Find the exact value for each trigonometric function
Final Answers:
- (a) sin(2π3)=32\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}
- (b) sin(4π3)=−32\sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}
- (c) sin(9π2)=1\sin\left(\frac{9\pi}{2}\right) = 1
Explanation
To solve these problems, we use the unit circle and the concept of reference angles. The sine function corresponds to the y-coordinate of a point on the unit circle.
(a) sin(2π3)\sin\left(\frac{2\pi}{3}\right) The angle 2π3\frac{2\pi}{3} lies in the second quadrant. The reference angle is π−2π3=π3\pi – \frac{2\pi}{3} = \frac{\pi}{3}. In the second quadrant, sine values are positive. Since sin(π3)=32\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}, we have:
sin(2π3)=32\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}
(b) sin(4π3)\sin\left(\frac{4\pi}{3}\right) The angle 4π3\frac{4\pi}{3} lies in the third quadrant. The reference angle is 4π3−π=π3\frac{4\pi}{3} – \pi = \frac{\pi}{3}. In the third quadrant, sine values are negative. So,
sin(4π3)=−32\sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}
(c) sin(9π2)\sin\left(\frac{9\pi}{2}\right) This angle is larger than 2π2\pi, so we reduce it by subtracting full rotations. Each full rotation is 2π2\pi, or 4π2\frac{4\pi}{2}.
9π2−4π=9π2−8π2=π2\frac{9\pi}{2} – 4\pi = \frac{9\pi}{2} – \frac{8\pi}{2} = \frac{\pi}{2}
Now we evaluate sin(π2)\sin\left(\frac{\pi}{2}\right), which equals 1. So,
sin(9π2)=1\sin\left(\frac{9\pi}{2}\right) = 1
These problems test your understanding of angle placement on the unit circle, quadrant rules, and reference angles. Mastery of these ideas is essential for solving trigonometric equations and analyzing periodic behavior in advanced mathematics.
