Question

Prove that `cosx >= 1-x^2/2 AA x in RR` ?

Answer 1

Please verify my solution below. Is there any other way or better explanation?

Explanation:

Let,
`y = f(x) = cosx`
`f(x) = cosx , f(0) = 0`
`f'(x) = -sinx, f'(0) = -1`
`f''(x) = -cosx, f''(0) = 0`
`f'''(x) = sinx, f'''(0) = 1`
`f^4(x) = cosx, f^4(0) = 0`
`f^5(x) = -sinx, f^5(0) = -1`

As we can see it repeats itself.

Now, for any value of `x, 1-x^2/2 <= -1`

Therefore, we can say,
`cosx >= 1-x^2/2`

Answer 2

Consider the following function:

`f(x) = cos(x) - 1 + x^2/2`

Then the assertion that `cosx ge 1-x^2/2` is identical to showing that:

`cosx -1 + x^2/2 ge 0 iff f(x) ge 0`

When `x=0` we have `f(0) = 0`

So to prove that `f(x) ge 0` we attempt to verify that the function `f(x)` is strictly increasing. To do this, take the derivative:

`f'(x) = -sin(x) + x`

If we show `f'(x)` is strictly positive for `x>0`, then we can apply the mean value theorem to show `f(x)` must be strictly increasing. Differentiating again we get:

`f''(x) = -cos(x) + 1`

Clearly as `cosx le 1 => 1-cosx ge 0`, and so we can use this (and the mean value theorem again) to prove `f'(x) > 0 AA x in RR`, hence the result follows.