home > s25 ma2090 - dambroise precalculus > assessment\nhomework #16 - inverse sine, cosine, tangent\nscore…

Explanation:

Step1: Use the periodicity of sine function

The sine - function (y = \sin(x)) has a period of (2\pi). So, (\sin\left(\frac{44\pi}{7}\right)=\sin\left(6\pi+\frac{2\pi}{7}\right)). Since (\sin(x + 2k\pi)=\sin(x)) for any real - number (x) and integer (k), and here (k = 3), (x=\frac{2\pi}{7}), we have (\sin\left(\frac{44\pi}{7}\right)=\sin\left(\frac{2\pi}{7}\right)).

Step2: Recall the property of inverse - sine function

The inverse - sine function (y=\arcsin(u)) has a domain (- 1\leq u\leq1) and range (\left[-\frac{\pi}{2},\frac{\pi}{2}\right]). And (\arcsin(\sin(x))=x) when (x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]). Since (\frac{2\pi}{7}\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]), then (\arcsin\left(\sin\left(\frac{44\pi}{7}\right)\right)=\arcsin\left(\sin\left(\frac{2\pi}{7}\right)\right)=\frac{2\pi}{7}).

Answer:

(\frac{2\pi}{7})