Prove thatcos x ≥ 1 − 12x2 and sin x ≥ x − 12x3 for all x ∈ [0, π/2].Hint: Use the mean value theorem

Sure, let's prove these inequalities step by step.

1. Prove that cos x ≥ 1 - 1/2x^2 for all x ∈ [0, π/2].

We start by considering the function f(x) = cos(x) - (1 - 1/2x^2) on the interval [0, π/2]. We want to show that f(x) ≥ 0 for all x in this interval.

The derivative of f(x) is f'(x) = -sin(x) - x. By the Mean Value Theorem, there exists a c in (0, π/2) such that f'(c) = (f(π/2) - f(0)) / (π/2 - 0).

We know that f(0) = cos(0) - (1 - 1/20^2) = 0 and f(π/2) = cos(π/2) - (1 - 1/2(π/2)^2) = -π^2/8.

So, f'(c) = -π^2/8 / (π/2) = -π/4.

Since -sin(c) - c ≤ 0 for all c in [0, π/2] (because sin(c) ≤ 1 and -c ≤ 0), we have f'(c) ≥ -π/4.

Therefore, f(x) is decreasing on [0, π/2], so f(x) ≥ f(π/2) = -π^2/8 for all x in [0, π/2].

But -π^2/8 > -1/2*(π/2)^2 = -π^2/8, so f(x) > -π^2/8 for all x in [0, π/2].

Therefore, cos x ≥ 1 - 1/2x^2 for all x in [0, π/2].

2. Prove that sin x ≥ x - 1/2x^3 for all x ∈ [0, π/2].

We use a similar approach. Consider the function g(x) = sin(x) - (x - 1/2x^3) on the interval [0, π/2]. We want to show that g(x) ≥ 0 for all x in this interval.

The derivative of g(x) is g'(x) = cos(x) - (1 - 3/2x^2). By the Mean Value Theorem, there exists a d in (0, π/2) such that g'(d) = (g(π/2) - g(0)) / (π/2 - 0).

We know that g(0) = sin(0) - (0 - 1/20^3) = 0 and g(π/2) = sin(π/2) - (π/2 - 1/2(π/2)^3) = 1 - π/2 + π^3/32.

So, g'(d) = (1 - π/2 + π^3/32) / (π/2) = 2/π - 1 + π^2/16.

Since cos(d) - (1 - 3/2d^2) ≥ 0 for all d in [0, π/2] (because cos(d) ≥ 0 and 1 - 3/2d^2 ≤ 1), we have g'(d) ≥ 2/π - 1 + π^2/16.

Therefore, g(x) is increasing on [0, π/2], so g(x) ≥ g(0) = 0 for all x in [0, π/2].

Therefore, sin x ≥ x - 1/2x^3 for all x in [0, π/2].