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Question Number 83591 by jagoll last updated on 04/Mar/20

3x^2 −x+(t^2 −4t+3) = 0 has a roots sin α and cos α. find (√(t^2 −4t+5))

$$\mathrm{3x}^{\mathrm{2}} −\mathrm{x}+\left(\mathrm{t}^{\mathrm{2}} −\mathrm{4t}+\mathrm{3}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{a}\:\mathrm{roots}\:\mathrm{sin}\:\alpha\:\mathrm{and}\:\mathrm{cos}\:\alpha. \\ $$$$\mathrm{find}\:\sqrt{\mathrm{t}^{\mathrm{2}} −\mathrm{4t}+\mathrm{5}} \\ $$

Commented by jagoll last updated on 04/Mar/20

sin α+cos α = (1/3) sin α×cos α = ((t^2 −4t+3)/3) ⇒ 1+2sin αcos α = (1/9) sin αcos α = −(4/9) ((t^2 −4t+3 )/3) = −(4/9) t^2 −4t+3 = −(4/3) t^2 −4t+5 = (2/3) (√(t^2 −4t+5)) = ((√6)/3)

$$\mathrm{sin}\:\alpha+\mathrm{cos}\:\alpha\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{sin}\:\alpha×\mathrm{cos}\:\alpha\:=\:\frac{\mathrm{t}^{\mathrm{2}} −\mathrm{4t}+\mathrm{3}}{\mathrm{3}} \\ $$$$\Rightarrow\:\mathrm{1}+\mathrm{2sin}\:\alpha\mathrm{cos}\:\alpha\:=\:\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$\mathrm{sin}\:\alpha\mathrm{cos}\:\alpha\:=\:−\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$\:\frac{\mathrm{t}^{\mathrm{2}} −\mathrm{4t}+\mathrm{3}\:}{\mathrm{3}}\:=\:−\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$\mathrm{t}^{\mathrm{2}} −\mathrm{4t}+\mathrm{3}\:=\:−\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\mathrm{t}^{\mathrm{2}} −\mathrm{4t}+\mathrm{5}\:=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\sqrt{\mathrm{t}^{\mathrm{2}} −\mathrm{4t}+\mathrm{5}}\:=\:\frac{\sqrt{\mathrm{6}}}{\mathrm{3}} \\ $$

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