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Question Number 64018 by Rio Michael last updated on 12/Jul/19

sin3θ=? cos3θ=? tan3θ=?

$${sin}\mathrm{3}\theta=? \\ $$$${cos}\mathrm{3}\theta=? \\ $$$${tan}\mathrm{3}\theta=? \\ $$

Commented by kaivan.ahmadi last updated on 12/Jul/19

sin3θ=sin(2θ+θ)=sin2θcosθ+cos2θsinθ= 2sinθcosθcosθ+(1−2sin^2 θ)sinθ= 2sinθ(1−sin^2 θ)+sinθ−2sin^3 θ= 3sinθ−4sin^3 θ

$${sin}\mathrm{3}\theta={sin}\left(\mathrm{2}\theta+\theta\right)={sin}\mathrm{2}\theta{cos}\theta+{cos}\mathrm{2}\theta{sin}\theta= \\ $$$$\mathrm{2}{sin}\theta{cos}\theta{cos}\theta+\left(\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \theta\right){sin}\theta= \\ $$$$\mathrm{2}{sin}\theta\left(\mathrm{1}−{sin}^{\mathrm{2}} \theta\right)+{sin}\theta−\mathrm{2}{sin}^{\mathrm{3}} \theta= \\ $$$$\mathrm{3}{sin}\theta−\mathrm{4}{sin}^{\mathrm{3}} \theta \\ $$

Commented by kaivan.ahmadi last updated on 12/Jul/19

cos3θ=cos(2θ+θ)=cos2θcosθ−sin2θsinθ= (2cos^2 θ−1)cosθ−2sinθcosθsinθ= 2cos^3 θ−cosθ−2cosθ(1−cos^2 θ)= 4cos^3 θ−3cosθ

$${cos}\mathrm{3}\theta={cos}\left(\mathrm{2}\theta+\theta\right)={cos}\mathrm{2}\theta{cos}\theta−{sin}\mathrm{2}\theta{sin}\theta= \\ $$$$\left(\mathrm{2}{cos}^{\mathrm{2}} \theta−\mathrm{1}\right){cos}\theta−\mathrm{2}{sin}\theta{cos}\theta{sin}\theta= \\ $$$$\mathrm{2}{cos}^{\mathrm{3}} \theta−{cos}\theta−\mathrm{2}{cos}\theta\left(\mathrm{1}−{cos}^{\mathrm{2}} \theta\right)= \\ $$$$\mathrm{4}{cos}^{\mathrm{3}} \theta−\mathrm{3}{cos}\theta \\ $$

Commented by kaivan.ahmadi last updated on 12/Jul/19

tan3θ=tan(2θ+θ)=((tan2θ+tanθ)/(1−tan2θtanθ))= ((((2tanθ)/(1−tan^2 θ))+tanθ)/(1−((2tanθ)/(1−tan^2 θ))tanθ))=((2tanθ+tanθ−tan^3 θ)/(1−tan^2 θ−2tan^2 θ))= ((3tanθ−tan^3 θ)/(1−3tan^2 θ))

$${tan}\mathrm{3}\theta={tan}\left(\mathrm{2}\theta+\theta\right)=\frac{{tan}\mathrm{2}\theta+{tan}\theta}{\mathrm{1}−{tan}\mathrm{2}\theta{tan}\theta}= \\ $$$$\frac{\frac{\mathrm{2}{tan}\theta}{\mathrm{1}−{tan}^{\mathrm{2}} \theta}+{tan}\theta}{\mathrm{1}−\frac{\mathrm{2}{tan}\theta}{\mathrm{1}−{tan}^{\mathrm{2}} \theta}{tan}\theta}=\frac{\mathrm{2}{tan}\theta+{tan}\theta−{tan}^{\mathrm{3}} \theta}{\mathrm{1}−{tan}^{\mathrm{2}} \theta−\mathrm{2}{tan}^{\mathrm{2}} \theta}= \\ $$$$\frac{\mathrm{3}{tan}\theta−{tan}^{\mathrm{3}} \theta}{\mathrm{1}−\mathrm{3}{tan}^{\mathrm{2}} \theta} \\ $$

Commented by Rio Michael last updated on 12/Jul/19

and sir sin^3 θ ≡?

