How do you solve $\sin \left( \dfrac{-5\pi }{2} \right)$ ? \[\]

Hint: We recall the unit circle definition of sine of an angle and the different values of sine in different quadrant $xy-$ Cartesian plane. Since we are given argument of sine in negative we use the identities $\sin \left( -\theta \right)=-\sin \theta $ and shift formula $\sin \left( 2\pi +\theta \right)=\sin \theta $ to find the result. \[\]

Complete step by step answer:
 Let us consider the below circle and the right angled triangle $OAB$.\[\]
Here we have sine of angle $\angle AOB$ as
\[\sin \theta =\dfrac{y}{h}\]
We see that the above sine values will be repeated after a full rotation $2\pi $ radian of OA in anti-clockwise. The value will be reflected with a negative sign with half-rotation $\pi $radian. We know that the sine is positive in the first and second quadrant and negative in the third-fourth quadrant. The repetition of values with periodicity of sine are called shift formula and it is given with arbitrary integer $k$ as
\[\begin{align}
  & \sin \left( \theta +2k\pi \right)=\sin \theta \\
 & \sin \left( \theta +k\pi \right)=-\sin \theta \\
\end{align}\]
We also know in the conventional sense angle is measured anti-clockwise direction. If it is measured in clockwise direction we have $\sin \left( -\theta \right)$. Using complementary angle relation we have
\[\begin{align}
  & \sin \left( \theta \right)=\cos \left( \dfrac{\pi }{2}-\theta \right) \\
 & \Rightarrow \sin \left( \theta \right)=\cos \left( \dfrac{\pi }{2}+\left( -\theta \right) \right) \\
\end{align}\]
We use the cosine sum of angles formula $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$ for $A=\dfrac{\pi }{2},B=\theta $ in the above step to have;
\[\begin{align}
  & \Rightarrow \sin \left( \theta \right)=\cos \left( \dfrac{\pi }{2} \right)\cos \left( -\theta \right)-\sin \left( \dfrac{\pi }{2} \right)\sin \left( -\theta \right) \\
 & \Rightarrow \sin \theta =0\cdot \cos \left( -\theta \right)-1\cdot \sin \left( -\theta \right) \\
 & \Rightarrow \sin \theta =-\sin \left( -\theta \right) \\
 & \Rightarrow \sin \left( -\theta \right)=-\sin \theta \\
\end{align}\]
We are asked to find the value of $\sin \left( \dfrac{-5\pi }{2} \right)$. We use the above trigonometric identity for $\theta =\dfrac{5\pi }{2}$ to have ;
\[\begin{align}
  & \sin \left( -\dfrac{5\pi }{2} \right)=-\sin \left( \dfrac{5\pi }{2} \right) \\
 & \Rightarrow \sin \left( -\dfrac{5\pi }{2} \right)=-\sin \left( 2\pi +\dfrac{\pi }{2} \right) \\
\end{align}\]
We us shift by full turn formula that is $\sin \left( 2\pi +\theta \right)=\sin \theta $ for $\theta =\dfrac{\pi }{2}$ in the above step to have;
\[\Rightarrow \sin \left( \dfrac{-5\pi }{2} \right)=-\sin \left( \dfrac{\pi }{2} \right)=-1\]

Note: We note from the trigonometric ratios in a right angled triangle the sine of any angle is given by the ratio of length of side opposite to the angle to length of hypotenuse. If $f\left( -x \right)=-f\left( -x \right)$ then we call $f$ an odd function and if $f\left( -x \right)=f\left( x \right)$ then we call $f$ an even function. . Here $\sin x$ is an odd function. Unlike sine, cosine is even function because $\cos \left( -x \right)=\cos x$.