Cos3x=4cos^3x-3cosx prove

1 answer:

6 0

$\bf cos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta)
\\\\\\
\textit{also recall that }cos(2\theta)=
\begin{cases}
cos^2(\theta)-sin^2(\theta)\\
1-2sin^2(\theta)\\
\boxed{2cos^2(\theta)-1}
\end{cases}
\\\\\\
and\qquad sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)=1-cos^2(\theta)\\\\
-------------------------------$

$\bf cos(3x)=4cos^3(x)-3cos(x)\\\\
-------------------------------\\\\
cos(3x)\implies cos(2x+x)\implies cos(2x)cos(x)-sin(2x)sin(x)
\\\\\\\
[2cos^2(x)-1]cos(x)-[2sin(x)cos(x)]sin(x)
\\\\\\
2cos^3(x)-cos(x)~~~~-~~~~2sin^2(x)cos(x)
\\\\\\
2cos^3(x)-cos(x)~~~~-~~~~2[1-cos^2(x)]cos(x)
\\\\\\
2cos^3(x)-cos(x)~~~~-~~~~[2cos(x)-2cos^3(x)]
\\\\\\
2cos^3(x)-cos(x)~~~~-~~~~2cos(x)+2cos^3(x)
\\\\\\
4cos^3(x)-3cos(x)$

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