*This is exactly how the question reads from the book
Using the fact that a complex equation is really two real equations, find the double angle formulas (for Sin2 , Cos2 ) by using this equation: (e^i )^n = (Cos + iSin )^n = Cos n + iSin n
I know if you put in 2 for n and just exapand it out I can get it but I figured that there might be some other way to prove it. Just seems a bit simple but any help or insight would be greatly appriciated.
Hello, gidget!
I'm truly grateful for this method of cranking out multiple-angle identities.Using the fact that a complex equation is really two real equations,
find the double-angle formulas for
by using this equation:
I know if you put in and just expand it out. I can get it.
But I figured that there might be some other way to prove it.
. . I don't know of a simpler way.
If we want a formula for , we'd normally proceed like this:
. . . .
. . . .
. . . .
. . . .
. . . .
Therefore: .
Then we must repeat the process for
Using Demoivre's Formula, we have:
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
. . . . . . . . . . . .
Equating real and imaginary components, we have:
. . . .
And we have both formulas . . . with much less work!