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For all x in [ 0 , 90 ] show that cos ( s i n ( x ) ) > s i n ( c o s ( x ) ) I understood the solution given in my book which said cos ( x ) + sin ( x ) ≤ 2 √ 90 cos ( x ) 90 − sin ( x ) But if sin ( x ) 90 − cos ( x ) , then when we take sin and cos of both sides respectively, we get two different opposite answers. Please explain to me where I have gone wrong.

For all x in [0,90] show that cos(sin(x))>sin(cos(x))

I understood the solution given in my book which said 

cos(x)+sin(x)290
cos(x)90sin(x)
But if sin(x)90cos(x), then when we take sin and cos of both sides respectively, 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
6 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)
It is more clear from the graph that

cos(sinx) > sin(cosx)
in [0, pi/2]

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