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Question Number 23304 by Tinkutara last updated on 28/Oct/17

If sin^4 x + cos^4 y + 2 = 4 sinx.cosy & 0 ≤ x,y ≤ (π/2) , then the value of sinx + cosy is equal to

$$\mathrm{If}\:\mathrm{sin}^{\mathrm{4}} \mathrm{x}\:+\:\mathrm{cos}^{\mathrm{4}} \mathrm{y}\:+\:\mathrm{2}\:=\:\mathrm{4}\:\mathrm{sinx}.\mathrm{cosy}\:\& \\ $$ $$\mathrm{0}\:\leqslant\:{x},{y}\:\leqslant\:\frac{\pi}{\mathrm{2}}\:,\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{sin}{x}\:+ \\ $$ $$\mathrm{cos}{y}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$

Commented byJoel577 last updated on 28/Oct/17

What′s the name of ur book?

$${What}'{s}\:{the}\:{name}\:{of}\:{ur}\:{book}? \\ $$

Commented byJoel577 last updated on 28/Oct/17

Owh, I think it was from ur textbook. Because u have asked so many questions I wonder how many assignments did u get from school everyday

$${Owh},\:{I}\:{think}\:{it}\:{was}\:{from}\:{ur}\:{textbook}. \\ $$ $${Because}\:{u}\:{have}\:{asked}\:{so}\:{many}\:{questions} \\ $$ $${I}\:{wonder}\:{how}\:{many}\:{assignments}\:{did}\:{u}\:{get}\:{from}\:{school} \\ $$ $${everyday} \\ $$

Answered by ajfour last updated on 28/Oct/17

let sin x+cos y=S sin x−cos y=D sin xcos y = P Then what is given converts to (sin^2 x−cos^2 y)^2 +2sin^2 xcos^2 y+2 = 4sin xcos y ⇒ S^( 2) D^2 +2P^( 2) +2−4P=0 ⇒ S^( 2) D^2 +2(P−1)^2 =0 S cannot be zero for 0≤x, y ≤(π/2) So D=0 and P =1 ⇒sin x=cos y and sin xcos y=1 this can be true only if x=(π/2) and y=0 So sin x+cos y =2 .

$${let}\:\mathrm{sin}\:{x}+\mathrm{cos}\:{y}={S} \\ $$ $$\:\:\:\:\:\:\:\mathrm{sin}\:{x}−\mathrm{cos}\:{y}={D} \\ $$ $$\:\:\:\:\:\:\:\mathrm{sin}\:{x}\mathrm{cos}\:{y}\:\:=\:{P} \\ $$ $${Then}\:{what}\:{is}\:{given}\:{converts}\:{to} \\ $$ $$\left(\mathrm{sin}\:^{\mathrm{2}} {x}−\mathrm{cos}\:^{\mathrm{2}} {y}\right)^{\mathrm{2}} +\mathrm{2sin}\:^{\mathrm{2}} {x}\mathrm{cos}\:^{\mathrm{2}} {y}+\mathrm{2} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{4sin}\:{x}\mathrm{cos}\:{y} \\ $$ $$\Rightarrow\:\:{S}^{\:\:\mathrm{2}} {D}^{\mathrm{2}} +\mathrm{2}{P}^{\:\mathrm{2}} +\mathrm{2}−\mathrm{4}{P}=\mathrm{0} \\ $$ $$\Rightarrow\:\:{S}^{\:\:\mathrm{2}} {D}^{\mathrm{2}} +\mathrm{2}\left({P}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$ $$\:{S}\:{cannot}\:{be}\:{zero}\:{for}\:\:\mathrm{0}\leqslant{x},\:{y}\:\leqslant\frac{\pi}{\mathrm{2}} \\ $$ $${So}\:\:{D}=\mathrm{0}\:\:{and}\:{P}\:=\mathrm{1} \\ $$ $$\Rightarrow\mathrm{sin}\:{x}=\mathrm{cos}\:{y}\:\:{and}\:\mathrm{sin}\:{x}\mathrm{cos}\:{y}=\mathrm{1} \\ $$ $${this}\:{can}\:{be}\:{true}\:{only}\:{if} \\ $$ $$\:\:\:{x}=\frac{\pi}{\mathrm{2}}\:\:{and}\:{y}=\mathrm{0} \\ $$ $${So}\:\:\mathrm{sin}\:{x}+\mathrm{cos}\:{y}\:=\mathrm{2}\:. \\ $$

Commented byTinkutara last updated on 29/Oct/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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