Defining Circular Functions - Questions - Class Mathematics School Portal

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Question 1

19030

Answer

\(-1\)

Worked Solution

\begin{align}
\sin \frac{3 \pi}{2} &=\sin \left(\pi+\frac{\pi}{2}\right) \\
&=-\sin \frac{\pi}{2} \\
&=-1
\end{align}

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