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Question Number 146866 by mathdanisur last updated on 16/Jul/21
((sin^6 𝛂 + cos^6 𝛂 - 1)/(sin^4 𝛂 - sin^2 𝛂)) = ?
$$\frac{{sin}^{\mathrm{6}} \boldsymbol{\alpha}\:+\:{cos}^{\mathrm{6}} \boldsymbol{\alpha}\:-\:\mathrm{1}}{{sin}^{\mathrm{4}} \boldsymbol{\alpha}\:-\:{sin}^{\mathrm{2}} \boldsymbol{\alpha}}\:=\:? \\ $$
Answered by Olaf_Thorendsen last updated on 16/Jul/21
a = sin^6 α+cos^6 α  a = (sin^2 α+cos^2 α)^3 −3sin^2 αcos^2 α(sin^2 α+cos^2 α)  a = 1−3sin^2 αcos^2 α = 1−(3/4)sin^2 2α  b = sin^4 α−cos^4 α  b = (sin^2 α−cos^2 α)(sin^2 α+cos^2 α)  b = −cos^2 2α  X = ((sin^6 α+cos^6 α−1)/(sin^4 α−cos^4 α))   X = ((a−1)/b)   X = ((1−(3/4)sin^2 2α−1)/(−2cos^2 2α))   X = (3/8)tan^2 2α
$${a}\:=\:\mathrm{sin}^{\mathrm{6}} \alpha+\mathrm{cos}^{\mathrm{6}} \alpha \\ $$$${a}\:=\:\left(\mathrm{sin}^{\mathrm{2}} \alpha+\mathrm{cos}^{\mathrm{2}} \alpha\right)^{\mathrm{3}} −\mathrm{3sin}^{\mathrm{2}} \alpha\mathrm{cos}^{\mathrm{2}} \alpha\left(\mathrm{sin}^{\mathrm{2}} \alpha+\mathrm{cos}^{\mathrm{2}} \alpha\right) \\ $$$${a}\:=\:\mathrm{1}−\mathrm{3sin}^{\mathrm{2}} \alpha\mathrm{cos}^{\mathrm{2}} \alpha\:=\:\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}\mathrm{sin}^{\mathrm{2}} \mathrm{2}\alpha \\ $$$${b}\:=\:\mathrm{sin}^{\mathrm{4}} \alpha−\mathrm{cos}^{\mathrm{4}} \alpha \\ $$$${b}\:=\:\left(\mathrm{sin}^{\mathrm{2}} \alpha−\mathrm{cos}^{\mathrm{2}} \alpha\right)\left(\mathrm{sin}^{\mathrm{2}} \alpha+\mathrm{cos}^{\mathrm{2}} \alpha\right) \\ $$$${b}\:=\:−\mathrm{cos}^{\mathrm{2}} \mathrm{2}\alpha \\ $$$$\mathrm{X}\:=\:\frac{\mathrm{sin}^{\mathrm{6}} \alpha+\mathrm{cos}^{\mathrm{6}} \alpha−\mathrm{1}}{\mathrm{sin}^{\mathrm{4}} \alpha−\mathrm{cos}^{\mathrm{4}} \alpha}\: \\ $$$$\mathrm{X}\:=\:\frac{{a}−\mathrm{1}}{{b}}\: \\ $$$$\mathrm{X}\:=\:\frac{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}\mathrm{sin}^{\mathrm{2}} \mathrm{2}\alpha−\mathrm{1}}{−\mathrm{2cos}^{\mathrm{2}} \mathrm{2}\alpha}\: \\ $$$$\mathrm{X}\:=\:\frac{\mathrm{3}}{\mathrm{8}}\mathrm{tan}^{\mathrm{2}} \mathrm{2}\alpha \\ $$
Commented by mathdanisur last updated on 23/Jul/21
Sir,  b=sin^4 α−sin^2 α
$${Sir},\:\:{b}={sin}^{\mathrm{4}} \alpha−{sin}^{\mathrm{2}} \alpha \\ $$
Commented by mathdanisur last updated on 16/Jul/21
thankyou Ser cool
$${thankyou}\:{Ser}\:{cool} \\ $$

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