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(2)2, we need to consider the points on the unit circle that have a y-coordinate of sqrt(2)2. Recall that sine is positive in Quadrants I and II.
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(2)2, the length of the leg which is not on the x-axis of each triangle is sqrt(2)2.
For both right triangles one of the legs is half the hypotenuse, so they are 45^(∘)-45^(∘)-90^(∘) triangles. In this type of triangle, both acute angles measure 45^(∘) or π4 radians. Knowing that a half turn measures π radians, we can calculate the desired angle measures. Since π4 is in the Quadrant I, it is already simplified. Therefore, we will subtract π4 from 2π to get the second angle.
We found two solutions for the equation sin θ= sqrt(2)2. These are also the solutions for the given equation.
θ= 3π4 radians and θ= π4 radians
