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Question Number 89193 by jagoll last updated on 16/Apr/20

cos x−sin x =(1/2) cos x sin x = (3/8) , π < x < 2π cos x + sin x =?

$$\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$ $$\mathrm{cos}\:{x}\:\mathrm{sin}\:{x}\:=\:\frac{\mathrm{3}}{\mathrm{8}}\:,\:\pi\:<\:{x}\:<\:\mathrm{2}\pi \\ $$ $$\mathrm{cos}\:{x}\:+\:\mathrm{sin}\:{x}\:=? \\ $$

Commented byTony Lin last updated on 16/Apr/20

∵π<x<2π ∴sinx<0 ∵cosxsinx=(3/8) ∴cosx<0 (cosx+sinx)^2 =(cosx−sinx)^2 +4cosxsinx =(1/4)+(3/2)=(7/4) ⇒cosx+sinx=−((√7)/2) { ((cosx+sinx=−((√7)/2))),((cosx−sinx=(1/2))) :} ⇒ { ((sinx=((−(√7)−1)/4))),((cosx=((−(√7)+1)/4))) :}

$$\because\pi<{x}<\mathrm{2}\pi \\ $$ $$\therefore{sinx}<\mathrm{0} \\ $$ $$\because{cosxsinx}=\frac{\mathrm{3}}{\mathrm{8}} \\ $$ $$\therefore{cosx}<\mathrm{0} \\ $$ $$\left({cosx}+{sinx}\right)^{\mathrm{2}} =\left({cosx}−{sinx}\right)^{\mathrm{2}} +\mathrm{4}{cosxsinx} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{7}}{\mathrm{4}} \\ $$ $$\Rightarrow{cosx}+{sinx}=−\frac{\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$ $$\begin{cases}{{cosx}+{sinx}=−\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}}\\{{cosx}−{sinx}=\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$ $$\Rightarrow\begin{cases}{{sinx}=\frac{−\sqrt{\mathrm{7}}−\mathrm{1}}{\mathrm{4}}}\\{{cosx}=\frac{−\sqrt{\mathrm{7}}+\mathrm{1}}{\mathrm{4}}}\end{cases} \\ $$

Commented byjohn santu last updated on 16/Apr/20

let cos x+sin x = p, p < 0 (1/2)p = cos 2x (i) (ii) 2sin xcos x = (3/4)⇒sin 2x=(3/4) cos 2x = −((√7)/4) ∴ (1/2)p = −((√7)/4) ⇒p = −((√7)/2) cos x+sin x = −((√7)/2)

$${let}\:\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\:=\:{p},\:{p}\:<\:\mathrm{0} \\ $$ $$\frac{\mathrm{1}}{\mathrm{2}}{p}\:=\:\mathrm{cos}\:\mathrm{2}{x}\:\:\:\left({i}\right) \\ $$ $$\left({ii}\right)\:\mathrm{2sin}\:{x}\mathrm{cos}\:{x}\:=\:\frac{\mathrm{3}}{\mathrm{4}}\Rightarrow\mathrm{sin}\:\mathrm{2}{x}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$ $$\mathrm{cos}\:\mathrm{2}{x}\:=\:−\frac{\sqrt{\mathrm{7}}}{\mathrm{4}} \\ $$ $$\therefore\:\frac{\mathrm{1}}{\mathrm{2}}{p}\:=\:−\frac{\sqrt{\mathrm{7}}}{\mathrm{4}}\:\Rightarrow{p}\:=\:−\frac{\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$ $$\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\:=\:−\frac{\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$

Commented byjagoll last updated on 16/Apr/20

thank you

$${thank}\:{you} \\ $$

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