After that, we will solve the obtained equation for cosθ. Then, we will draw the unit circle on the coordinate plane. Let's do these things one at a time.
Rewriting the Equation
Note that we will need to rearrange the terms of the Pythagorean Identity so that we can use it for our expression.
sin2θ+cos2θ=1⇕sin2θ=1−cos2θ
We will start by rewriting sin2θ as 1−cos2θ. Then we will simplify the equation. Let's do it!
To further simplify the equation, we need to separate the right-hand side into two cases.
z=41±3
z=41+3
z=41−3
z=44
z=4-2
z=1
z=-21
Since z=cosθ and we found the values of z, we can rewrite the results in terms of cosθ.
{z=1z=-21⇒{cosθ=1cosθ=-21(I)(II)
We obtained two values for cosθ. Recall that the cosine of an angle in standard position is the x-coordinate of the point of intersection P of its terminal side and the unit circle.
P(x,y)=(cosθ,sinθ)
Let's solve the two equations one at a time.
cosθ=1
To solve the first equation, cosθ=1, we need to consider the points on the unit circle that have an x-coordinate of 1.
This is the only solution to the first equation on the unit circle. Therefore, all the solutions to the equation cosθ=1 are 0 plus a multiple of 2π.
0+2πnradians, wherenisanyinteger
cosθ=-21
To solve the second equation, cosθ=-21, we need to consider the points on the unit circle that have an x-coordinate of -21. Recall that cosine is negative in Quadrants II and III.
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 x-coordinate of both points is -21, the length of the leg which is on the x-axis of each triangle is 21.
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 angle adjacent the shorter leg measures 60∘ or 3πradians. With this information and knowing that half a turn measures π radians, we can calculate the desired angle measures. To do so, we will do two things.
Subtract 3π from π.
Add 3π to π.
Let's do it!
We found two solutions to the given equation, θ=32π and θ=34π. Finally, keep in mind that if we add or subtract a multiple of 2π radians, the terminal side of the angle will be in the same position. Therefore, the resulting angles will also be solutions to the given equation.
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