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For all θ in [ 0 , 2 π ] , cos ( sin θ ) □ sin ( cos θ )
This problem is part of the set Trigonometry .
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We have
cos θ + sin θ = 2 ( 2 1 cos θ + 2 1 sin θ )
cos θ sin θ = 2 ( sin 4 π cos θ + cos 4 π sin θ )
cos θ + sin θ = 2 sin ( 4 π + θ )
∴ cos θ + sin θ ≤ 2 < 2 π
( a s 2 ≈ 1 . 4 1 4 1 , 2 π ≈ 1 . 5 7 )
∴ cos θ + sin θ < 2 π
cos θ < 2 π − sin θ
sin ( cos θ ) < sin ( 2 π − sin θ )
sin ( cos θ ) < cos ( sin θ )
cos ( sin θ ) > sin ( cos θ )
Note that we do not have an equality, because if we did, then we would have the result cos θ + sin θ > 2 , which is not possible because cos θ + sin θ = sin ( 4 π + θ ) .
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For theta [0,pi/2] ,sin increases from 0 to 1 , while for cos it decreases from 1 to 0. Hence cos(sin theta) and sin(cos theta) both are decreasing functions with initial values at theta = 0 is 1 and sin1 and value at theta = pi/2 are cos1 and 0 respectively. Therefore we can say that the left hand side decreases from 1 -> cos1 while right hand side decreases from sin1 -> 0. hence '>' is the answer!