If sin(πcosθ)=cos(πsinθ),then sin2θ may take value(s)

a.3/4

b.-3/4

c.1/4

d.None of these

Hi Menka,

Here cos(Πsinx)=sin(Πcosx)

So cos(Πsinx) = cos(Π/2-Πcosx)

So Πsinx = (2nΠ) +/- (Π/2-Πcosx)

So sinx = (2n) +/- (1/2 - cosx), where n is any integer.

So n can be 0,1,-1 (Other values of n would mean either sinx>1 or sinx<-1 both of which is not possible).

n=0

sinx = 1/2-cosx or sinx=cosx-1/2

ie sinx+cos=1/2 or cosx-sinx=1/2

by squaring both sides, 1+sin2x = 1/4 or 1-sin2x=1/4

sin2x = -3/4 or sin2x=3/4.

n=1

sinx = 2-1/2+cosx

sinx-cosx = 3/2 (Not possible) as squaring would give sin2x = 1-9/4 = -5/4 (less than -1).

Similarly you can check for n=-1, also not possible.

Hence sin2x = 3/4 or -3/4

Option (A)/(B).

Hope that helps.

Best Regards,

Ashwin (IIT Madras).

Approved