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Question Number 83430 by jagoll last updated on 02/Mar/20
prove that? sin 3θ sin^3 θ + cos 3θ cos^3 θ = cos^3 (2θ)
$$\mathrm{prove}\:\mathrm{that}? \\ $$$$\mathrm{sin}\:\mathrm{3}\theta\:\mathrm{sin}^{\mathrm{3}} \:\theta\:+\:\mathrm{cos}\:\mathrm{3}\theta\:\mathrm{cos}\:^{\mathrm{3}} \theta\:=\:\: \\ $$$$\mathrm{cos}\:^{\mathrm{3}} \:\left(\mathrm{2}\theta\right) \\ $$
Answered by mind is power last updated on 02/Mar/20
sin(3θ)sin^3 (θ)+cos(3θ)cos^3 (θ) =sin(θ)sin(3θ)sin^2 (θ)+cos(θ)cos(3θ)(cos^2 (θ)) =sin(θ)sin(3θ)(1−cos^2 (θ))+cos(θ)cos(3θ)cos^2 (θ) =sin(θ)sin(3θ)+cos^2 (θ)(cos(θ)cos(3θ)−sin(θ)sin(3θ)) =((cos(2θ)−cos(4θ))/2)+cos^2 (θ)(cos(4θ)) =cos^2 (θ)−cos^2 (2θ)+cos^2 (θ)(2cos^2 (2θ)−1) =((1+cos(2θ))/2)−cos^2 (2θ)+(((1+cos(2θ))/2))(2cos^2 (2θ)−1) =((1+cos(2θ))/2)−cos^2 (2θ)−(1/2)−((cos(2θ))/2)+cos^3 (2θ)+cos^2 (2θ) =cos^3 (2θ)
$${sin}\left(\mathrm{3}\theta\right){sin}^{\mathrm{3}} \left(\theta\right)+{cos}\left(\mathrm{3}\theta\right){cos}^{\mathrm{3}} \left(\theta\right) \\ $$$$={sin}\left(\theta\right){sin}\left(\mathrm{3}\theta\right){sin}^{\mathrm{2}} \left(\theta\right)+{cos}\left(\theta\right){cos}\left(\mathrm{3}\theta\right)\left({cos}^{\mathrm{2}} \left(\theta\right)\right) \\ $$$$={sin}\left(\theta\right){sin}\left(\mathrm{3}\theta\right)\left(\mathrm{1}−{cos}^{\mathrm{2}} \left(\theta\right)\right)+{cos}\left(\theta\right){cos}\left(\mathrm{3}\theta\right){cos}^{\mathrm{2}} \left(\theta\right) \\ $$$$={sin}\left(\theta\right){sin}\left(\mathrm{3}\theta\right)+{cos}^{\mathrm{2}} \left(\theta\right)\left({cos}\left(\theta\right){cos}\left(\mathrm{3}\theta\right)−{sin}\left(\theta\right){sin}\left(\mathrm{3}\theta\right)\right) \\ $$$$=\frac{{cos}\left(\mathrm{2}\theta\right)−{cos}\left(\mathrm{4}\theta\right)}{\mathrm{2}}+{cos}^{\mathrm{2}} \left(\theta\right)\left({cos}\left(\mathrm{4}\theta\right)\right) \\ $$$$={cos}^{\mathrm{2}} \left(\theta\right)−{cos}^{\mathrm{2}} \left(\mathrm{2}\theta\right)+{cos}^{\mathrm{2}} \left(\theta\right)\left(\mathrm{2}{cos}^{\mathrm{2}} \left(\mathrm{2}\theta\right)−\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}−{cos}^{\mathrm{2}} \left(\mathrm{2}\theta\right)+\left(\frac{\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}\right)\left(\mathrm{2}{cos}^{\mathrm{2}} \left(\mathrm{2}\theta\right)−\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}−{cos}^{\mathrm{2}} \left(\mathrm{2}\theta\right)−\frac{\mathrm{1}}{\mathrm{2}}−\frac{{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}+{cos}^{\mathrm{3}} \left(\mathrm{2}\theta\right)+{cos}^{\mathrm{2}} \left(\mathrm{2}\theta\right) \\ $$$$={cos}^{\mathrm{3}} \left(\mathrm{2}\theta\right) \\ $$$$ \\ $$
Commented by mind is power last updated on 02/Mar/20
withe pleasur
$${withe}\:{pleasur} \\ $$
Commented by jagoll last updated on 02/Mar/20
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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