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Question Number 59246 by hovea cw last updated on 06/May/19
Commented by MJS last updated on 06/May/19
(1) doesn′t seem to have a “nice” solution I get (√3)tan 10° cos 40° ≈.233956
$$\left(\mathrm{1}\right)\:\mathrm{doesn}'\mathrm{t}\:\mathrm{seem}\:\mathrm{to}\:\mathrm{have}\:\mathrm{a}\:“\mathrm{nice}''\:\mathrm{solution} \\ $$$$\mathrm{I}\:\mathrm{get}\:\sqrt{\mathrm{3}}\mathrm{tan}\:\mathrm{10}°\:\mathrm{cos}\:\mathrm{40}°\:\approx.\mathrm{233956} \\ $$
Commented by maxmathsup by imad last updated on 07/May/19
let A =sin(10^o )+cos(70^o ).tan(190^o ) let transform to radians π→180 x→10 ° ⇒180x=10π ⇒x =((10π)/(180)) =(π/(18)) also 70 =((70π)/(180)) =((7π)/(18)) 190 =((190π)/(180)) =((19π)/(18)) ⇒A =sin(((5π)/(90)))+cos(((7π)/(18))).tan(π+(π/(18))) =sin((π/(18))) +cos(((7π)/(18)))tan((π/(18))) but cos(((7π)/(18))) =sin((π/2) −((7π)/(18))) =sin(((2π)/(18)))=sin((π/9)) ⇒cos(((7π)/(18)))tan((π/(18)))=2sin((π/(18)))cos((π/(18))) ((sin((π/(18))))/(cos((π/(18))))) =2sin^2 ((π/(18)))=1−cos((π/9)) ⇒ A =sin((π/(18)))+1−cos((π/9)) =2sin^2 ((π/(18)))+sin((π/(18))) rest to calculate sin((π/(18)))...
$${let}\:{A}\:={sin}\left(\mathrm{10}^{{o}} \right)+{cos}\left(\mathrm{70}^{{o}} \right).{tan}\left(\mathrm{190}^{{o}} \right)\:\:\:{let}\:{transform}\:{to}\:{radians} \\ $$$$\pi\rightarrow\mathrm{180} \\ $$$${x}\rightarrow\mathrm{10}\:°\:\Rightarrow\mathrm{180}{x}=\mathrm{10}\pi\:\Rightarrow{x}\:=\frac{\mathrm{10}\pi}{\mathrm{180}}\:=\frac{\pi}{\mathrm{18}}\:{also}\:\:\:\:\mathrm{70}\:=\frac{\mathrm{70}\pi}{\mathrm{180}}\:=\frac{\mathrm{7}\pi}{\mathrm{18}} \\ $$$$\mathrm{190}\:=\frac{\mathrm{190}\pi}{\mathrm{180}}\:=\frac{\mathrm{19}\pi}{\mathrm{18}}\:\Rightarrow{A}\:={sin}\left(\frac{\mathrm{5}\pi}{\mathrm{90}}\right)+{cos}\left(\frac{\mathrm{7}\pi}{\mathrm{18}}\right).{tan}\left(\pi+\frac{\pi}{\mathrm{18}}\right) \\ $$$$={sin}\left(\frac{\pi}{\mathrm{18}}\right)\:+{cos}\left(\frac{\mathrm{7}\pi}{\mathrm{18}}\right){tan}\left(\frac{\pi}{\mathrm{18}}\right)\:\:{but}\:\:{cos}\left(\frac{\mathrm{7}\pi}{\mathrm{18}}\right)\:={sin}\left(\frac{\pi}{\mathrm{2}}\:−\frac{\mathrm{7}\pi}{\mathrm{18}}\right)\:={sin}\left(\frac{\mathrm{2}\pi}{\mathrm{18}}\right)={sin}\left(\frac{\pi}{\mathrm{9}}\right) \\ $$$$\Rightarrow{cos}\left(\frac{\mathrm{7}\pi}{\mathrm{18}}\right){tan}\left(\frac{\pi}{\mathrm{18}}\right)=\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{18}}\right){cos}\left(\frac{\pi}{\mathrm{18}}\right)\:\frac{{sin}\left(\frac{\pi}{\mathrm{18}}\right)}{{cos}\left(\frac{\pi}{\mathrm{18}}\right)}\:=\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{18}}\right)=\mathrm{1}−{cos}\left(\frac{\pi}{\mathrm{9}}\right)\:\Rightarrow \\ $$$${A}\:={sin}\left(\frac{\pi}{\mathrm{18}}\right)+\mathrm{1}−{cos}\left(\frac{\pi}{\mathrm{9}}\right)\:=\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{18}}\right)+{sin}\left(\frac{\pi}{\mathrm{18}}\right)\:\:{rest}\:{to}\:{calculate}\:{sin}\left(\frac{\pi}{\mathrm{18}}\right)… \\ $$
Commented by maxmathsup by imad last updated on 07/May/19
2) its eazy ((cos^2 θ−sin^2 θ)/(cosθ +sinθ)) =(((cosθ−sinθ)(cosθ +sinθ))/(cosθ +sinθ)) =cosθ −sinθ and θ ≠−(π/4) +2kπ
$$\left.\mathrm{2}\right)\:{its}\:{eazy}\:\:\:\:\frac{{cos}^{\mathrm{2}} \theta−{sin}^{\mathrm{2}} \theta}{{cos}\theta\:+{sin}\theta}\:=\frac{\left({cos}\theta−{sin}\theta\right)\left({cos}\theta\:+{sin}\theta\right)}{{cos}\theta\:+{sin}\theta}\:={cos}\theta\:−{sin}\theta \\ $$$${and}\:\theta\:\:\neq−\frac{\pi}{\mathrm{4}}\:+\mathrm{2}{k}\pi \\ $$
Commented by maxmathsup by imad last updated on 07/May/19
cos(3a) =cos(2a+a) =cos(2a)cos(a)−sin(2a)sin(a) =(2cos^2 a−1)cos(a)−2cosasin^2 a =2cos^3 a−cosa −2cosa(1−cos^2 a) =2cos^3 a−cosa−2cosa +2cos^3 a =4cos^3 a−3cosa ⇒ cos(3a) =4cos^3 a−3cosa .
$${cos}\left(\mathrm{3}{a}\right)\:={cos}\left(\mathrm{2}{a}+{a}\right)\:={cos}\left(\mathrm{2}{a}\right){cos}\left({a}\right)−{sin}\left(\mathrm{2}{a}\right){sin}\left({a}\right) \\ $$$$=\left(\mathrm{2}{cos}^{\mathrm{2}} {a}−\mathrm{1}\right){cos}\left({a}\right)−\mathrm{2}{cosasin}^{\mathrm{2}} {a}\:=\mathrm{2}{cos}^{\mathrm{3}} {a}−{cosa}\:−\mathrm{2}{cosa}\left(\mathrm{1}−{cos}^{\mathrm{2}} {a}\right) \\ $$$$=\mathrm{2}{cos}^{\mathrm{3}} {a}−{cosa}−\mathrm{2}{cosa}\:+\mathrm{2}{cos}^{\mathrm{3}} {a}\:=\mathrm{4}{cos}^{\mathrm{3}} {a}−\mathrm{3}{cosa}\:\Rightarrow \\ $$$${cos}\left(\mathrm{3}{a}\right)\:=\mathrm{4}{cos}^{\mathrm{3}} {a}−\mathrm{3}{cosa}\:. \\ $$
Answered by MJS last updated on 06/May/19
(2) is easy ((c^2 −s^2 )/(c+s))=(((c−s)(c+s))/(c+s))=c−s
$$\left(\mathrm{2}\right)\:\mathrm{is}\:\mathrm{easy} \\ $$$$\frac{\mathrm{c}^{\mathrm{2}} −\mathrm{s}^{\mathrm{2}} }{\mathrm{c}+\mathrm{s}}=\frac{\left(\mathrm{c}−\mathrm{s}\right)\left(\mathrm{c}+\mathrm{s}\right)}{\mathrm{c}+\mathrm{s}}=\mathrm{c}−\mathrm{s} \\ $$
Answered by MJS last updated on 06/May/19
(3) sin (α+β) =cos α sin β +sin α cos β cos (α+β) =cos α cos β −sin α sin β cos 3α =cos (α+2α) =cos α cos 2α −sin α sin 2α = =cos α cos (α+α) −sin α sin (α+α) = =cos α (cos^2 α −sin^2 α)−sin α (2cos α sin α)= =cos α (cos^2 α −(1−cos^2 α))−2cos α (1−cos^2 α)= =2cos^3 α −cos α −2cos α +2cos^3 α = =4cos^3 α −3cos α
