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