HATLEY1997
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- Oct 24, 2023
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Complex numbers
- Thread starter HATLEY1997
- Start date
If you compare your result with the required one it is apparent you made a mistake at the start of the derivation. I recommend using the binomial theorem combined with the exponential form of complex numbers.I have got this down to the real numbers (I think). But struggling with what to do at the end here to get the required trig identity?
any help?
de Moivre says [imath] (\cos(x) +i \sin(x))^n =\cos(nx) + i \sin(nx) [/imath] and we are interested in [imath] \cos(5x). [/imath] Hence,
[math]\begin{array}{lll} \cos(5x)&=(\cos(x) +i \sin(x))^5-i \sin(5x)\\ &=\cos^5(x) -10\cos^3(x)\sin^2(x)+5\cos(x)\sin^4(x) + i f(\cos(x),\sin(x))\\ &=\cos^5(x)-10\cos^3(x)(1-\cos^2(x))+5\cos(x)(1-\cos^2(x))^2 + i f(\cos(x),\sin(x))\\ &=\ldots \end{array}[/math]with some function [imath] f(\cos(x),\sin(x)) [/imath] which we are not interested in since it has to be zero anyway.
[math]\begin{array}{lll} \cos(5x)&=(\cos(x) +i \sin(x))^5-i \sin(5x)\\ &=\cos^5(x) -10\cos^3(x)\sin^2(x)+5\cos(x)\sin^4(x) + i f(\cos(x),\sin(x))\\ &=\cos^5(x)-10\cos^3(x)(1-\cos^2(x))+5\cos(x)(1-\cos^2(x))^2 + i f(\cos(x),\sin(x))\\ &=\ldots \end{array}[/math]with some function [imath] f(\cos(x),\sin(x)) [/imath] which we are not interested in since it has to be zero anyway.