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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).
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