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Let-x-4sin-2-10-o-4sin-2-50-o-cos20-o-cos80-o-and-y-cos-2-pi-5-cos-2-2pi-15-cos-2-8pi-15-find-x-y-




Question Number 29581 by math solver last updated on 10/Feb/18
Let x = 4sin^2 10^o +4sin^2 50^o cos20^o +cos80^o   and y = cos^2  (π/5)+cos^2 ((2π)/(15))+cos^2 ((8π)/(15)).  find x+y ?
$${Let}\:{x}\:=\:\mathrm{4}{sin}^{\mathrm{2}} \mathrm{10}^{{o}} +\mathrm{4}{sin}^{\mathrm{2}} \mathrm{50}^{{o}} {cos}\mathrm{20}^{{o}} +{cos}\mathrm{80}^{{o}} \\ $$$${and}\:{y}\:=\:{cos}^{\mathrm{2}} \:\frac{\pi}{\mathrm{5}}+{cos}^{\mathrm{2}} \frac{\mathrm{2}\pi}{\mathrm{15}}+{cos}^{\mathrm{2}} \frac{\mathrm{8}\pi}{\mathrm{15}}. \\ $$$${find}\:{x}+{y}\:? \\ $$
Answered by ajfour last updated on 10/Feb/18
x=2+(1/2)   ;  y=1+(1/2)  ;  x+y = 4  x=2(1−cos 20°)+2(1+cos 80°)cos 20°                         +cos 80°  x=2+cos 80°+2cos 80°cos 20°     = 2+cos 80°+cos 100°+cos 60°     =2+cos 80°−sin 10°+(1/2)    ⇒   x = 2+(1/2)       .......................................   y=(1/2)[1+cos ((2π)/5)+1+cos ((4π)/(15))+1+cos ((16π)/(15))]    =(3/2)+(1/2)[cos ((6π)/(15))+cos ((4π)/(15))−cos (π/(15))]    =(3/2)+(1/2)[2cos (π/3)cos (π/(15))−cos (π/(15))]  ⇒  y = (3/2)+0   ⇒    x+y = 4 .
$${x}=\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}\:\:\:;\:\:{y}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:\:;\:\:{x}+{y}\:=\:\mathrm{4} \\ $$$${x}=\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{20}°\right)+\mathrm{2}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{80}°\right)\mathrm{cos}\:\mathrm{20}° \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{cos}\:\mathrm{80}° \\ $$$${x}=\mathrm{2}+\mathrm{cos}\:\mathrm{80}°+\mathrm{2cos}\:\mathrm{80}°\mathrm{cos}\:\mathrm{20}° \\ $$$$\:\:\:=\:\mathrm{2}+\mathrm{cos}\:\mathrm{80}°+\mathrm{cos}\:\mathrm{100}°+\mathrm{cos}\:\mathrm{60}° \\ $$$$\:\:\:=\mathrm{2}+\mathrm{cos}\:\mathrm{80}°−\mathrm{sin}\:\mathrm{10}°+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\Rightarrow\:\:\:\boldsymbol{{x}}\:=\:\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:………………………………… \\ $$$$\:{y}=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{5}}+\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{15}}+\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{16}\pi}{\mathrm{15}}\right] \\ $$$$\:\:=\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{15}}+\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{15}}−\mathrm{cos}\:\frac{\pi}{\mathrm{15}}\right] \\ $$$$\:\:=\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{2cos}\:\frac{\pi}{\mathrm{3}}\mathrm{cos}\:\frac{\pi}{\mathrm{15}}−\mathrm{cos}\:\frac{\pi}{\mathrm{15}}\right] \\ $$$$\Rightarrow\:\:\boldsymbol{{y}}\:=\:\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{0}\: \\ $$$$\Rightarrow\:\:\:\:\boldsymbol{{x}}+\boldsymbol{{y}}\:=\:\mathrm{4}\:. \\ $$
Commented by math solver last updated on 11/Feb/18
thank you sir.
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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