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Question
find reference angle sin(-45)
Step1: Use the property of sine function
Since \(\sin(-\alpha)=-\sin\alpha\), for \(\alpha = 45^{\circ}\), we have \(\sin(-45^{\circ})=-\sin(45^{\circ})\)
Step2: Calculate the value of \(\sin(45^{\circ})\)
We know that \(\sin(45^{\circ})=\frac{\sqrt{2}}{2}\)
Step3: Find the value of \(\sin(-45^{\circ})\)
Substituting the value of \(\sin(45^{\circ})\) into \(\sin(-45^{\circ})=-\sin(45^{\circ})\), we get \(\sin(-45^{\circ})=-\frac{\sqrt{2}}{2}\)
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\(-\frac{\sqrt{2}}{2}\)