Question Number 20704 by tammi last updated on 01/Sep/17
$$\mathrm{4cos}\:\theta\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}+\theta\right)\mathrm{cos}\:\left(\frac{\mathrm{4}\pi}{\mathrm{3}}+\theta\right)=\mathrm{cos}\:\mathrm{3}\theta \\ $$
Commented by myintkhaing last updated on 01/Sep/17
![L.H.S=2cosθ[cos(2π+θ)+cos((2π)/3)] =2cosθ[cos2θ−cos(π/3)]=2cosθ[2cos^2 θ−1−(1/2)] =4cos^3 θ−3cosθ=cos3θ=R.H.S proved](Q20711.png)
$$\mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{2cos}\theta\left[\mathrm{cos}\left(\mathrm{2}\pi+\theta\right)+\mathrm{cos}\frac{\mathrm{2}\pi}{\mathrm{3}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2cos}\theta\left[\mathrm{cos2}\theta−\mathrm{cos}\frac{\pi}{\mathrm{3}}\right]=\mathrm{2cos}\theta\left[\mathrm{2cos}^{\mathrm{2}} \theta−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4cos}^{\mathrm{3}} \theta−\mathrm{3cos}\theta=\mathrm{cos3}\theta=\mathrm{R}.\mathrm{H}.\mathrm{S} \\ $$$$\mathrm{proved} \\ $$
Commented by $@ty@m last updated on 01/Sep/17
$${Please}\:{use}\:``{Add}\:{an}\:{answer}\:{option}'' \\ $$$${in}\:{place}\:{of}\:``{Add}\:{a}\:{comment}''. \\ $$
Answered by myintkhaing last updated on 01/Sep/17
Commented by $@ty@m last updated on 01/Sep/17
$${Thanks}= \\ $$$${Now}\:{it}\:{disappeared}\:{from}\:``{Unanswered}\:{question}'' \\ $$