Is this a valid approach? (1 Viewer)

laters · Oct 20, 2015

Hi, in the 2007 HSC question 7 a) i) says "show that sin x < x for x > 0"

The method I used I have seen many times before but not in this particular question, I said that for some u>0, cos u <= 1 using a graph and then integrated with borders 0 to x. Pretty simple.

The reason I am asking this is because the marking guidelines did not mention this method - I thought many more people would have done such a thing - so I am unsure whether my method is valid. I'm pretty sure it is but just making sure.

Thanks!

kawaiipotato · Oct 20, 2015

It should be valid. Make sure you integrated from 0 to x to get rid of constant

InteGrand · Oct 20, 2015

laters said:
Hi, in the 2007 HSC question 7 a) i) says "show that sin x < x for x > 0"

The method I used I have seen many times before but not in this particular question, I said that for some u>0, cos u <= 1 using a graph and then integrated with borders 0 to x. Pretty simple.

The reason I am asking this is because the marking guidelines did not mention this method - I thought many more people would have done such a thing - so I am unsure whether my method is valid. I'm pretty sure it is but just making sure.

Thanks!

$Yes, that's valid: $\sin x = \int _0 ^x \cos t \text{ d}t < \int _0 ^x 1 \text{ d}t = x$, for $x>0$. In general, if $f(t) \leq g(t)$ for all $t$ in the interval $a\leq t \leq b$ (so $a\leq b$), we have $\int _a ^b f(t) \text{ d}t \leq \int _a ^b g(t) \text{ d}t$.$

laters · Oct 20, 2015

InteGrand said:
$Yes, that's valid: $\sin x = \int _0 ^x \cos t \text{ d}t < \int _0 ^x 1 \text{ d}t = x$. In general, if $f(t) \leq g(t)$ for all $t$ in the interval $a\leq t \leq b$, we have $\int _a ^b f(t) \text{ d}t \leq \int _a ^b g(t) \text{ d}t$.$

Thanks!

Is this a valid approach? (1 Viewer)

laters

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kawaiipotato

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InteGrand

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laters

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