Menu Close

sin-10-x-cos-10-x-29-16-cos-4-2x-find-x-

Tinku Tara 0 Comments
Pin




Question Number 200565 by hardmath last updated on 20/Nov/23
sin^(10) x + cos^(10) x = ((29)/(16)) cos^4 2x find: x = ?
$$\mathrm{sin}^{\mathrm{10}} \mathrm{x}\:\:+\:\:\mathrm{cos}^{\mathrm{10}} \mathrm{x}\:\:=\:\:\frac{\mathrm{29}}{\mathrm{16}}\:\mathrm{cos}^{\mathrm{4}} \mathrm{2x} \\ $$$$\mathrm{find}:\:\:\:\mathrm{x}\:=\:? \\ $$
Commented by Harnada last updated on 21/Nov/23
Q200565
$${Q}\mathrm{200565} \\ $$
Commented by Harnada last updated on 21/Nov/23
Q168467 Q378000
$${Q}\mathrm{168467}\:{Q}\mathrm{378000} \\ $$
Answered by witcher3 last updated on 20/Nov/23
cos^2 (x)+sin^2 (x)=1 sin^2 (x)cos^2 (x)=((sin^2 (2x))/4) (cos^2 (x)+sin^2 (x))^5 =1= Σ_(k=0) ^5 cos^(2k) (x)sin^(2(5−k)) (x) =cos^(10) (x)+sin^(10) (x)+5cos^8 (x)sin^2 (x)+5sin^8 (x)cos^2 (x) +10sin^6 (x)cos^4 (x)+10sin^4 (x)cos^6 (x) =cos^(10) (x)+sin^(10) (x)+10sin^4 (x)cos^4 (x) +5cos^2 (x)sin^2 (x)(cos^6 (x)+sin^6 (x)) sin^6 (x)+cos^6 (x)=(cos^2 (x)+sin^2 (x))(cos^4 (x)+sin^4 (x)−sin^2 (x)cos^2 (x)) =(1−3sin^2 (x)cos^2 (x) =1−(3/4)sin^2 (2x) 1=cos^(10) (x)+sin^(10) (x)+(5/4)sin^2 (2x)(1−(3/4)sin^2 (2x)) +((10)/(16))sin^4 (2x);sin^2 (2x)=a cos^(10) (x)+sin^(10) (x)=1−(5/4)a(1−(3/4)a)+((10a^2 )/(16)) =1−((5a)/4)+((29)/(16))a^2 =((29cos^4 (2x))/(16)) cos^4 (2x)=(1−a)^2 ⇔1−((5a)/4)+((29a^2 )/(16))=((29)/(16))(a^2 −2a+1) ((19a)/8)=((13)/(16)) a=((13)/(38))=sin^2 (2x)
$$\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)=\mathrm{1} \\ $$$$\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)=\frac{\mathrm{sin}^{\mathrm{2}} \left(\mathrm{2x}\right)}{\mathrm{4}} \\ $$$$\left(\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)\right)^{\mathrm{5}} =\mathrm{1}= \\ $$$$\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}\mathrm{cos}^{\mathrm{2k}} \left(\mathrm{x}\right)\mathrm{sin}^{\mathrm{2}\left(\mathrm{5}−\mathrm{k}\right)} \left(\mathrm{x}\right) \\ $$$$=\mathrm{cos}^{\mathrm{10}} \left(\mathrm{x}\right)+\mathrm{sin}^{\mathrm{10}} \left(\mathrm{x}\right)+\mathrm{5cos}^{\mathrm{8}} \left(\mathrm{x}\right)\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{5sin}^{\mathrm{8}} \left(\mathrm{x}\right)\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right) \\ $$$$+\mathrm{10sin}^{\mathrm{6}} \left(\mathrm{x}\right)\mathrm{cos}^{\mathrm{4}} \left(\mathrm{x}\right)+\mathrm{10sin}^{\mathrm{4}} \left(\mathrm{x}\right)\mathrm{cos}^{\mathrm{6}} \left(\mathrm{x}\right) \\ $$$$=\mathrm{cos}^{\mathrm{10}} \left(\mathrm{x}\right)+\mathrm{sin}^{\mathrm{10}} \left(\mathrm{x}\right)+\mathrm{10sin}^{\mathrm{4}} \left(\mathrm{x}\right)\mathrm{cos}^{\mathrm{4}} \left(\mathrm{x}\right) \\ $$$$+\mathrm{5cos}^{\mathrm{2}} \left(\mathrm{x}\right)\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)\left(\mathrm{cos}^{\mathrm{6}} \left(\mathrm{x}\right)+\mathrm{sin}^{\mathrm{6}} \left(\mathrm{x}\right)\right) \\ $$$$\mathrm{sin}^{\mathrm{6}} \left(\mathrm{x}\right)+\mathrm{cos}^{\mathrm{6}} \left(\mathrm{x}\right)=\left(\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)\right)\left(\mathrm{cos}^{\mathrm{4}} \left(\mathrm{x}\right)+\mathrm{sin}^{\mathrm{4}} \left(\mathrm{x}\right)−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)\right) \\ $$$$=\left(\mathrm{1}−\mathrm{3sin}^{\mathrm{2}} \left(\mathrm{x}\right)\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)\right. \\ $$$$=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}\mathrm{sin}^{\mathrm{2}} \left(\mathrm{2x}\right) \\ $$$$\mathrm{1}=\mathrm{cos}^{\mathrm{10}} \left(\mathrm{x}\right)+\mathrm{sin}^{\mathrm{10}} \left(\mathrm{x}\right)+\frac{\mathrm{5}}{\mathrm{4}}\mathrm{sin}^{\mathrm{2}} \left(\mathrm{2x}\right)\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}\mathrm{sin}^{\mathrm{2}} \left(\mathrm{2x}\right)\right) \\ $$$$+\frac{\mathrm{10}}{\mathrm{16}}\mathrm{sin}^{\mathrm{4}} \left(\mathrm{2x}\right);\mathrm{sin}^{\mathrm{2}} \left(\mathrm{2x}\right)=\mathrm{a} \\ $$$$\mathrm{cos}^{\mathrm{10}} \left(\mathrm{x}\right)+\mathrm{sin}^{\mathrm{10}} \left(\mathrm{x}\right)=\mathrm{1}−\frac{\mathrm{5}}{\mathrm{4}}\mathrm{a}\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}\mathrm{a}\right)+\frac{\mathrm{10a}^{\mathrm{2}} }{\mathrm{16}} \\ $$$$=\mathrm{1}−\frac{\mathrm{5a}}{\mathrm{4}}+\frac{\mathrm{29}}{\mathrm{16}}\mathrm{a}^{\mathrm{2}} =\frac{\mathrm{29cos}^{\mathrm{4}} \left(\mathrm{2x}\right)}{\mathrm{16}} \\ $$$$\mathrm{cos}^{\mathrm{4}} \left(\mathrm{2x}\right)=\left(\mathrm{1}−\mathrm{a}\right)^{\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{1}−\frac{\mathrm{5a}}{\mathrm{4}}+\frac{\mathrm{29a}^{\mathrm{2}} }{\mathrm{16}}=\frac{\mathrm{29}}{\mathrm{16}}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{2a}+\mathrm{1}\right) \\ $$$$\frac{\mathrm{19a}}{\mathrm{8}}=\frac{\mathrm{13}}{\mathrm{16}} \\ $$$$\mathrm{a}=\frac{\mathrm{13}}{\mathrm{38}}=\mathrm{sin}^{\mathrm{2}} \left(\mathrm{2x}\right) \\ $$
Commented by hardmath last updated on 20/Nov/23
thank you dear professor, but, x = ?
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor},\:\mathrm{but},\:\mathrm{x}\:=\:? \\ $$
Commented by witcher3 last updated on 20/Nov/23
its easy to finish from here
$$\mathrm{its}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{finish}\:\mathrm{from}\:\mathrm{here} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *