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Question Number 123160 by liberty last updated on 23/Nov/20
Given sin ρ + cos ρ = (4/3) , where 0 < ρ < (π/4). Find the value of cos ρ−sin ρ.
$$\:{Given}\:\mathrm{sin}\:\rho\:+\:\mathrm{cos}\:\rho\:=\:\frac{\mathrm{4}}{\mathrm{3}}\:,\:{where}\: \\ $$$$\:\mathrm{0}\:<\:\rho\:<\:\frac{\pi}{\mathrm{4}}.\:{Find}\:{the}\:{value}\:{of}\:\mathrm{cos}\:\rho−\mathrm{sin}\:\rho.\: \\ $$
Answered by benjo_mathlover last updated on 23/Nov/20
⇒(sin ρ+cos ρ)^2 =((16)/9) ⇒1+2sin ρcos ρ=((16)/9) ⇒2sin ρcos ρ=(7/9) since 0<ρ<(π/4) then cos ρ>sin ρ so cos ρ−sin ρ>0 consider cos ρ−sin ρ=(√(1−2sin ρcos ρ)) cos ρ−sin ρ=(√(1−(7/9))) = ((√2)/3).
$$\Rightarrow\left(\mathrm{sin}\:\rho+\mathrm{cos}\:\rho\right)^{\mathrm{2}} =\frac{\mathrm{16}}{\mathrm{9}} \\ $$$$\Rightarrow\mathrm{1}+\mathrm{2sin}\:\rho\mathrm{cos}\:\rho=\frac{\mathrm{16}}{\mathrm{9}} \\ $$$$\Rightarrow\mathrm{2sin}\:\rho\mathrm{cos}\:\rho=\frac{\mathrm{7}}{\mathrm{9}} \\ $$$${since}\:\mathrm{0}<\rho<\frac{\pi}{\mathrm{4}}\:{then}\:\mathrm{cos}\:\rho>\mathrm{sin}\:\rho \\ $$$${so}\:\mathrm{cos}\:\rho−\mathrm{sin}\:\rho>\mathrm{0} \\ $$$${consider}\:\mathrm{cos}\:\rho−\mathrm{sin}\:\rho=\sqrt{\mathrm{1}−\mathrm{2sin}\:\rho\mathrm{cos}\:\rho} \\ $$$$\mathrm{cos}\:\rho−\mathrm{sin}\:\rho=\sqrt{\mathrm{1}−\frac{\mathrm{7}}{\mathrm{9}}}\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}. \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 23/Nov/20
(sinp+cosp)^2 +(cosp−sinp)^2 =2 (cosp−sinp)=(√(2−((16)/9))) =((√2)/3) (As 0<p<(π/4))
$$\left({sinp}+{cosp}\right)^{\mathrm{2}} +\left({cosp}−{sinp}\right)^{\mathrm{2}} =\mathrm{2} \\ $$$$\left({cosp}−{sinp}\right)=\sqrt{\mathrm{2}−\frac{\mathrm{16}}{\mathrm{9}}}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}\:\:\:\left({As}\:\mathrm{0}<{p}<\frac{\pi}{\mathrm{4}}\right) \\ $$$$ \\ $$

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