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The-value-of-the-expression-3-sin-cos-4-6-sin-cos-2-4-sin-6-cos-6-is-




Question Number 21549 by x² – y²@gmail.com last updated on 27/Sep/17
The value of the expression   3(sin θ−cos θ)^4 +6(sin θ+cos θ)^2                              +4(sin^6 θ+cos^6 θ)  is
$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{expression}\: \\ $$$$\mathrm{3}\left(\mathrm{sin}\:\theta−\mathrm{cos}\:\theta\right)^{\mathrm{4}} +\mathrm{6}\left(\mathrm{sin}\:\theta+\mathrm{cos}\:\theta\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{4}\left(\mathrm{sin}^{\mathrm{6}} \theta+\mathrm{cos}^{\mathrm{6}} \theta\right)\:\:\mathrm{is} \\ $$
Answered by Tikufly last updated on 27/Sep/17
Sol  3[(sinθ−cosθ)^2 ]^2 +6(1+2sinθcosθ)                                       +4[(sin^2 θ)^3 +(cos^2 θ)^3 ]  ⇒3(1−2sinθcosθ)^2 +6(1+sin2θ)      +4[(sin^2 θ+cos^2 θ)^3 −3sin^2 θcos^2 θ(sin^2 θ+cos^2 θ)]  ⇒3(1−sin2θ)^2 +6(1+sin2θ)+4(1−3sin^2 θcos^2 θ)  ⇒3[(1+sin2θ)^2 −4sin2θ]+6(1+sin2θ)+4(1−(3/4)sin^2 2θ)  ⇒3(1+sin2θ)^2 −12sin2θ+6+6sin2θ+4−3sin^2 2θ  ⇒3(1+sin2θ)^2 +3−3sin^2 2θ+7−6sin2θ  ⇒3(1+sin2θ)(1+sin2θ+1−sin2θ)+7−6sin2θ  ⇒6(1+sin2θ)+7−6sin2θ  ⇒6+6sin2θ+7−6sin2θ  ⇒13
$$\mathrm{Sol} \\ $$$$\mathrm{3}\left[\left(\mathrm{sin}\theta−\mathrm{cos}\theta\right)^{\mathrm{2}} \right]^{\mathrm{2}} +\mathrm{6}\left(\mathrm{1}+\mathrm{2sin}\theta\mathrm{cos}\theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{4}\left[\left(\mathrm{sin}^{\mathrm{2}} \theta\right)^{\mathrm{3}} +\left(\mathrm{cos}^{\mathrm{2}} \theta\right)^{\mathrm{3}} \right] \\ $$$$\Rightarrow\mathrm{3}\left(\mathrm{1}−\mathrm{2sin}\theta\mathrm{cos}\theta\right)^{\mathrm{2}} +\mathrm{6}\left(\mathrm{1}+\mathrm{sin2}\theta\right) \\ $$$$\:\:\:\:+\mathrm{4}\left[\left(\mathrm{sin}^{\mathrm{2}} \theta+\mathrm{cos}^{\mathrm{2}} \theta\right)^{\mathrm{3}} −\mathrm{3sin}^{\mathrm{2}} \theta\mathrm{cos}^{\mathrm{2}} \theta\left(\mathrm{sin}^{\mathrm{2}} \theta+\mathrm{cos}^{\mathrm{2}} \theta\right)\right] \\ $$$$\Rightarrow\mathrm{3}\left(\mathrm{1}−\mathrm{sin2}\theta\right)^{\mathrm{2}} +\mathrm{6}\left(\mathrm{1}+\mathrm{sin2}\theta\right)+\mathrm{4}\left(\mathrm{1}−\mathrm{3sin}^{\mathrm{2}} \theta\mathrm{cos}^{\mathrm{2}} \theta\right) \\ $$$$\Rightarrow\mathrm{3}\left[\left(\mathrm{1}+\mathrm{sin2}\theta\right)^{\mathrm{2}} −\mathrm{4sin2}\theta\right]+\mathrm{6}\left(\mathrm{1}+\mathrm{sin2}\theta\right)+\mathrm{4}\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}\mathrm{sin}^{\mathrm{2}} \mathrm{2}\theta\right) \\ $$$$\Rightarrow\mathrm{3}\left(\mathrm{1}+\mathrm{sin2}\theta\right)^{\mathrm{2}} −\mathrm{12sin2}\theta+\mathrm{6}+\mathrm{6sin2}\theta+\mathrm{4}−\mathrm{3sin}^{\mathrm{2}} \mathrm{2}\theta \\ $$$$\Rightarrow\mathrm{3}\left(\mathrm{1}+\mathrm{sin2}\theta\right)^{\mathrm{2}} +\mathrm{3}−\mathrm{3sin}^{\mathrm{2}} \mathrm{2}\theta+\mathrm{7}−\mathrm{6sin2}\theta \\ $$$$\Rightarrow\mathrm{3}\left(\mathrm{1}+\mathrm{sin2}\theta\right)\left(\mathrm{1}+\mathrm{sin2}\theta+\mathrm{1}−\mathrm{sin2}\theta\right)+\mathrm{7}−\mathrm{6sin2}\theta \\ $$$$\Rightarrow\mathrm{6}\left(\mathrm{1}+\mathrm{sin2}\theta\right)+\mathrm{7}−\mathrm{6sin2}\theta \\ $$$$\Rightarrow\mathrm{6}+\mathrm{6sin2}\theta+\mathrm{7}−\mathrm{6sin2}\theta \\ $$$$\Rightarrow\mathrm{13} \\ $$
