Question

4sin2xcos2x

Answer 1

So the expression is

\[\Large 4\sin(2x)\cos(2x)\]

what are they asking you to do?

Answer 2

\(\normalsize\color{blue}{ 4~\sin(2x)~\cos(2x) }\)

APPLY, \(\normalsize\color{black}{ \sin(2x)= \sin(x+x)=\sin(x)\cos(x) +\sin(x)\cos(x)=2\sin(x)\cos(x) }\)

and \(\normalsize\color{black}{ \cos(2x)= \cos(x+x)=\cos(x)\cos(x) -\sin(x)\sin(x)=\cos^2(x)-\sin^2(x) }\)


THUS, \(\normalsize\color{blue}{ 4~\sin(2x)~\cos(2x) ~~= }\)

\(\normalsize\color{blue}{ 4 \times~2\sin x~\cos x \times~(\cos^2x-\sin^2x) ~~= }\)

\(\normalsize\color{blue}{ 8 \times\sin x~\cos x \times~(\cos^2x-\sin^2x) ~~= }\)

\(\normalsize\color{blue}{ 8 \times~\cos^3x~\sin x-\sin^3x~\cos x ~~= }\)

\(\normalsize\color{blue}{ 8~\cos^3x~\sin x-\sin^3x~\cos x ~~= }\)

this is to write it in terms of one x.