Daily Graph 12-08-09

[sin x] = [cos x]

2 Answers

106

Asish Mahapatra ·2009-10-07 03:38:54

[sinx] = 0 in x = [0,pi/2) U (pi/2,pi]
= 1 in x = pi/2
= -1 in x = (pi,2pi)

[cosx] = 1 in x = 0
= 0 in x = (0, pi/2]
= -1 in x = (pi/2, 3pi/2)
= 0 in x = [3pi/2, 2pi]

So [sinx]=[cosx] when x = (0,pi/2) U (pi/3pi/2)

The rest points are decided by periodicity

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