“All you need is sine”: 
	Identities Between ArcCos and ArcSin, 
	and ArcTan and ArcSin.
Andy Long, 1/28/2019

ArcCos in terms of ArcSin

As we know, “All you need is sine”: Cosine is just a shifted Sine:

Plot[{Cos[x],Sin[x]},{x,-2Pi,2Pi}]

In particular, we could just shift Sine left by Pi/2:

Plot[{Cos[x],Sin[x+Pi/2]},{x,-2Pi,2Pi}]

There are trickier ways: we could flip Sine vertically and shift right by Pi/2:

Plot[{Cos[x],-Sin[x-Pi/2]},{x,-2Pi,2Pi}]

Now, to talk about inverses, we need to restrict the domains of Sine and Cosine:

I’ll do this by making the Coss and Sinn functions equal to Cos and Sin, respectively on their restricted domains, and imaginary outside of those:

Sinn[x_]:=If[x>Pi/2|| x<-Pi/2,I,Sin[x]]
Coss[x_]:=If[x>Pi|| x<0,I,Cos[x]]
Plot[{Sinn[x],Coss[x]},{x,-Pi/2,Pi},AspectRatio->1/2]

It’s important to realize that these choices are arbitrary: there are infinitely many “branches” of each function that could serve. But it’s tradition to restrict Sine to [-Pi/2,Pi/2] (and that makes sense, since the restriction has an extra property -- is odd -- if we choose that particular “branch”). 

Cosine is a little trickier: the obvious two choices are [0,Pi] and [-Pi,0] -- one to the right, the other to the left of the origin. Traditionally [0,Pi] is chosen (so that the arccos will give positive angles, I presume).

Plot[{ArcSin[x],ArcCos[x]},{x,-2,2},AspectRatio->1]

Expressing ArcCos in terms of ArcSin

From the picture above, it’s clear how one want to do this: flip ArcSin about the x-axis, then lift it by Pi/2:

ArcCos[x]=Pi/2-ArcSin[x]

There is another way of seeing this, however, rather than simply by allowing Mathematica to plot it:

y = Coss[x] 
iff
x=ArcCos[y]

y = Coss[x] = -Sinn[x-Pi/2]

so

ArcSin[Coss[x]] = ArcSin[ -Sinn[x-Pi/2]]
but since ArcSin is odd,
ArcSin[ -Sinn[x-Pi/2]] = -ArcSin[ Sinn[x-Pi/2]] = -(x-Pi/2))

but 

ArcSin[ -Sinn[x-Pi/2]] = ArcSin[ y ]

so

ArcSin[ y ] = -(x-Pi/2))

and

x=Pi/2 - ArcSin[y]

But x = ArcCos[y], so

ArcCos[y] = Pi/2 - ArcSin[y]

Here’s everyone plotted happily together on the same graph. Looks like some kind of insect to me...:)

Plot[{ArcSin[x],ArcCos[x],Coss[x],Sinn[x],x},{x,-Pi/2,Pi},AspectRatio->1,PlotStyle->{Black,Black,Red,Red, Dashed}]

ArcTan in terms of ArcSin

Tangent, too, can now be written in terms of Sine (and a shifted Sine, since Cosine is a shifted Sine) --  and we could rewrite all the other trig functions using only Sines:

Tann[x_]:=Sin[x]/Sin[x+Pi/2]
Plot[Tann[x],{x,-2Pi,2Pi}]

Now to talk about inverses, we need to restrict the domains of Sine and Cosine:

I’ll do this by making the Tann function equal to Tan on its restricted domains, and imaginary outside:

Tann[x_]:=If[x>Pi/2|| x<-Pi/2,I,Sin[x]/Sin[x+Pi/2]]
Plot[Tann[x],{x,-Pi,Pi},AspectRatio->1/2]

It’s important to realize that this choice is arbitrary: there are infinitely many “branches” of the tangent function that could serve. But it’s tradition to restrict Tan to [-Pi/2,Pi/2] (and that makes sense, since the restriction has an extra property -- is odd -- if we choose that particular “branch”).

Plot[{ArcTan[x],Tann[x]},{x,-Pi,Pi},AspectRatio->1]

Expressing ArcTan in terms of ArcSin

y = ArcTan[x] 
iff
x=Tann[y]

Since Sin^2[y]+Cos^2[y] = 1, 
Tann^2[y] +1 = 1/Cos^2[y]

so

x^2 +1 = 1/Cos^2[y]

and

Cos^2[y] = 1/(x^2+1)

but 

Cos^2[y] = 1 - Sin^2[y],

so 
 
 Cos^2[y] = 1/(x^2+1)= 1 - Sin^2[y],
 
 and therefore
 
Sin^2[y] = 1 - 1/(x^2+1) = x^2/(x^2+1) 

Now take square roots, and realize that Tann[y]=x iff their signs agree: so we choose the “plus” root, and 
 
Sin[y] =  x/Sqrt[x^2+1]

Now  apply ArcSin to both sides, and 

y = ArcTan[x]  = ArcSin[ x/Sqrt[x^2+1]]

Here’s everyone plotted happily together on the same graph. Everything here can be expressed in terms of Sines!

Plot[{ArcSin[x/Sqrt[x^2+1]],Tann[x],x},{x,-2Pi,2Pi},AspectRatio->1,PlotStyle->{Black,Red,Dashed}]