Question

show that cos(x)> 1-(x^2/2)

Answer 1

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Answer 2

How about proving that it is false?! Try x = 0.

Answer 3

Do we get the Taylor Series?

Answer 4

Since this is differentiable for all R, let's apply the Mean Value Theorem
to f for all x between 0 and t:

f(t) - f(0) = f '(c) (t - 0) for some c between 0 and t
==> cos t - 1 + t^2/2 = (c - sin c) * t.

However, c ≥ sin c for all c ≥ 0.
==> (c - sin c) * t ≥ 0 for t ≥ 0.

Similarly, c ≤ sin c for all c < 0 (by using the above inequality and multiplying both sides by -1)
==> (c - sin c) * t ≥ 0 for t < 0.

Therefore, cos t - 1 + t^2/2 ≥ 0 for all t
==> cos t ≥ 1 - t^2/2 for all t.

I hope this helps!

Answer 5

First, we must establish that \(\ge\) was intended and not \(>\).