De Moivre's Theorem
((2cosx)^3)*((2jsinx)^4)=(z+z^-1)^3*(z-z^-1)^4
I have this down in my notes as a way of solving for cos^3xsin^4x.
I understand you can use binomial expansion, then replace certain parts with their respective multiple angle formula. But I remember my teacher mentioning another less cumbersome way of doing it, which I didn't write down :/ Does anyone know what they might have been talking about, or just a different way of doing it?
Thanks in advance.
I have this down in my notes as a way of solving for cos^3xsin^4x.
I understand you can use binomial expansion, then replace certain parts with their respective multiple angle formula. But I remember my teacher mentioning another less cumbersome way of doing it, which I didn't write down :/ Does anyone know what they might have been talking about, or just a different way of doing it?
Thanks in advance.
Original post by Chromatic
((2cosx)^3)*((2jsinx)^4)=(z+z^-1)^3*(z-z^-1)^4
I have this down in my notes as a way of solving for cos^3xsin^4x.
I understand you can use binomial expansion, then replace certain parts with their respective multiple angle formula. But I remember my teacher mentioning another less cumbersome way of doing it, which I didn't write down :/ Does anyone know what they might have been talking about, or just a different way of doing it?
Thanks in advance.
I have this down in my notes as a way of solving for cos^3xsin^4x.
I understand you can use binomial expansion, then replace certain parts with their respective multiple angle formula. But I remember my teacher mentioning another less cumbersome way of doing it, which I didn't write down :/ Does anyone know what they might have been talking about, or just a different way of doing it?
Thanks in advance.
What is there to solve?
Original post by NotNotBatman
What is there to solve?
Probably didn't make it clear enough. I'm not solving, I'm simplifying (z+1/z)^3(z-1/z) ^4 into multiple angle form, ultimately to get equations for cos cubed x and sin ^4 x in terms of multiple angles.
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