Trig help please!?

Question

Prove by counter example, that the statement:
cosec x - sin x > 0 for all values in the interval 0< x < Pi
is false

The wording of the question is confusing me!

Am I proving that it is false and doing that by giving a value which makes the equation equal below zero?

Answer 1

For x = pi /2, sin x =1 and cosec x = 1,

so : for x = pi /2 where 0< pi/2< pi, cosec x - sin x = 0, which contradicts: cosec x - sin x> 0
So the statement:
cosec x - sin x > 0 for all values in the interval 0< x < Pi
is false

Answer 2

You've to prove that the statement is false, citing an example.

x = Pi/2 should do the trick in which case,

cosec x - sin x = 0

Answer 3

Let x=¶/2
cosec x - sinx = 1/sinx - sinx
= 1/sin(¶/2) - sin(¶/2) = 1/1 -1 = 0
This is therefore a counter example (because 0 is not greater than 0)

Answer 4

consider left hand side

cosecx=1/sinx
cosecx-sinx=1/sinx-sinx
=1-square of sinx
=square of cosx
when x =0.5,we get the square of cosx =0.9999>0 .
when x=1 the answer becomes 0.9997>0
when x=2 the answer becomes 0.9988.
therefore cosecx-sinx>0 for all values in the interval 0 note the values 0.5,1and 2 are all in the required range 0

Answer 5

it means find a number between 0 and pi where sin(x) is greater than csc(x). Easiest way would be grab a graphing calculator and graph them both to find a point where that happens.