if sinx+siny≥cosα.cosx for every x belongs to R,then siny+cosα is equal to ________?

Hi Menka,

This is a good conceptual Question.

Since the inequality is true for every x belonging to R.

It is true even for x = -(pie)/2

So, (-1) + sinY ≥ 0.

So sinY ≥ 1.

But SinY can never be greater than 1. So sinY=1 (always).

And so, SinY+Cos(Alpha) = 1+cos(Alpha) = 2cos²(Alpha/2).

Best Regards,

Ashwin (IIT Madras).

as x belongs to r

let x=0

then

siny>=cosa

y>=90-a

let 

y=90-a

then siny+cosa=2siny=2cosa<=2

if y>90-a

siny+cosa<2 for ex:a=0

then y>90 then siny<1 and cosa=1 therefore

siny+cosa<=2

therefore siny+cosa<=2