First, we use the important identity sin2theta=2sinthetacosthetasin2θ=2sinθcosθ:
sin2theta-costheta=0sin2θ−cosθ=0
->2sinthetacostheta-costheta=0→2sinθcosθ−cosθ=0
Now we can factor out costhetacosθ:
2sinthetacostheta-costheta=02sinθcosθ−cosθ=0
->costheta(2sintheta-1)=0→cosθ(2sinθ−1)=0
And using the zero product property, we obtain solutions of:
costheta=0" and "2sintheta-1=0->sintheta=1/2cosθ=0 and 2sinθ−1=0→sinθ=12
So, when does costheta=0cosθ=0 on the interval -pi/2<=theta<=(3pi)/2−π2≤θ≤3π2? The solutions can be found by using the unit circle and a property of the cosine function:
cos(-theta)=costhetacos(−θ)=cosθ
If theta=pi/2θ=π2, then:
cos(-pi/2)=cos(pi/2)cos(−π2)=cos(π2)
From the unit circle, we know that cos(pi/2)=0cos(π2)=0, which also means cos(-pi/2)=0cos(−π2)=0; so two solutions are -pi/2−π2 and pi/2π2. Also, the unit circle tells us that cos((3pi)/2)=0cos(3π2)=0, so we have another solution there.
Now, onto sintheta=1/2sinθ=12. Again, we will need the unit circle to find our solutions.
We know from the unit circle that sin(pi/6)=1/2sin(π6)=12, and sin((5pi)/6)=1/2sin(5π6)=12, so we add pi/6π6 and (5pi)/65π6 to the list of solutions.
Finally, we put all of our solutions together: theta=-pi/2, pi/6, pi/2, (5pi)/6θ=−π2,π6,π2,5π6, and (3pi)/23π2.
The Unit Circle
![From Math is Fun]()
