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Question Number 118078 by bemath last updated on 15/Oct/20
3(cos x+sin x)^4 +6(sin x−cos x)^2 −  3(sin^6 x+cos^6 x) = ?
$$\mathrm{3}\left(\mathrm{cos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{4}} +\mathrm{6}\left(\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} − \\ $$$$\mathrm{3}\left(\mathrm{sin}\:^{\mathrm{6}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{6}} \mathrm{x}\right)\:=\:?\: \\ $$
Answered by bobhans last updated on 15/Oct/20
(1) (sin x+cos x)^4 =(1+sin 2x)^2                                             =1+2sin 2x+sin^2 2x  (2)(sin x−cos x)^2 =1−sin 2x  (3)sin^6 x+cos^6 x = (sin^2 x)^3 +(cos^2 x)^3         = 1(sin^4 x−sin^2 xcos^2 x+cos^4 x)        =(sin^2 x)^2 +(cos^2 x)^2 −(1/4)sin^2 2x         = 1−2sin^2 xcos^2 x−(1/4)sin^2 2x         = 1−(3/4)sin^2 2x  Now we want compute 3×(1)+6×(2)−3×(3)  ⇒3+6sin 2x+3sin^2 2x+6−6sin 2x−  (3−(9/4)sin^2 2x) = 9+3sin^2 2x−3+(9/4)sin^2 2x  = 6+((21)/4)sin^2 2x
$$\left(\mathrm{1}\right)\:\left(\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{4}} =\left(\mathrm{1}+\mathrm{sin}\:\mathrm{2x}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}+\mathrm{2sin}\:\mathrm{2x}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{2x} \\ $$$$\left(\mathrm{2}\right)\left(\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} =\mathrm{1}−\mathrm{sin}\:\mathrm{2x} \\ $$$$\left(\mathrm{3}\right)\mathrm{sin}\:^{\mathrm{6}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{6}} \mathrm{x}\:=\:\left(\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\right)^{\mathrm{3}} +\left(\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:=\:\mathrm{1}\left(\mathrm{sin}\:^{\mathrm{4}} \mathrm{x}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{xcos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}\right) \\ $$$$\:\:\:\:\:\:=\left(\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\right)^{\mathrm{2}} +\left(\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:^{\mathrm{2}} \mathrm{2x} \\ $$$$\:\:\:\:\:\:\:=\:\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{xcos}\:^{\mathrm{2}} \mathrm{x}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:^{\mathrm{2}} \mathrm{2x} \\ $$$$\:\:\:\:\:\:\:=\:\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}\mathrm{sin}\:^{\mathrm{2}} \mathrm{2x} \\ $$$$\mathrm{Now}\:\mathrm{we}\:\mathrm{want}\:\mathrm{compute}\:\mathrm{3}×\left(\mathrm{1}\right)+\mathrm{6}×\left(\mathrm{2}\right)−\mathrm{3}×\left(\mathrm{3}\right) \\ $$$$\Rightarrow\mathrm{3}+\mathrm{6sin}\:\mathrm{2x}+\mathrm{3sin}\:^{\mathrm{2}} \mathrm{2x}+\mathrm{6}−\mathrm{6sin}\:\mathrm{2x}− \\ $$$$\left(\mathrm{3}−\frac{\mathrm{9}}{\mathrm{4}}\mathrm{sin}\:^{\mathrm{2}} \mathrm{2x}\right)\:=\:\mathrm{9}+\mathrm{3sin}\:^{\mathrm{2}} \mathrm{2x}−\mathrm{3}+\frac{\mathrm{9}}{\mathrm{4}}\mathrm{sin}\:^{\mathrm{2}} \mathrm{2x} \\ $$$$=\:\mathrm{6}+\frac{\mathrm{21}}{\mathrm{4}}\mathrm{sin}\:^{\mathrm{2}} \mathrm{2x}\: \\ $$
Answered by MJS_new last updated on 15/Oct/20
3(c+s)^4 +6(s−c)^2 −3(s^6 +c^6 )=  =−3(c^6 +s^6 −c^4 −s^4 −4c^3 s−4cs^3 −6c^2 s^2 −2c^2 −2s^2 +4cs)=       [c=(√(1−s^2 ))]  =−3(7(s^4 −s^2 )−2)=  =3(7s^2 (1−s^2 )+2)=  =3(7s^2 c^2 +2)=  =6+21sin^2  x cos^2  x =  =6+((21)/4)sin^2  2x =  =((69)/8)−((21)/8)cos 4x
$$\mathrm{3}\left({c}+{s}\right)^{\mathrm{4}} +\mathrm{6}\left({s}−{c}\right)^{\mathrm{2}} −\mathrm{3}\left({s}^{\mathrm{6}} +{c}^{\mathrm{6}} \right)= \\ $$$$=−\mathrm{3}\left({c}^{\mathrm{6}} +{s}^{\mathrm{6}} −{c}^{\mathrm{4}} −{s}^{\mathrm{4}} −\mathrm{4}{c}^{\mathrm{3}} {s}−\mathrm{4}{cs}^{\mathrm{3}} −\mathrm{6}{c}^{\mathrm{2}} {s}^{\mathrm{2}} −\mathrm{2}{c}^{\mathrm{2}} −\mathrm{2}{s}^{\mathrm{2}} +\mathrm{4}{cs}\right)= \\ $$$$\:\:\:\:\:\left[{c}=\sqrt{\mathrm{1}−{s}^{\mathrm{2}} }\right] \\ $$$$=−\mathrm{3}\left(\mathrm{7}\left({s}^{\mathrm{4}} −{s}^{\mathrm{2}} \right)−\mathrm{2}\right)= \\ $$$$=\mathrm{3}\left(\mathrm{7}{s}^{\mathrm{2}} \left(\mathrm{1}−{s}^{\mathrm{2}} \right)+\mathrm{2}\right)= \\ $$$$=\mathrm{3}\left(\mathrm{7}{s}^{\mathrm{2}} {c}^{\mathrm{2}} +\mathrm{2}\right)= \\ $$$$=\mathrm{6}+\mathrm{21sin}^{\mathrm{2}} \:{x}\:\mathrm{cos}^{\mathrm{2}} \:{x}\:= \\ $$$$=\mathrm{6}+\frac{\mathrm{21}}{\mathrm{4}}\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}{x}\:= \\ $$$$=\frac{\mathrm{69}}{\mathrm{8}}−\frac{\mathrm{21}}{\mathrm{8}}\mathrm{cos}\:\mathrm{4}{x} \\ $$
Commented by bemath last updated on 15/Oct/20
short cut way
$$\mathrm{short}\:\mathrm{cut}\:\mathrm{way} \\ $$

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