Bobosharif S. answered • 01/15/18
Tutor
4.4
(32)
Mathematics/Statistics Tutor
cos(Θ+i*Φ)=
= cos(Θ) cosh(Φ) -i sin(Θ) sinh(Φ)=cos(α)+i sin(α)
Now take absolute value from both sides of
cos(Θ) cosh(Φ) -i sin(Θ) sinh(Φ)=cos(α)+i sin(α)
(cos(Θ) cosh(Φ))2 +(sin(Θ) sinh(Φ))2=1
(1/2)*(cos(2Θ)+cosh(2Φ))=1. This is your second equation.
You get the first equation by dividing real and imaginary parts.
Bobosharif S.
Alright, we have
cos(Θ) cosh(Φ) -i sin(Θ) sinh(Φ)=cos(α)+i sin(α)
Form here
sin(α)=sin(Θ) sinh(Φ) <---- (1)
and
cos(α)=cos(Θ) cosh(Φ) <-----(2)
Square (1) and (2)
sin2(α)=sin2(Θ) sinh2(Φ) <----(1')
and
cos2(α)=cos2(Θ) cosh2(Φ) <----(2')
and
cos2(α)=cos2(Θ) cosh2(Φ) <----(2')
Now from (1')
sin2(α)=sin2(Θ) sinh2(Φ)=
=sin2(Θ) (cosh2(Φ)-1)=
=|cosh2(Φ)=cos2(α)/cos2(Θ) from (2')|=
=sin2(Θ) (cos2(α)/cos2(Θ)-1)=
=tan2(Θ) *cos2(α)-sin2(Θ)=
=tan2(Θ)-tan2(Θ) *sin2(α)-sin2(Θ).
So, what we have is
sin2(α)=tan2(Θ)-tan2(Θ) *sin2(α)-sin2(Θ).
Now
sin2(α)(1+tan2(Θ))=tan2(Θ)-sin2(Θ)
sin2(α)=cos2(Θ)tan2(Θ)-cos2(Θ)sin2(Θ)=sin4(Θ). (**)
From (**) you get
sin2(Θ)=±sin(α). This is (i).
Hope it helps. It is hard to type here. For me it is much easier to write on a paper.
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01/16/18
Ken T.
01/16/18