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Question Number 109625 by qwerty111 last updated on 24/Aug/20

((sin 2𝛂+2sin 𝛂∙cos 2𝛂)/(1+cos 𝛂+cos2 𝛂+cos3 𝛂))

$$\frac{\mathrm{sin}\:\mathrm{2}\boldsymbol{\alpha}+\mathrm{2sin}\:\boldsymbol{\alpha}\centerdot\mathrm{cos}\:\mathrm{2}\boldsymbol{\alpha}}{\mathrm{1}+\mathrm{cos}\:\boldsymbol{\alpha}+\mathrm{cos2}\:\boldsymbol{\alpha}+\mathrm{cos3}\:\boldsymbol{\alpha}} \\ $$

Commented by nimnim last updated on 24/Aug/20

tanα

$${tan}\alpha \\ $$

Commented by kaivan.ahmadi last updated on 24/Aug/20

((2sinαcosα+2sinα(2cos^2 α−1))/(1+cosα+(2cos^2 α−1)+(4cos^3 α−3cosα)))=  ((2sinα(2cos^2 α+cosα−1))/(4cos^3 α+2cos^2 α−2cosα))=((2sinα(2cos^2 α+cosα−1))/(2cosα(2cos^2 α+cosα−1)))=((sinα)/(cosα))  tgα

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

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