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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 But then Cosx Then take sin and cos both sides resp we get two different opposite answers. Please explain to me where I 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
But then
Cosx
Then take sin and cos both sides resp we get two different opposite answers.
Please explain to me where I I have gone wrong.

Grade:12

1 Answers

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

I am assuming you meant 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 the value from 0 to 1 in increasing order.
cos(sin1)\leq f_{1}(x)\leq 1
f_{2}(x) = sin(cosx)
cosx will take the value from 1 to 0.
0\leq f_{2}(x)\leq sin(cos1)
You can draw the graph of sinx & cosx, it is clearly seen that
cos(sinx) > sin(cosx), x\in [0, \frac{\pi }{2}],

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