$$\left(\mathrm{3}\right) \\ $$$$\mathrm{sin}\:\left(\alpha+\beta\right)\:=\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta\:+\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta \\ $$$$\mathrm{cos}\:\left(\alpha+\beta\right)\:=\mathrm{cos}\:\alpha\:\mathrm{cos}\:\beta\:−\mathrm{sin}\:\alpha\:\mathrm{sin}\:\beta \\ $$$$ \\ $$$$\mathrm{cos}\:\mathrm{3}\alpha\:=\mathrm{cos}\:\left(\alpha+\mathrm{2}\alpha\right)\:=\mathrm{cos}\:\alpha\:\mathrm{cos}\:\mathrm{2}\alpha\:−\mathrm{sin}\:\alpha\:\mathrm{sin}\:\mathrm{2}\alpha\:= \\ $$$$=\mathrm{cos}\:\alpha\:\mathrm{cos}\:\left(\alpha+\alpha\right)\:−\mathrm{sin}\:\alpha\:\mathrm{sin}\:\left(\alpha+\alpha\right)\:= \\ $$$$=\mathrm{cos}\:\alpha\:\left(\mathrm{cos}^{\mathrm{2}} \:\alpha\:−\mathrm{sin}^{\mathrm{2}} \:\alpha\right)−\mathrm{sin}\:\alpha\:\left(\mathrm{2cos}\:\alpha\:\mathrm{sin}\:\alpha\right)= \\ $$$$=\mathrm{cos}\:\alpha\:\left(\mathrm{cos}^{\mathrm{2}} \:\alpha\:−\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\alpha\right)\right)−\mathrm{2cos}\:\alpha\:\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\alpha\right)= \\ $$$$=\mathrm{2cos}^{\mathrm{3}} \:\alpha\:−\mathrm{cos}\:\alpha\:−\mathrm{2cos}\:\alpha\:+\mathrm{2cos}^{\mathrm{3}} \:\alpha\:= \\ $$$$=\mathrm{4cos}^{\mathrm{3}} \:\alpha\:−\mathrm{3cos}\:\alpha \\ $$
Answered by tanmay last updated on 07/May/19
1)sin10^o +cos70^o .tan(190^o ) =sin10^o +sin20^o ×tan(180^o +10^o ) =sin10^o +sin20^o ×tan10^o =sin10^o (1+((sin20^o )/(cos10^o ))) =sin10^o (((cos10^o +sin20^o )/(cos10^o ))) =sin10^0 ×(((cos10^o +cos70^o )/(cos10^o ))) =sin10^o ×((2cos40^o ×cos30^o )/(cos10^o )) =(√3) ×tan10^o cos40^o
$$\left.\mathrm{1}\right){sin}\mathrm{10}^{{o}} +{cos}\mathrm{70}^{{o}} .{tan}\left(\mathrm{190}^{{o}} \right) \\ $$$$={sin}\mathrm{10}^{{o}} +{sin}\mathrm{20}^{{o}} ×{tan}\left(\mathrm{180}^{{o}} +\mathrm{10}^{{o}} \right) \\ $$$$={sin}\mathrm{10}^{{o}} +{sin}\mathrm{20}^{{o}} ×{tan}\mathrm{10}^{{o}} \\ $$$$={sin}\mathrm{10}^{{o}} \left(\mathrm{1}+\frac{{sin}\mathrm{20}^{{o}} }{{cos}\mathrm{10}^{{o}} }\right) \\ $$$$={sin}\mathrm{10}^{{o}} \left(\frac{{cos}\mathrm{10}^{{o}} +{sin}\mathrm{20}^{{o}} }{{cos}\mathrm{10}^{{o}} }\right) \\ $$$$={sin}\mathrm{10}^{\mathrm{0}} ×\left(\frac{{cos}\mathrm{10}^{{o}} +{cos}\mathrm{70}^{{o}} }{{cos}\mathrm{10}^{{o}} }\right) \\ $$$$={sin}\mathrm{10}^{{o}} ×\frac{\mathrm{2}{cos}\mathrm{40}^{{o}} ×{cos}\mathrm{30}^{{o}} }{{cos}\mathrm{10}^{{o}} } \\ $$$$=\sqrt{\mathrm{3}}\:×{tan}\mathrm{10}^{{o}} {cos}\mathrm{40}^{{o}} \\ $$

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