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Trigonometry> For all x in [0,90] show that cos(sinx)>s...

question mark

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.

Jai Mahajan , 10 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 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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