We obtained the value for sin θ. Recall that the sine of an angle in standard position is the y-coordinate of the point of intersection P of its terminal side and the unit circle.
P(x,y)=(cosθ,sinθ)
To solve the equation sin θ = - sqrt(3)2, we need to consider the points on the unit circle that have a y-coordinate of - sqrt(3)2. Recall that sine is negative in Quadrants III and IV.
We can now draw two congruent right triangles, each with a leg on the x-axis. Since the radius of the unit circle is 1, the hypotenuse of each triangle is also 1. Furthermore, since the y-coordinate of both points is - sqrt(3)2, the length of the leg which is not on the x-axis of each triangle is sqrt(3)2.
For both right triangles one of the legs is half the hypotenuse, so they are 30^(∘)-60^(∘)-90^(∘) triangles. In this type of triangle, the medium angle measures 60^(∘), or π3 radians. With this information and knowing that a half turn measures π radians and a full turn measures 2π radians, we can calculate the desired angle measures. We will add π3 to π, and subtract π3 from 2π.
We found two solutions for the equation sin θ=- sqrt(3)2. These are also the solutions for the given equation.
θ= 4π3 radians and θ= 5π3 radians
