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If-tan-equals-the-integral-solution-of-the-inequality-4x-2-16x-15-lt-0-and-cos-equals-to-the-slope-of-the-bisector-of-the-first-quadrant-then-sin-sin-is-equal-to-

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Question Number 12169 by indreshpatelindresh@435gmail.i last updated on 15/Apr/17
If tan α equals the integral solution of the inequality 4x^2 −16x+15<0 and cos β equals to the slope of the bisector of the first quadrant, then sin (α+β) sin (α−β) is equal to
$$\mathrm{If}\:\:\mathrm{tan}\:\alpha\:\mathrm{equals}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{solution}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{inequality}\:\:\mathrm{4}{x}^{\mathrm{2}} −\mathrm{16}{x}+\mathrm{15}<\mathrm{0}\:\mathrm{and}\: \\ $$$$\mathrm{cos}\:\beta\:\:\mathrm{equals}\:\mathrm{to}\:\mathrm{the}\:\mathrm{slope}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bisector} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{quadrant},\:\mathrm{then}\: \\ $$$$\mathrm{sin}\:\left(\alpha+\beta\right)\:\mathrm{sin}\:\left(\alpha−\beta\right)\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$
Answered by mrW1 last updated on 15/Apr/17
4x^2 −16x+15=4(x^2 −4x+4)−1 =4(x−2)^2 −1=(2x−3)(2x−5)=0 x_1 =(3/2)=1.5 x_2 =(5/2)=2.5 ⇒tan α=2 cos β=tan 45°=1 ⇒sin β=0 sin (α+β)sin (α−β) =(sin αcos β+cos αsin β)(sin αcos β−cos αsin β) =(sin α)(sin α) =sin^2 α=((tan^2 α)/(1+tan^2 α))=(2^2 /(1+2^2 ))=(4/5)
$$\mathrm{4}{x}^{\mathrm{2}} −\mathrm{16}{x}+\mathrm{15}=\mathrm{4}\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}\right)−\mathrm{1} \\ $$$$=\mathrm{4}\left({x}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{1}=\left(\mathrm{2}{x}−\mathrm{3}\right)\left(\mathrm{2}{x}−\mathrm{5}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{1}.\mathrm{5} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{2}}=\mathrm{2}.\mathrm{5} \\ $$$$\Rightarrow\mathrm{tan}\:\alpha=\mathrm{2} \\ $$$$ \\ $$$$\mathrm{cos}\:\beta=\mathrm{tan}\:\mathrm{45}°=\mathrm{1} \\ $$$$\Rightarrow\mathrm{sin}\:\beta=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{sin}\:\left(\alpha+\beta\right)\mathrm{sin}\:\left(\alpha−\beta\right) \\ $$$$=\left(\mathrm{sin}\:\alpha\mathrm{cos}\:\beta+\mathrm{cos}\:\alpha\mathrm{sin}\:\beta\right)\left(\mathrm{sin}\:\alpha\mathrm{cos}\:\beta−\mathrm{cos}\:\alpha\mathrm{sin}\:\beta\right) \\ $$$$=\left(\mathrm{sin}\:\alpha\right)\left(\mathrm{sin}\:\alpha\right) \\ $$$$=\mathrm{sin}^{\mathrm{2}} \:\alpha=\frac{\mathrm{tan}^{\mathrm{2}} \:\alpha}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\alpha}=\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{1}+\mathrm{2}^{\mathrm{2}} }=\frac{\mathrm{4}}{\mathrm{5}} \\ $$

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