Question

The value of $\sin \left( {60^\circ + \theta } \right) - \cos \left( {30^\circ - \theta } \right)$ is equal to
A.$2\cos \theta $
B.$2\sin \theta $
C.0
D.1

Answer 1

Hint- In this particular type of question we need to use the trigonometric formula for sin(a+b) and cos(a-b). To further solve it we need to expand and simplify to get the desired answer.

Complete step-by-step answer:
Since
$
  \sin \left( {a + b} \right) = \sin a.\cos b + \cos a.\sin b \\
  and{\text{ cos}}\left( {a - b} \right) = \cos a.\cos b + \sin a.\sin b \\
$
\[
  \sin \left( {60^\circ + \theta } \right) - \cos \left( {30^\circ - \theta } \right) \\
   = \sin 60^\circ .\cos \theta + \cos 60^\circ .sin\theta - \left( {\cos 30^\circ .\cos \theta + \sin 30^\circ .\sin \theta } \right) \\
   = \dfrac{{\sqrt 3 }}{2}.\cos \theta + \dfrac{1}{2}.\sin \theta - \dfrac{{\sqrt 3 }}{2}.\cos \theta - \dfrac{1}{2}\sin \theta \\
   = 0 \\
\]
Note-
Remember to recall the basic trigonometric formulas while solving such types of questions. Note that this question could also be solved by directly converting $\sin \left( {60^\circ + \theta } \right)$ into $\sin \left( {90^\circ - \left( {30^\circ - \theta } \right)} \right)$ and further into $\cos \left( {30^\circ - \theta } \right)$ , thus cancelling out both the terms to get 0.