$${and}\:{sir}\:\:{sin}^{\mathrm{3}} \theta\:\equiv? \\ $$

Commented by Rio Michael last updated on 12/Jul/19

thanks so much

$${thanks}\:{so}\:{much} \\ $$

Commented by Rio Michael last updated on 12/Jul/19

cos^3 θ≡? tan^3 θ≡?

$${cos}^{\mathrm{3}} \theta\equiv? \\ $$$${tan}^{\mathrm{3}} \theta\equiv? \\ $$

Commented by kaivan.ahmadi last updated on 12/Jul/19

are you serious?

$${are}\:{you}\:{serious}? \\ $$

Commented by Joel122 last updated on 12/Jul/19

just mulptiply by itself 3 times

$$\mathrm{just}\:\mathrm{mulptiply}\:\mathrm{by}\:\mathrm{itself}\:\:\mathrm{3}\:\mathrm{times} \\ $$

Commented by MJS last updated on 12/Jul/19

sin^3 x =(3/4)sin x −(1/4)sin 3x cos^3 x =(3/4)cos x +(1/4)cos 3x tan^3 x =((sin^3 x)/(cos^3 x)) sin^4 x =(1/8)cos 4x −(1/2)cos 2x +(3/8) cos^4 x =(1/8)cos 4x +(1/2)cos 2x +(3/8) tan^4 x =((sin^4 x)/(cos^4 x))

$$\mathrm{sin}^{\mathrm{3}} \:{x}\:=\frac{\mathrm{3}}{\mathrm{4}}\mathrm{sin}\:{x}\:−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\mathrm{3}{x} \\ $$$$\mathrm{cos}^{\mathrm{3}} \:{x}\:=\frac{\mathrm{3}}{\mathrm{4}}\mathrm{cos}\:{x}\:+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\:\mathrm{3}{x} \\ $$$$\mathrm{tan}^{\mathrm{3}} \:{x}\:=\frac{\mathrm{sin}^{\mathrm{3}} \:{x}}{\mathrm{cos}^{\mathrm{3}} \:{x}} \\ $$$$\mathrm{sin}^{\mathrm{4}} \:{x}\:=\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cos}\:\mathrm{4}{x}\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}{x}\:+\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$\mathrm{cos}^{\mathrm{4}} \:{x}\:=\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cos}\:\mathrm{4}{x}\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2}{x}\:+\frac{\mathrm{3}}{\mathrm{8}} \\ $$$$\mathrm{tan}^{\mathrm{4}} \:{x}\:=\frac{\mathrm{sin}^{\mathrm{4}} \:{x}}{\mathrm{cos}^{\mathrm{4}} \:{x}} \\ $$

Commented by Rio Michael last updated on 12/Jul/19

yeah i know, but anyway thanks. a textbook almlst confused me

$${yeah}\:{i}\:{know},\:{but}\:{anyway}\:{thanks}.\:{a}\:{textbook}\:{almlst}\:{confused} \\ $$$${me} \\ $$

Commented by Tony Lin last updated on 12/Jul/19

sin3θ=3sinθ−4sin^3 θ =4sin((π/3)−θ)θ((π/3)+θ) cos3θ=4cos^3 −3cosθ =4cos((π/3)−θ)θ((π/3)+θ) tan3θ=((3tanθ−tan^3 θ)/(1−3tan^2 θ)) =tan((π/3)−θ)θ((π/3)+θ)

$${sin}\mathrm{3}\theta=\mathrm{3}{sin}\theta−\mathrm{4}{sin}^{\mathrm{3}} \theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4}{sin}\left(\frac{\pi}{\mathrm{3}}−\theta\right)\theta\left(\frac{\pi}{\mathrm{3}}+\theta\right) \\ $$$${cos}\mathrm{3}\theta=\mathrm{4}{cos}^{\mathrm{3}} −\mathrm{3}{cos}\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4}{cos}\left(\frac{\pi}{\mathrm{3}}−\theta\right)\theta\left(\frac{\pi}{\mathrm{3}}+\theta\right) \\ $$$${tan}\mathrm{3}\theta=\frac{\mathrm{3}{tan}\theta−{tan}^{\mathrm{3}} \theta}{\mathrm{1}−\mathrm{3}{tan}^{\mathrm{2}} \theta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:={tan}\left(\frac{\pi}{\mathrm{3}}−\theta\right)\theta\left(\frac{\pi}{\mathrm{3}}+\theta\right) \\ $$

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