Trigonometry! #34

Geometry Level 3

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For all θ \theta in [ 0 , π 2 ] [0,\frac {\pi}{2}] , cos ( sin θ ) sin ( cos θ ) \cos(\sin \theta) \Box \sin (\cos \theta)

This problem is part of the set Trigonometry .

> < =

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2 solutions

Ramesh Goenka
Jan 31, 2015

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!

Omkar Kulkarni
Feb 13, 2015

We have

cos θ + sin θ = 2 ( 1 2 cos θ + 1 2 sin θ ) \cos\theta+\sin\theta=\sqrt{2}\left(\frac{1}{\sqrt{2}}\cos\theta+\frac{1}{\sqrt{2}}\sin\theta\right)

cos θ sin θ = 2 ( sin π 4 cos θ + cos π 4 sin θ ) \cos\theta\sin\theta= \sqrt{2}\left(\sin\frac{\pi}{4}\cos\theta+\cos\frac{\pi}{4}\sin\theta\right)

cos θ + sin θ = 2 sin ( π 4 + θ ) \cos\theta+\sin\theta= \sqrt{2}\sin\left(\frac{\pi}{4}+\theta\right)

cos θ + sin θ 2 < π 2 \therefore \cos\theta+\sin\theta\le\sqrt{2}<\frac{\pi}{2}

( a s 2 1.4141 , π 2 1.57 ) \left(as~\sqrt{2}\approx1.4141,~\frac{\pi}{2}\approx1.57\right)

cos θ + sin θ < π 2 \therefore \cos\theta+\sin\theta<\frac{\pi}{2}

cos θ < π 2 sin θ \cos\theta<\frac{\pi}{2}-\sin\theta

sin ( cos θ ) < sin ( π 2 sin θ ) \sin(\cos\theta)<\sin\left(\frac{\pi}{2}-\sin\theta\right)

sin ( cos θ ) < cos ( sin θ ) \sin(\cos\theta)<\cos(\sin\theta)

cos ( sin θ ) > sin ( cos θ ) \boxed{\cos(\sin\theta)>\sin(\cos\theta)}

Note that we do not have an equality, because if we did, then we would have the result cos θ + sin θ > 2 \cos\theta+\sin\theta>\sqrt{2} , which is not possible because cos θ + sin θ = sin ( π 4 + θ ) \cos\theta+\sin\theta=\sin\left(\frac{\pi}{4}+\theta\right) .

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