First notice that we can deduct some things from the equation:
cos(cos^-1(-pi/3))cos(cos−1(−π3))
We can't have cos^-1(-pi/3)cos−1(−π3) , it is outside the domain of the function f(x)=cos^-1(x)f(x)=cos−1(x).
Howeve, because cos^-1(cos(x))=xcos−1(cos(x))=x , the only way I see to move on, is to apply the same concept.
That means the equation from above tells us, that at some point, some angle was converted into -pi/3−π3 radiens.
Imagine -pi/3−π3 radiens on the unit circle. This angle is not within 0<=theta<=pi0≤θ≤π . In fact this translates to the upper half of the unit circle (see illustrations).![enter image source here]()
You can see that:
Cos(-pi/3)=cos(2pi-pi/3)cos(−π3)=cos(2π−π3)
Now calculate cos(2pi-pi/3)=0.5cos(2π−π3)=0.5
Using your prefered graph tool, you can also see that cos(x)=0.5cos(x)=0.5 has many more solutions. One particular solution exist in between cos(-pi/3)cos(−π3) and cos(2pi-pi/3)cos(2π−π3), that is cos(pi/3)cos(π3).
This means:
cos(-pi/3)=cos(2pi-pi/3)=cos(pi/3)cos(−π3)=cos(2π−π3)=cos(π3)
As you can see:
only one solution satisfies, 0<=theta<=pi0≤θ≤π , that is theta = pi/3θ=π3
Imagine theta=pi/3= cos^-1(cos(pi/3))=cos^-1(cos(-pi/3))θ=π3=cos−1(cos(π3))=cos−1(cos(−π3)) , because cos(pi/3)=cos(-pi/3)cos(π3)=cos(−π3)
and so if cos(cos^-1(u))=ucos(cos−1(u))=u then we can write:
u=-pi/3u=−π3
cos^-1(cos(-pi/3))=cos^-1(cos(u))=u=cos(cos^-1(u))=cos(cos^-1(-pi/3))cos−1(cos(−π3))=cos−1(cos(u))=u=cos(cos−1(u))=cos(cos−1(−π3))
To conclude, when theta=pi/3θ=π3 , it satisfies 0<=theta<=pi0≤θ≤π , and that can be rewritten to form the equation cos(cos^-1(-pi/3))=pi/3cos(cos−1(−π3))=π3 if cos(cos^-1(u))=ucos(cos−1(u))=u
Idon't know... It makes sense for me to do it this way, but I would like someone to double check it for fundamental flaws :)
