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  sqrt{(2 + 2 cos theta)} 

By inspection from the above diagram:

    2theta = arctan( {sin(theta)}/{cos(theta)} )  = 2arctan({sin(theta)} / {1+cos(theta)})       

 

Looking at this with the double angle tangent formula

    tan(2theta) = {2tan(theta)}/{1+tan^2(theta)}      

    y = {2x}/{1+x^2}        – the above is of this form.  Solving for x

    yx^2 + 2x - y = 0      

    x = { -1 +  sqrt(1+y^2)} / {y}      

   tan(theta) = { -1 +  sqrt(1+tan(2theta)^2)} / {tan(2theta)}      

   tan(theta) = { -1 +   1/(cos(2theta))} / {tan(2theta)}      

   tan(theta) = { -1 + cos(2theta)} / {sin(2theta)}      

Example 1: 

Using:      tan(theta) = { -1 +  sqrt(1+tan(2theta)^2)} / {tan(2theta)}        with   tan(2theta) = 1       

   tan(theta) = { -1 +  sqrt(1+tan(2theta)^2)} / {tan(2theta)}      

   tan(theta) = { -1 +  sqrt(1+1^2)} / {1} = 0.4142          so      tan(22.5) = 0.4142      

Example 2: 

Using:      atan( {0.3420}/{9397} ) = 19.9999          with     sin(20) = 0.3420    cos(20) = 0.9397      

 

Which brings me to questions about the multiplication of vectors

   x_2(3,1) + y_2(-1,3) = (5,5)        

   tan^-1(1/3) = 18.4349    

   tan^-1(1/2) = 26.565    

   tan^-1(1/3) + tan^-1(1/2) = 45    

 

 

 

 

 

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