Question

Cos3x=4cos^3x-3cosx prove

Answer 1

`\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)`