How do you derive the multiple angles formula?

1 Answer
Apr 12, 2015

One approach is to use repeated applications of the sum formulas.

#trig(2a) = trig (a+a)#

#trig(3a) = trig (2a+a)#

#trig(4a) = trig (3a+a)# (or #=trig(2a+2a)#)

Another approach is to rewrite using Euler's Formula: #e^(i n x) = cos(nx)+i sin(nx)#

So #sin(nx) = (e^( i n x) - e^( - i n x))/(2 i) = ((e^( i x))^n - (e^( - i x))^n)/(2 i)#

#= ((cosx+isinx)^n - (cosx-isinx)^n)/(2 i)#.

Now expand using the binomial theorem, and simplify.

Here's an outside link with the details:

Multiple Angle Formulas