Answer to Question #114625 in Calculus for ANJU JAYACHANDRAN

Question

Prove that cos x graterthan or equal to 1(-x^2/2)for all x in the interval [0, 2pi]

Answer 1

$We \ apply \ the \ mean \ value \ theorem \ to\\ f(x)=\frac{x^2}{2} +cos x \ on \ the \ interval \ [0, 2\pi].\\ Show \ that \ cos x > 1 - \frac{x^2}{2}.\\ We \ know \ that \ \frac{df(x)}{dx} = x - sin x, for \ x>0. \\ By \ the \ MVT, if \ x > 0, then \\ f(x) - f(0) = (x - 0)f\rq(c) \ for \ some \ c > 0. \\ f(x) - f(0) = (x -0)f\rq(c) = x(c - sin c). \\ Let \ 𝑔(x)= x − sin x .\\ We \ can \ show \ that \\ 𝑔\rq(x) = 1-cos x \\ which \ is \ always \ greater \ than \ 0 \ due \ to \ bounded \ nature \ of \ cos x. \\ As \ 𝑔(0) = 0 \ and \ it \ is \ an \ increasing \ function, \\ { 𝑔\rq(x) > 0 \ ∀ x > 0 } , thus \ 𝑔(x) > 0 \ ∀ x > 0. \\ So, f(x) - f(0) > 0 \ ∀ x > 0 \ as \ x > 0 \ and \\ c - sin c > 0 \ ∀ c > 0 \ (0 < c < x). \\ As \ f(x) > f(0) = 1, then \ \frac{x^2}{2 }+ cos x > 1 \\ cos x > 1 - \frac{x^2}{2} \ for \ x∈[0,2\pi]$