Question

Simplify the given expression.
 $ \sin \left( {\pi - \theta } \right) + \cos \left( {\dfrac{{3\pi }}{2} + \theta } \right) + \sin \left( {\pi + \theta } \right) + \cos \left( {\dfrac{{3\pi }}{2} - \theta } \right) $

Answer 1

Hint: In this particular type of question we need to simplify the trigonometric functions by using the formulas of $ \sin \left( {\pi - \theta } \right) = \sin \theta ,\cos \left( {\dfrac{{3\pi }}{2} + \theta } \right) = \cos \theta ,\sin \left( {\pi + \theta } \right) = - \sin \theta {\text{ and }}\cos \left( {\dfrac{{3\pi }}{2} - \theta } \right) = - \cos \theta $ keeping in mind their signs in different quadrants. After simplifying we need to solve the trigonometric equation to get the desired answer.



Complete step-by-step answer:
The expression given is,
 $ \sin \left( {\pi - \theta } \right) + \cos \left( {\dfrac{{3\pi }}{2} + \theta } \right) + \sin \left( {\pi + \theta } \right) + \cos \left( {\dfrac{{3\pi }}{2} - \theta } \right) $
Now we will substitute the terms by the trigonometry formula. i.e.
  $ \sin \left( {\pi - \theta } \right) = \sin \theta ,\cos \left( {\dfrac{{3\pi }}{2} + \theta } \right) = \cos \theta ,\sin \left( {\pi + \theta } \right) = - \sin \theta {\text{ and }}\cos \left( {\dfrac{{3\pi }}{2} - \theta } \right) = - \cos \theta $

 $ \begin{gathered}
   \Rightarrow \sin \theta + \cos \theta - \sin \theta - \cos \theta \\
   = 0 \\
\end{gathered} $
( Since sin $ \theta $ is positive in $ \pi - \theta $ and negative in $ \pi + \theta $ and cos $ \theta $ is positive in $ \dfrac{{3\pi }}{2} + \theta $ and negative in $ \dfrac{{3\pi }}{2} - \theta $ )

Note-
It is important to recall that Sin and Cos are positive in the first - second quadrant and the first fourth - quadrant respectively. Note that Sin and Cos are both positive in the first quadrant but differ in the others.