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.