Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 7587 by ten last updated on 05/Sep/16

cos4xsin4x

$${cos}\mathrm{4}{xsin}\mathrm{4}{x} \\ $$

Answered by Rasheed Soomro last updated on 05/Sep/16

cos4xsin4x  cos α sin β=(1/2)[sin(α+β)−sin(α−β)]    cos 4x sin 4x=(1/2)[sin(4x+4x)−sin(4x−4x)]                              =(1/2)sin 8x  Or  cos 4x sin 4x=(1/2)(2cos 4x sin 4x)              2 sin θ cos θ  =sin 2θ                             =(1/2)sin 8x

$${cos}\mathrm{4}{xsin}\mathrm{4}{x} \\ $$$$\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta=\left(\mathrm{1}/\mathrm{2}\right)\left[\mathrm{sin}\left(\alpha+\beta\right)−\mathrm{sin}\left(\alpha−\beta\right)\right]\:\: \\ $$$$\mathrm{cos}\:\mathrm{4}{x}\:\mathrm{sin}\:\mathrm{4}{x}=\left(\mathrm{1}/\mathrm{2}\right)\left[\mathrm{sin}\left(\mathrm{4}{x}+\mathrm{4}{x}\right)−\mathrm{sin}\left(\mathrm{4}{x}−\mathrm{4}{x}\right)\right]\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{1}/\mathrm{2}\right)\mathrm{sin}\:\mathrm{8}{x} \\ $$$${Or} \\ $$$$\mathrm{cos}\:\mathrm{4}{x}\:\mathrm{sin}\:\mathrm{4}{x}=\left(\mathrm{1}/\mathrm{2}\right)\left(\mathrm{2cos}\:\mathrm{4}{x}\:\mathrm{sin}\:\mathrm{4}{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\:\:=\mathrm{sin}\:\mathrm{2}\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{1}/\mathrm{2}\right)\mathrm{sin}\:\mathrm{8}{x} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com