have,
$ \Rightarrow \sin \left( {2 \times {{30}^o}} \right) = 2\sin {30^o}\cos {30^o}$
So this is the required simplification of given trigonometric equation $\sin \left( {{{60}^o}} \right)$
Hence option (A) is the correct answer.

Note – The second method is more based upon simplification. As we know the value of $\sin \left( {{{60}^o}} \right)$ which is $\dfrac{{\sqrt 3 }}{2}$, the value of $\sin \left( {{{30}^o}} \right)$ is $\dfrac{1}{2}$, $\cos \left( {{{60}^o}} \right)$ is $\dfrac{1}{2}$ and $\cos \left( {{{30}^o}} \right)$ is $\dfrac{{\sqrt 3 }}{2}$, so if we simplify each and every option statement and see that which option simplification is equal to the question simplification than that option will correspond to the right answer. However it is a much lengthy and time taking process so it is better to use the first method only.