We consider tan(3A) as tan(2A+A) and then
we use the formula for tan(a+b) which is this:
and we substitute 2A for a and A for b:
equation 1:
But equation 1 is not enough, for we must now find tan(2A).
We find that by considering 2A as A+A and use this formula again
substituting A for a and also A for b:
Next we must substitute for tan(2A) in equation 1:
To simplify that compound fraction multiply top and bottom
by (1-tan2A)
Edwin
