To relate (π)/(3) = cos^(-1)((1)/(2)) with the angle whose cosine is (3)/(5), we consider the function f(x) = cos^(-1)(x) on the closed interval [(1)/(2), (3)/(5)]. Notice that (1)/(2) = 0.5 and (3)/(5) = 0.6, so (1)/(2) < (3)/(5).

To relate (π)/(3) = cos^(-1)((1)/(2)) with the angle whose cosine is (3)/(5), we consider the function f(x) = cos^(-1)(x) on the closed interval [(1)/(2), (3)/(5)]. Notice that (1)/(2) = 0.5 and (3)/(5) = 0.6, so (1)/(2) < (3)/(5).

Since f(x) = cos^(-1)(x) is continuous on [(1)/(2), (3)/(5)] and differentiable on ((1)/(2), (3)/(5)) with f'(x) = -(1)/(√(1-x²)), by Lagrange's Mean Value Theorem there exists some c ((1)/(2), (3)/(5)) satisfying this equation.

Since f(x) = cos^(-1)(x) is continuous on [(1)/(2), (3)/(5)] and differentiable on ((1)/(2), (3)/(5)) with f'(x) = -(1)/(√(1-x²)), by Lagrange's Mean Value Theorem there exists some c ((1)/(2), (3)/(5)) satisfying this equation.

Evaluating the denominator, we have (3)/(5) - (1)/(2) = (1)/(10), and cos^(-1)((1)/(2)) = (π)/(3). Multiplying both sides by (1)/(10) gives cos^(-1)((3)/(5)) - (π)/(3) = -(1)/(10√(1-c²)).

Since (1)/(2) < c < (3)/(5), squaring gives (1)/(4) < c² < (9)/(25). Subtracting from 1 gives 1 - (9)/(25) < 1 - c² < 1 - (1)/(4), so (16)/(25) < 1 - c² < (3)/(4). Taking square roots yields (4)/(5) < √(1-c²) < (√3)/(2).

Since (1)/(2) < c < (3)/(5), squaring gives (1)/(4) < c² < (9)/(25). Subtracting from 1 gives 1 - (9)/(25) < 1 - c² < 1 - (1)/(4), so (16)/(25) < 1 - c² < (3)/(4). Taking square roots yields (4)/(5) < √(1-c²) < (√3)/(2).

Taking reciprocals and multiplying by (1)/(10), we obtain (1)/(10 · (√3)/(2)) < (1)/(10√(1-c²)) < (1)/(10 · (4)/(5)), which simplifies to (1)/(5√3) < (1)/(10√(1-c²)) < (1)/(8). Negating and adding (π)/(3) completes the proof.

Taking reciprocals and multiplying by (1)/(10), we obtain (1)/(10 · (√3)/(2)) < (1)/(10√(1-c²)) < (1)/(10 · (4)/(5)), which simplifies to (1)/(5√3) < (1)/(10√(1-c²)) < (1)/(8). Negating and adding (π)/(3) completes the proof.

Solve: Prove using Lagrange's Mean Value theorem. pi/3 - 1/(5sqrt(3)) > 3/5 > pi/3 - 1/8

Answer: $\frac{\pi}{3} - \frac{1}{5\sqrt{3}} > \cos^{-1}\left(\frac{3}{5}\right) > \frac{\pi}{3} - \frac{1}{8}$

Step-by-step solution

Step 1: Define the function and interval

To relate $\frac{\pi}{3} = \cos^{-1}\left(\frac{1}{2}\right)$ with the angle whose cosine is $\frac{3}{5}$ , we consider the function $f(x) = \cos^{-1}(x)$ on the closed interval $\left[\frac{1}{2}, \frac{3}{5}\right]$ . Notice that $\frac{1}{2} = 0.5$ and $\frac{3}{5} = 0.6$ , so $\frac{1}{2} < \frac{3}{5}$ .

Step 2: Apply Lagrange's Mean Value Theorem

Since $f(x) = \cos^{-1}(x)$ is continuous on $\left[\frac{1}{2}, \frac{3}{5}\right]$ and differentiable on $\left(\frac{1}{2}, \frac{3}{5}\right)$ with $f'(x) = -\frac{1}{\sqrt{1-x^2}}$ , by Lagrange's Mean Value Theorem there exists some $c \in \left(\frac{1}{2}, \frac{3}{5}\right)$ satisfying this equation.

Step 3: Simplify the Mean Value expression

Evaluating the denominator, we have $\frac{3}{5} - \frac{1}{2} = \frac{1}{10}$ , and $\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}$ . Multiplying both sides by $\frac{1}{10}$ gives $\cos^{-1}\left(\frac{3}{5}\right) - \frac{\pi}{3} = -\frac{1}{10\sqrt{1-c^2}}$ .

Step 4: Bound the derivative term

Since $\frac{1}{2} < c < \frac{3}{5}$ , squaring gives $\frac{1}{4} < c^2 < \frac{9}{25}$ . Subtracting from 1 gives $1 - \frac{9}{25} < 1 - c^2 < 1 - \frac{1}{4}$ , so $\frac{16}{25} < 1 - c^2 < \frac{3}{4}$ . Taking square roots yields $\frac{4}{5} < \sqrt{1-c^2} < \frac{\sqrt{3}}{2}$ .

Step 5: Establish the final inequality

Taking reciprocals and multiplying by $\frac{1}{10}$ , we obtain $\frac{1}{10 \cdot \frac{\sqrt{3}}{2}} < \frac{1}{10\sqrt{1-c^2}} < \frac{1}{10 \cdot \frac{4}{5}}$ , which simplifies to $\frac{1}{5\sqrt{3}} < \frac{1}{10\sqrt{1-c^2}} < \frac{1}{8}$ . Negating and adding $\frac{\pi}{3}$ completes the proof.

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