Question Number 10458 by paonky last updated on 10/Feb/17

what is the maximum and minimum  value of   sinx + cosx + sinxcosx

Answered by mrW1 last updated on 10/Feb/17

sinx +cos x  =(√2)(sin x×((√2)/2)+cos x×((√2)/2))  =(√2)(sin x×cos (π/4)+cos x×sin (π/4))  =(√2)sin (x+(π/4))  sin xcos x=(1/2)sin 2x=−(1/2)cos (2x+(π/2))  =−(1/2)cos 2(x+(π/4))=−(1/2)[1−2sin^2  (x+(π/4))]  =sin^2  (x+(π/4))−(1/2)    f=sinx + cosx + sinxcosx  =(√2)sin (x+(π/4))+sin^2  (x+(π/4))−(1/2)  =(((√2)/2))^2 +2×((√2)/2)sin (x+(π/4))+sin^2  (x+(π/4))−1  =[((√2)/2)+sin (x+(π/4))]^2 −1    when sin (x+(π/4))=−((√2)/2) , i.e. x=π+2iπ or x=((3π)/2)+2iπ  ⇒ f_(min) =−1  when sin (x+(π/4))=1, i.e. x=(π/4)+2iπ  ⇒ f_(max) =(((√2)/2)+1)^2 −1=(1/2)+(√2)≈1.914

Commented bypaonky last updated on 10/Feb/17

thank you sir