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