Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 162649 by tounghoungko last updated on 31/Dec/21

sin^(10) (x)+cos^(10) (x)=((11)/(36)) sin^(12) (x)+cos^(12) (x)=?

$$\:\mathrm{sin}\:^{\mathrm{10}} \left({x}\right)+\mathrm{cos}\:^{\mathrm{10}} \left({x}\right)=\frac{\mathrm{11}}{\mathrm{36}} \\ $$$$\:\mathrm{sin}\:^{\mathrm{12}} \left({x}\right)+\mathrm{cos}\:^{\mathrm{12}} \left({x}\right)=? \\ $$

Answered by Ar Brandon last updated on 31/Dec/21

S^(12) +C^(12) =S^(10) (1−C^2 )+C^(10) (1−S^2 ) =S^(10) +C^(10) −S^(10) C^2 −C^(10) S^2 =S^(10) +C^(10) −S^2 C^2 (S^8 +C^8 ) =S^(10) +C^(10) −S^2 C^2 [(S^4 +C^4 )^2 −2S^4 C^4 ] =S^(10) +C^(10) −S^2 C^2 [((S^2 +C^2 )^2 −2S^2 C^2 )^2 −2S^4 C^4 ] =S^(10) +C^(10) −S^2 C^2 [1−4S^2 C^2 +2S^4 C^4 ] =...

$$\mathrm{S}^{\mathrm{12}} +\mathrm{C}^{\mathrm{12}} =\mathrm{S}^{\mathrm{10}} \left(\mathrm{1}−\mathrm{C}^{\mathrm{2}} \right)+\mathrm{C}^{\mathrm{10}} \left(\mathrm{1}−\mathrm{S}^{\mathrm{2}} \right) \\ $$$$=\mathrm{S}^{\mathrm{10}} +\mathrm{C}^{\mathrm{10}} −\mathrm{S}^{\mathrm{10}} \mathrm{C}^{\mathrm{2}} −\mathrm{C}^{\mathrm{10}} \mathrm{S}^{\mathrm{2}} \\ $$$$=\mathrm{S}^{\mathrm{10}} +\mathrm{C}^{\mathrm{10}} −\mathrm{S}^{\mathrm{2}} \mathrm{C}^{\mathrm{2}} \left(\mathrm{S}^{\mathrm{8}} +\mathrm{C}^{\mathrm{8}} \right) \\ $$$$=\mathrm{S}^{\mathrm{10}} +\mathrm{C}^{\mathrm{10}} −\mathrm{S}^{\mathrm{2}} \mathrm{C}^{\mathrm{2}} \left[\left(\mathrm{S}^{\mathrm{4}} +\mathrm{C}^{\mathrm{4}} \right)^{\mathrm{2}} −\mathrm{2S}^{\mathrm{4}} \mathrm{C}^{\mathrm{4}} \right] \\ $$$$=\mathrm{S}^{\mathrm{10}} +\mathrm{C}^{\mathrm{10}} −\mathrm{S}^{\mathrm{2}} \mathrm{C}^{\mathrm{2}} \left[\left(\left(\mathrm{S}^{\mathrm{2}} +\mathrm{C}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2S}^{\mathrm{2}} \mathrm{C}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2S}^{\mathrm{4}} \mathrm{C}^{\mathrm{4}} \right] \\ $$$$=\mathrm{S}^{\mathrm{10}} +\mathrm{C}^{\mathrm{10}} −\mathrm{S}^{\mathrm{2}} \mathrm{C}^{\mathrm{2}} \left[\mathrm{1}−\mathrm{4S}^{\mathrm{2}} \mathrm{C}^{\mathrm{2}} +\mathrm{2S}^{\mathrm{4}} \mathrm{C}^{\mathrm{4}} \right] \\ $$$$=... \\ $$

Answered by mindispower last updated on 31/Dec/21

c^2 +s^2 =1 c^(10) +s^(10) =((11)/(36)) (c^2 +s^2 )^5 =c^(10) +s^(10) +5c^8 s^2 +5c^2 s^8 +10c^4 s^6 +10s^4 c^6 1=((11)/(36))+5c^2 s^2 (c^6 +s^6 )+10c^4 s^4 ....(E) c^6 +s^6 =1−3s^2 c^2 ,X=(sc)^2 ..(E)⇔ ((25)/(36))=5X(1−3X)+10X^2 5X^2 −5X+((25)/(36))=0 X=25−((125)/9)=((100)/9) ((5+((10)/3))/(10))2p=((25)/(30))>(1/4),((5−((10)/3))/(10))=(5/(30))=(1/6)<(1/4) (sc)^2 =((sin^2 (2x))/4)≤(1/4) (sc)^2 =(1/6) (s^2 +c^2 )^6 =S^(12) +C^(12) +6S^2 C^2 (S^8 +C^8 )+15S^4 C^4 (S^4 +C^4 ) +20C^6 S^6 S^8 +C^8 =(s^2 +c^2 )^4 −4s^2 c^2 (s^4 +c^4 )−6s^4 s^4 =1−(2/3).(2/3)−(1/6)=(7/(18)) (S^4 +C^4 )=(s^2 +c^2 )^2 −2c^2 s^2 =1−(1/3)=(2/3) c^(12) +s^(12) =1−(7/(18))−((15)/(36)).(2/3)−20.(1/(216)) tchek calculus i have problemes withe my phone horrible to use tactil

$${c}^{\mathrm{2}} +{s}^{\mathrm{2}} =\mathrm{1} \\ $$$${c}^{\mathrm{10}} +{s}^{\mathrm{10}} =\frac{\mathrm{11}}{\mathrm{36}} \\ $$$$\left({c}^{\mathrm{2}} +{s}^{\mathrm{2}} \right)^{\mathrm{5}} ={c}^{\mathrm{10}} +{s}^{\mathrm{10}} +\mathrm{5}{c}^{\mathrm{8}} {s}^{\mathrm{2}} +\mathrm{5}{c}^{\mathrm{2}} {s}^{\mathrm{8}} +\mathrm{10}{c}^{\mathrm{4}} {s}^{\mathrm{6}} +\mathrm{10}{s}^{\mathrm{4}} {c}^{\mathrm{6}} \\ $$$$\mathrm{1}=\frac{\mathrm{11}}{\mathrm{36}}+\mathrm{5}{c}^{\mathrm{2}} {s}^{\mathrm{2}} \left({c}^{\mathrm{6}} +{s}^{\mathrm{6}} \right)+\mathrm{10}{c}^{\mathrm{4}} {s}^{\mathrm{4}} ....\left({E}\right) \\ $$$${c}^{\mathrm{6}} +{s}^{\mathrm{6}} =\mathrm{1}−\mathrm{3}{s}^{\mathrm{2}} {c}^{\mathrm{2}} ,{X}=\left({sc}\right)^{\mathrm{2}} ..\left({E}\right)\Leftrightarrow \\ $$$$\frac{\mathrm{25}}{\mathrm{36}}=\mathrm{5}{X}\left(\mathrm{1}−\mathrm{3}{X}\right)+\mathrm{10}{X}^{\mathrm{2}} \\ $$$$\mathrm{5}{X}^{\mathrm{2}} −\mathrm{5}{X}+\frac{\mathrm{25}}{\mathrm{36}}=\mathrm{0} \\ $$$${X}=\mathrm{25}−\frac{\mathrm{125}}{\mathrm{9}}=\frac{\mathrm{100}}{\mathrm{9}} \\ $$$$\frac{\mathrm{5}+\frac{\mathrm{10}}{\mathrm{3}}}{\mathrm{10}}\mathrm{2}{p}=\frac{\mathrm{25}}{\mathrm{30}}>\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{5}−\frac{\mathrm{10}}{\mathrm{3}}}{\mathrm{10}}=\frac{\mathrm{5}}{\mathrm{30}}=\frac{\mathrm{1}}{\mathrm{6}}<\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\left({sc}\right)^{\mathrm{2}} =\frac{{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right)}{\mathrm{4}}\leqslant\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\left({sc}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\left({s}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)^{\mathrm{6}} ={S}^{\mathrm{12}} +{C}^{\mathrm{12}} +\mathrm{6}{S}^{\mathrm{2}} {C}^{\mathrm{2}} \left({S}^{\mathrm{8}} +{C}^{\mathrm{8}} \right)+\mathrm{15}{S}^{\mathrm{4}} {C}^{\mathrm{4}} \left({S}^{\mathrm{4}} +{C}^{\mathrm{4}} \right) \\ $$$$+\mathrm{20}{C}^{\mathrm{6}} {S}^{\mathrm{6}} \\ $$$${S}^{\mathrm{8}} +{C}^{\mathrm{8}} =\left({s}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)^{\mathrm{4}} −\mathrm{4}{s}^{\mathrm{2}} {c}^{\mathrm{2}} \left({s}^{\mathrm{4}} +{c}^{\mathrm{4}} \right)−\mathrm{6}{s}^{\mathrm{4}} {s}^{\mathrm{4}} \\ $$$$=\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}.\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{6}}=\frac{\mathrm{7}}{\mathrm{18}} \\ $$$$\left({S}^{\mathrm{4}} +{C}^{\mathrm{4}} \right)=\left({s}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}{c}^{\mathrm{2}} {s}^{\mathrm{2}} =\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${c}^{\mathrm{12}} +{s}^{\mathrm{12}} =\mathrm{1}−\frac{\mathrm{7}}{\mathrm{18}}−\frac{\mathrm{15}}{\mathrm{36}}.\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{20}.\frac{\mathrm{1}}{\mathrm{216}} \\ $$$${tchek}\:{calculus}\:{i}\:{have}\:{problemes}\:{withe}\:{my} \\ $$$${phone}\:{horrible}\:{to}\:{use}\:{tactil} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Answered by MJS_new last updated on 31/Dec/21

s^(10) +c^(10) −((11)/(36))=0 [c=(√(1−s^2 ))] 5s^8 −10s^6 +10s^4 −5s^2 +((25)/(36))=0 (s^2 )^4 −2(s^2 )^3 +2(s^2 )^2 −(s^2 )+(5/(36))=0 ((s^2 )^2 −(s^2 )+(1/6))((s^2 )^2 −(s^2 )+(5/6))=0 (s^2 )=(1/2)±((√3)/6) (s^2 )^6 +((√(1−(s^2 ))))^(12) =((13)/(54))

$${s}^{\mathrm{10}} +{c}^{\mathrm{10}} −\frac{\mathrm{11}}{\mathrm{36}}=\mathrm{0} \\ $$$$\:\:\:\:\:\left[{c}=\sqrt{\mathrm{1}−{s}^{\mathrm{2}} }\right] \\ $$$$\mathrm{5}{s}^{\mathrm{8}} −\mathrm{10}{s}^{\mathrm{6}} +\mathrm{10}{s}^{\mathrm{4}} −\mathrm{5}{s}^{\mathrm{2}} +\frac{\mathrm{25}}{\mathrm{36}}=\mathrm{0} \\ $$$$\left({s}^{\mathrm{2}} \right)^{\mathrm{4}} −\mathrm{2}\left({s}^{\mathrm{2}} \right)^{\mathrm{3}} +\mathrm{2}\left({s}^{\mathrm{2}} \right)^{\mathrm{2}} −\left({s}^{\mathrm{2}} \right)+\frac{\mathrm{5}}{\mathrm{36}}=\mathrm{0} \\ $$$$\left(\left({s}^{\mathrm{2}} \right)^{\mathrm{2}} −\left({s}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{6}}\right)\left(\left({s}^{\mathrm{2}} \right)^{\mathrm{2}} −\left({s}^{\mathrm{2}} \right)+\frac{\mathrm{5}}{\mathrm{6}}\right)=\mathrm{0} \\ $$$$\left({s}^{\mathrm{2}} \right)=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{6}} \\ $$$$\left({s}^{\mathrm{2}} \right)^{\mathrm{6}} +\left(\sqrt{\mathrm{1}−\left({s}^{\mathrm{2}} \right)}\right)^{\mathrm{12}} =\frac{\mathrm{13}}{\mathrm{54}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com