We obtained two equations for θ. Let's solve them one at a time.
cosθ=0
Let's solve the first equation.
cosθ=0
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)
We want to find the values of θ which are the solutions to our equation, cos θ =0. To do so, we need to consider the points on the unit circle that have the x-coordinate of 0.
We found two points that have 0 as the x-coordinate. Knowing that full turn measure is 360^(∘), we can find the desired angle measures.
We found two solutions to the equation cosθ=0.
θ = 90^(∘) and θ= 270^(∘)
tanθ=1
Now, let's solve the second equation.
tanθ=1
Suppose (x,y) is the point of intersection at which the terminal side of the angle θ intersects the unit circle. Then the ratio yx is the tangent of θ.
y/x=tan θ
We want to find the values of θ which are the solutions to our equation, tan θ = 1.
tan θ =1 ⇔ y/x=1⇔ y=x
To do so, we need to consider the points on the unit circle with equal x- and y-coordinates.
We can see that there are two points on a unit circle that have the same equal x- and y-coordinates. Now, let's 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- and y-coordinates are the same, the lengths of the legs are equal.
For both right triangles, the legs have equal lengths. Therefore, they are 45^(∘)-45^(∘)-90^(∘) triangles. In this type of triangle, both acute angles measure 45^(∘) or π4 radians. With this information, plus knowing that a half turn measures 180 ^(∘) and a full turn measures 360^(∘), we can calculate the desired angle measures.
We found two solutions for the equation tan θ=1.
θ= 45^(∘) and θ= 225^(∘)
All Angles
In total we obtained four solutions to the original equation.
θ= 90^(∘), θ= 270^(∘), θ= 45^(∘), andθ= 225^(∘)
Keep in mind that if we add or subtract a multiple of 360^(∘), the terminal side of the angle will be in the same position. This means that resulting angles will also be the solutions to the original equation. Therefore, we can now write all angles which are solutions to the original equation.
All Solutions
45^(∘)+ k * 360^(∘)
270^(∘)+ k * 360^(∘)
90^(∘)+ k * 360^(∘)
225^(∘)+ k * 360^(∘)
In the solutions above, k is any integer. Let's try to simplify this, starting with 45^(∘)+ k * 360^(∘).
Notice that 2k is even. Therefore, the obtained expression is the same as the following.
45^(∘)+ even integer * 180^(∘)
Now, let's simplify 225^(∘)+ k * 360^(∘).
Notice that 2k+1 is not even. This time the obtained expression is the same as the following.
45^(∘)+ odd integer * 180^(∘)
We can write both obtained expressions as a single expression.
45^(∘)+ odd integer * 180^(∘) 45^(∘)+ even integer * 180^(∘) ⇕ 45^(∘)+ any integer* 180 ^(∘)
To make this expression even simpler we will call that integer n. Let's summarize what we managed to simplify.
45^(∘)+ k * 360^(∘) 225^(∘)+ k * 360^(∘) ⇒ 45^(∘)+n* 180 ^(∘)
We can do the same with the other two expressions.
90^(∘)+ k * 360^(∘) 270^(∘)+ k * 360^(∘) ⇒ 90^(∘)+n* 180 ^(∘)
Finally, we can state the simplified set of all solutions to the original equation.
