Menu Close

factorise-cos-cos3-cos5-cos7-

Tinku Tara 0 Comments
Pin




Question Number 63460 by Rio Michael last updated on 04/Jul/19
factorise cosθ−cos3θ−cos5θ +cos7θ
$${factorise} \\ $$$${cos}\theta−{cos}\mathrm{3}\theta−{cos}\mathrm{5}\theta\:+{cos}\mathrm{7}\theta \\ $$
Commented by Tony Lin last updated on 04/Jul/19
cosθ−cos3θ−(cos5θ−cos7θ) =−2sin2θsin(−θ)−[−2sin6θsin(−θ)] =2sin2θsinθ−2sin6θsinθ =2sinθ(sin2θ−sin6θ) =2sinθ[2cos4θsin(−2θ)] =−4sinθcos4θsin2θ
$${cos}\theta−{cos}\mathrm{3}\theta−\left({cos}\mathrm{5}\theta−{cos}\mathrm{7}\theta\right) \\ $$$$=−\mathrm{2}{sin}\mathrm{2}\theta{sin}\left(−\theta\right)−\left[−\mathrm{2}{sin}\mathrm{6}\theta{sin}\left(−\theta\right)\right] \\ $$$$=\mathrm{2}{sin}\mathrm{2}\theta{sin}\theta−\mathrm{2}{sin}\mathrm{6}\theta{sin}\theta \\ $$$$=\mathrm{2}{sin}\theta\left({sin}\mathrm{2}\theta−{sin}\mathrm{6}\theta\right) \\ $$$$=\mathrm{2}{sin}\theta\left[\mathrm{2}{cos}\mathrm{4}\theta{sin}\left(−\mathrm{2}\theta\right)\right] \\ $$$$=−\mathrm{4}{sin}\theta{cos}\mathrm{4}\theta{sin}\mathrm{2}\theta \\ $$
Commented by Tony Lin last updated on 04/Jul/19
sinα+sinβ=2sin((α+β)/2)cos((a−β)/2) sinα−sinβ=2cos((α+β)/2)sin((α−β)/2) cosα+cosβ=2cos((α+β)/2)cos((α−β)/2) cosα−cosβ=−2sin((α+β)/2)sin((α−β)/2)
$${sin}\alpha+{sin}\beta=\mathrm{2}{sin}\frac{\alpha+\beta}{\mathrm{2}}{cos}\frac{{a}−\beta}{\mathrm{2}} \\ $$$${sin}\alpha−{sin}\beta=\mathrm{2}{cos}\frac{\alpha+\beta}{\mathrm{2}}{sin}\frac{\alpha−\beta}{\mathrm{2}} \\ $$$${cos}\alpha+{cos}\beta=\mathrm{2}{cos}\frac{\alpha+\beta}{\mathrm{2}}{cos}\frac{\alpha−\beta}{\mathrm{2}} \\ $$$${cos}\alpha−{cos}\beta=−\mathrm{2}{sin}\frac{\alpha+\beta}{\mathrm{2}}{sin}\frac{\alpha−\beta}{\mathrm{2}} \\ $$
Answered by MJS last updated on 04/Jul/19
cos θ −cos 3θ −cos 5θ +cos 7θ = cos θ +4sin^2 θ cos θ −cos θ +16sin^2 θ cos^3 θ +4cos^3 θ −5cos θ −64sin^2 θ cos^5 θ −48cos^5 θ +56cos^3 θ −7cos θ = 8cos θ ((8sin^4 θ +1)cos^2 θ −1)
$$\mathrm{cos}\:\theta \\ $$$$\:\:\:\:\:−\mathrm{cos}\:\mathrm{3}\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:−\mathrm{cos}\:\mathrm{5}\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{cos}\:\mathrm{7}\theta \\ $$$$= \\ $$$$\mathrm{cos}\:\theta \\ $$$$\:\:\:\:\:+\mathrm{4sin}^{\mathrm{2}} \:\theta\:\mathrm{cos}\:\theta\:−\mathrm{cos}\:\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:+\mathrm{16sin}^{\mathrm{2}} \:\theta\:\mathrm{cos}^{\mathrm{3}} \:\theta\:+\mathrm{4cos}^{\mathrm{3}} \:\theta\:−\mathrm{5cos}\:\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{64sin}^{\mathrm{2}} \:\theta\:\mathrm{cos}^{\mathrm{5}} \:\theta\:−\mathrm{48cos}^{\mathrm{5}} \:\theta\:+\mathrm{56cos}^{\mathrm{3}} \:\theta\:−\mathrm{7cos}\:\theta \\ $$$$= \\ $$$$\mathrm{8cos}\:\theta\:\left(\left(\mathrm{8sin}^{\mathrm{4}} \:\theta\:+\mathrm{1}\right)\mathrm{cos}^{\mathrm{2}} \:\theta\:−\mathrm{1}\right) \\ $$
Commented by Rio Michael last updated on 05/Jul/19
i don′t get these method.
$${i}\:{don}'{t}\:{get}\:{these}\:{method}. \\ $$
Commented by MJS last updated on 05/Jul/19
you need to know sin (α+β) =cos α sin β +sin α cos β cos (α+β) =cos α cos β −sin α sin β cos 3θ =cos (θ+2θ) =cos θ cos 2θ −sin θ sin 2θ = cos 2θ =cos (θ+θ) =cos^2 θ −sin^2 θ sin 2θ =sin (θ+θ) =2sin θ cos θ =cos θ (cos^2 θ −sin^2 θ)−sin θ (2sin θ cos θ)= =c^3 −cs^2 −2cs^2 =c(c^2 −3s^2 )=c(c^2 +s^2 −4s^2 )= =c(1−4s^2 )=cos θ −4sin^2 θ cos θ and so on
$$\mathrm{you}\:\mathrm{need}\:\mathrm{to}\:\mathrm{know} \\ $$$$\mathrm{sin}\:\left(\alpha+\beta\right)\:=\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta\:+\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta \\ $$$$\mathrm{cos}\:\left(\alpha+\beta\right)\:=\mathrm{cos}\:\alpha\:\mathrm{cos}\:\beta\:−\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta \\ $$$$\mathrm{cos}\:\mathrm{3}\theta\:=\mathrm{cos}\:\left(\theta+\mathrm{2}\theta\right)\:=\mathrm{cos}\:\theta\:\mathrm{cos}\:\mathrm{2}\theta\:−\mathrm{sin}\:\theta\:\mathrm{sin}\:\mathrm{2}\theta\:= \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{cos}\:\mathrm{2}\theta\:=\mathrm{cos}\:\left(\theta+\theta\right)\:=\mathrm{cos}^{\mathrm{2}} \:\theta\:−\mathrm{sin}^{\mathrm{2}} \:\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{sin}\:\mathrm{2}\theta\:=\mathrm{sin}\:\left(\theta+\theta\right)\:=\mathrm{2sin}\:\theta\:\mathrm{cos}\:\theta \\ $$$$=\mathrm{cos}\:\theta\:\left(\mathrm{cos}^{\mathrm{2}} \:\theta\:−\mathrm{sin}^{\mathrm{2}} \:\theta\right)−\mathrm{sin}\:\theta\:\left(\mathrm{2sin}\:\theta\:\mathrm{cos}\:\theta\right)= \\ $$$$={c}^{\mathrm{3}} −{cs}^{\mathrm{2}} −\mathrm{2}{cs}^{\mathrm{2}} ={c}\left({c}^{\mathrm{2}} −\mathrm{3}{s}^{\mathrm{2}} \right)={c}\left({c}^{\mathrm{2}} +{s}^{\mathrm{2}} −\mathrm{4}{s}^{\mathrm{2}} \right)= \\ $$$$={c}\left(\mathrm{1}−\mathrm{4}{s}^{\mathrm{2}} \right)=\mathrm{cos}\:\theta\:−\mathrm{4sin}^{\mathrm{2}} \:\theta\:\mathrm{cos}\:\theta \\ $$$$\mathrm{and}\:\mathrm{so}\:\mathrm{on} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *