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cos-3-x-sin-2-x-1-16-2cos-x-cos-3x-cos-5x-

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Question Number 29111 by gyugfeet last updated on 04/Feb/18
cos^3 x.sin^2 x=1/16(2cos x−cos 3x−cos 5x)
$$\mathrm{cos}^{\mathrm{3}} {x}.\mathrm{sin}\:^{\mathrm{2}} {x}=\mathrm{1}/\mathrm{16}\left(\mathrm{2cos}\:{x}−\mathrm{cos}\:\mathrm{3}{x}−\mathrm{cos}\:\mathrm{5}{x}\right) \\ $$
Commented by abdo imad last updated on 04/Feb/18
another method cos^3 x sin^2 x=cosx (cosxsinx)^2 =(1/4)cosx(sin(2x))^2 =(1/4)cosx (((1−cos(4x))/2))=(1/8)(cosx −cosx cos(4x)) but cosx cos(4x)= (1/2)( cos(5x) +cos(3x)) ⇒ cos^3 x sin^2 x = (1/8)(cosx −(1/2)cos(3x)−(1/2)cos(5x)) = (1/(16))( 2cosx −cos(3x) −cos(5x)) .
$${another}\:{method} \\ $$$${cos}^{\mathrm{3}} {x}\:{sin}^{\mathrm{2}} {x}={cosx}\:\left({cosxsinx}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}{cosx}\left({sin}\left(\mathrm{2}{x}\right)\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{cosx}\:\left(\frac{\mathrm{1}−{cos}\left(\mathrm{4}{x}\right)}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{8}}\left({cosx}\:−{cosx}\:{cos}\left(\mathrm{4}{x}\right)\right)\:{but} \\ $$$${cosx}\:{cos}\left(\mathrm{4}{x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\:{cos}\left(\mathrm{5}{x}\right)\:+{cos}\left(\mathrm{3}{x}\right)\right)\:\Rightarrow \\ $$$${cos}^{\mathrm{3}} {x}\:{sin}^{\mathrm{2}} {x}\:=\:\frac{\mathrm{1}}{\mathrm{8}}\left({cosx}\:−\frac{\mathrm{1}}{\mathrm{2}}{cos}\left(\mathrm{3}{x}\right)−\frac{\mathrm{1}}{\mathrm{2}}{cos}\left(\mathrm{5}{x}\right)\right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{16}}\left(\:\mathrm{2}{cosx}\:−{cos}\left(\mathrm{3}{x}\right)\:−{cos}\left(\mathrm{5}{x}\right)\right)\:. \\ $$
Answered by ajfour last updated on 04/Feb/18
cos 3x=4cos^3 x−3cos x 2sin^2 x=1−cos 2x so (4cos^3 x)(2sin^2 x)=(3cos x+cos 3x)(1−cos 2x) =3cos x−3cos xcos 2x +cos 3x−cos 3xcos 2x =3cos x−(3/2)(cos x+cos 3x) +cos 3x−(1/2)(cos x+cos 5x) =cos x−(1/2)cos 3x−(1/2)cos 5x ⇒ cos^3 xsin^2 x = (1/(16))(2cos x−cos 3x−cos 5x) .
$$\mathrm{cos}\:\mathrm{3}{x}=\mathrm{4cos}\:^{\mathrm{3}} {x}−\mathrm{3cos}\:{x} \\ $$$$\mathrm{2sin}\:^{\mathrm{2}} {x}=\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x} \\ $$$${so} \\ $$$$\left(\mathrm{4cos}\:^{\mathrm{3}} {x}\right)\left(\mathrm{2sin}\:^{\mathrm{2}} {x}\right)=\left(\mathrm{3cos}\:{x}+\mathrm{cos}\:\mathrm{3}{x}\right)\left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}\right) \\ $$$$=\mathrm{3cos}\:{x}−\mathrm{3cos}\:{x}\mathrm{cos}\:\mathrm{2}{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{cos}\:\mathrm{3}{x}−\mathrm{cos}\:\mathrm{3}{x}\mathrm{cos}\:\mathrm{2}{x} \\ $$$$=\mathrm{3cos}\:{x}−\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{cos}\:{x}+\mathrm{cos}\:\mathrm{3}{x}\right) \\ $$$$\:\:\:+\mathrm{cos}\:\mathrm{3}{x}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\:{x}+\mathrm{cos}\:\mathrm{5}{x}\right) \\ $$$$=\mathrm{cos}\:{x}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{3}{x}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{5}{x} \\ $$$$\Rightarrow\:\:\mathrm{cos}\:^{\mathrm{3}} {x}\mathrm{sin}\:^{\mathrm{2}} {x}\:= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{2cos}\:{x}−\mathrm{cos}\:\mathrm{3}{x}−\mathrm{cos}\:\mathrm{5}{x}\right)\:. \\ $$

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