Commented by x² – y²@gmail.com last updated on 27/Sep/17
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$$$ \\ $$
Answered by $@ty@m last updated on 20/Oct/17
3(sin θ−cos θ)^4 =3{(sin θ−cos θ)^2 }^2            =3(1−2sin θcos θ)^2            =3(1+4sin^2 θcos^2 θ−4sin θcos θ)  −−−(1)  6(sin θ+cos θ)^2 =6(1+2sinθcos θ)   −−−(2)  4(sin^6 θ+cos^6 θ)=4{(sin^2 θ)^3 +(cos^2 θ)^3 }           =4(sin^2 θ+cos^2 θ)(sin^4 θ+cos^4 θ−sin^2 θcos^2 θ)           =4{(sin^2 θ+cos^2 θ)^2 −2sin^2 θcos^2 θ−sin^2 θcos^2 θ}           =4(1−3sin^2 θcos^2 θ)  −−(3)  Adding (1), (2) and(3)  The given expression  =3+6+4+(12−12)sin^2 θcos^2 θ+(12−12)sin θcos θ  =13
$$\mathrm{3}\left(\mathrm{sin}\:\theta−\mathrm{cos}\:\theta\right)^{\mathrm{4}} =\mathrm{3}\left\{\left(\mathrm{sin}\:\theta−\mathrm{cos}\:\theta\right)^{\mathrm{2}} \right\}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{3}\left(\mathrm{1}−\mathrm{2sin}\:\theta\mathrm{cos}\:\theta\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{3}\left(\mathrm{1}+\mathrm{4sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:^{\mathrm{2}} \theta−\mathrm{4sin}\:\theta\mathrm{cos}\:\theta\right)\:\:−−−\left(\mathrm{1}\right) \\ $$$$\mathrm{6}\left(\mathrm{sin}\:\theta+\mathrm{cos}\:\theta\right)^{\mathrm{2}} =\mathrm{6}\left(\mathrm{1}+\mathrm{2sin}\theta\mathrm{cos}\:\theta\right)\:\:\:−−−\left(\mathrm{2}\right) \\ $$$$\mathrm{4}\left(\mathrm{sin}^{\mathrm{6}} \theta+\mathrm{cos}^{\mathrm{6}} \theta\right)=\mathrm{4}\left\{\left(\mathrm{sin}\:^{\mathrm{2}} \theta\right)^{\mathrm{3}} +\left(\mathrm{cos}\:^{\mathrm{2}} \theta\right)^{\mathrm{3}} \right\} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{4}\left(\mathrm{sin}\:^{\mathrm{2}} \theta+\mathrm{cos}\:^{\mathrm{2}} \theta\right)\left(\mathrm{sin}\:^{\mathrm{4}} \theta+\mathrm{cos}\:^{\mathrm{4}} \theta−\mathrm{sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:^{\mathrm{2}} \theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{4}\left\{\left(\mathrm{sin}\:^{\mathrm{2}} \theta+\mathrm{cos}\:^{\mathrm{2}} \theta\right)^{\mathrm{2}} −\mathrm{2sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:^{\mathrm{2}} \theta−\mathrm{sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:^{\mathrm{2}} \theta\right\} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{4}\left(\mathrm{1}−\mathrm{3sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:^{\mathrm{2}} \theta\right)\:\:−−\left(\mathrm{3}\right) \\ $$$${Adding}\:\left(\mathrm{1}\right),\:\left(\mathrm{2}\right)\:{and}\left(\mathrm{3}\right) \\ $$$${The}\:{given}\:{expression} \\ $$$$=\mathrm{3}+\mathrm{6}+\mathrm{4}+\left(\mathrm{12}−\mathrm{12}\right)\mathrm{sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:^{\mathrm{2}} \theta+\left(\mathrm{12}−\mathrm{12}\right)\mathrm{sin}\:\theta\mathrm{cos}\:\theta \\ $$$$=\mathrm{13} \\ $$
Commented by $@ty@m last updated on 27/Sep/17
Thanks  Corrected....
$${Thanks} \\ $$$${Corrected}…. \\ $$
Commented by x² – y²@gmail.com last updated on 27/Sep/17
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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