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