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For all x in [0,90] show that cos(sinx)>sin(cosx) I understood the solution given in my book which said Cosx +sinx Cosx But if Sinx Then when we take sin and cos both sides resp we get two different opposite answers. Please explain to me where I have gone wrong.

For all x in [0,90] show that cos(sinx)>sin(cosx)
I understood the solution given in my book which said 
Cosx +sinx
Cosx
But if 
Sinx
Then when we take sin and cos both sides resp we get two different opposite answers.
Please explain to me where I have gone wrong.

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
Hello Student,
Please find answer to your question below

I am assuming 90 to be pi/2.
For
x\in [0, \frac{\pi }{2}]
0\leq sinx\leq 1
0\leq cosx\leq 1
f_{1}(x) = cos(sinx)
sinx will take value from 0 to 1 in increasing order.
cos(sin1)\leq f_{1}(x)\leq 1
f_{2}(x) = sin(cosx)
cosx will take value from 1 to 0.
0\leq f_{2}(x)\leq sin(cos1)
If you can draw the graph, it will be more clear.

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