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Trigonometry-If-tan-2-and-tan-2-are-the-roots-of-the-equation-8x-2-26x-15-0-then-what-is-cos-equal-to-




Question Number 134788 by bramlexs22 last updated on 07/Mar/21
Trigonometry  If tan^2(α) and tan^2(β) are the roots of the equation 8x^2−26x+15=0, then what is cos(α+β) equal to?
$$\mathrm{Trigonometry} \\ $$If tan^2(α) and tan^2(β) are the roots of the equation 8x^2−26x+15=0, then what is cos(α+β) equal to?
Answered by john_santu last updated on 07/Mar/21
If tan^2 α and tan^2 β are the roots of  the equation 8x^2 −26x+15=0  then what is cos (α+β) equal to?  Let  { ((tan α)),((tan β)) :} are the roots of equation  8((√u) )^2 −26((√u))+15 = 0  or 64u^2 −436u +225 = 0  By Vieta′s theorem     { ((tan α+tan β = ((436)/(64)) ...(i))),((tan α. tan β = ((225)/(64))...(ii))) :}  so we get tan (α+β) = (((436)/(64))/(1−((225)/(64)))) = −((436)/(161))  then sec^2 (α+β) = 1+(−((436)/(161)))^2 =((216 017)/(25 921))  ∴ cos (α+β)=(1/(sec (α+β))) = −(√((25 921)/(216 017)))  cos (α+β) ≈ −0.3464  Note : we use negative root because  by eq(i) and (ii) α,β are acute and  α+β is in 2^(nd)  quadrant.
$${If}\:\mathrm{tan}\:^{\mathrm{2}} \alpha\:{and}\:\mathrm{tan}\:^{\mathrm{2}} \beta\:{are}\:{the}\:{roots}\:{of} \\ $$$${the}\:{equation}\:\mathrm{8}{x}^{\mathrm{2}} −\mathrm{26}{x}+\mathrm{15}=\mathrm{0} \\ $$$${then}\:{what}\:{is}\:\mathrm{cos}\:\left(\alpha+\beta\right)\:{equal}\:{to}? \\ $$$${Let}\:\begin{cases}{\mathrm{tan}\:\alpha}\\{\mathrm{tan}\:\beta}\end{cases}\:{are}\:{the}\:{roots}\:{of}\:{equation} \\ $$$$\mathrm{8}\left(\sqrt{{u}}\:\right)^{\mathrm{2}} −\mathrm{26}\left(\sqrt{{u}}\right)+\mathrm{15}\:=\:\mathrm{0} \\ $$$${or}\:\mathrm{64}{u}^{\mathrm{2}} −\mathrm{436}{u}\:+\mathrm{225}\:=\:\mathrm{0} \\ $$$${By}\:{Vieta}'{s}\:{theorem}\: \\ $$$$\:\begin{cases}{\mathrm{tan}\:\alpha+\mathrm{tan}\:\beta\:=\:\frac{\mathrm{436}}{\mathrm{64}}\:…\left({i}\right)}\\{\mathrm{tan}\:\alpha.\:\mathrm{tan}\:\beta\:=\:\frac{\mathrm{225}}{\mathrm{64}}…\left({ii}\right)}\end{cases} \\ $$$${so}\:{we}\:{get}\:\mathrm{tan}\:\left(\alpha+\beta\right)\:=\:\frac{\frac{\mathrm{436}}{\mathrm{64}}}{\mathrm{1}−\frac{\mathrm{225}}{\mathrm{64}}}\:=\:−\frac{\mathrm{436}}{\mathrm{161}} \\ $$$${then}\:\mathrm{sec}\:^{\mathrm{2}} \left(\alpha+\beta\right)\:=\:\mathrm{1}+\left(−\frac{\mathrm{436}}{\mathrm{161}}\right)^{\mathrm{2}} =\frac{\mathrm{216}\:\mathrm{017}}{\mathrm{25}\:\mathrm{921}} \\ $$$$\therefore\:\mathrm{cos}\:\left(\alpha+\beta\right)=\frac{\mathrm{1}}{\mathrm{sec}\:\left(\alpha+\beta\right)}\:=\:−\sqrt{\frac{\mathrm{25}\:\mathrm{921}}{\mathrm{216}\:\mathrm{017}}} \\ $$$$\mathrm{cos}\:\left(\alpha+\beta\right)\:\approx\:−\mathrm{0}.\mathrm{3464} \\ $$$${Note}\::\:{we}\:{use}\:{negative}\:{root}\:{because} \\ $$$${by}\:{eq}\left({i}\right)\:{and}\:\left({ii}\right)\:\alpha,\beta\:{are}\:{acute}\:{and} \\ $$$$\alpha+\beta\:{is}\:{in}\:\mathrm{2}^{{nd}} \:{quadrant}.\: \\ $$

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