Question

find the exact value.
sin \frac{9\pi}{4}
type + or -
? \frac{\sqrt{}}{ }

Explanation:

Step1: Find the coterminal angle

Since \(\frac{9\pi}{4}=2\pi+\frac{\pi}{4}\), and \(\sin(x + 2\pi)=\sin x\) (periodicity of sine function), so \(\sin\frac{9\pi}{4}=\sin(2\pi+\frac{\pi}{4})\).
By the periodicity formula \(\sin(x + 2k\pi)=\sin x,k\in\mathbb{Z}\), here \(k = 1\), then \(\sin(2\pi+\frac{\pi}{4})=\sin\frac{\pi}{4}\).

Step2: Evaluate \(\sin\frac{\pi}{4}\)

We know that for the unit - circle definition, in a right - triangle with an angle of \(\frac{\pi}{4}\) (or \(45^{\circ}\)), if the hypotenuse \(r = 1\) and the opposite side \(y\) and adjacent side \(x\) satisfy \(x=y\) (because \(\tan\frac{\pi}{4}=1=\frac{y}{x}\)) and \(x^{2}+y^{2}=r^{2}=1\). Then \(2y^{2}=1\), \(y=\frac{\sqrt{2}}{2}\). And \(\sin\theta=\frac{y}{r}\), so \(\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}\).

Answer:

\(+\frac{\sqrt{2}}{2}\)