Question

using lagrages mean value theorem prove that
π/6+1/8>sin^-1(3/5)>π/6+1/5√3​

Answer 1

Answer:

sin^(-1)(3/5) - sin^(-1)(0) = f'(d)(3/5 - 0)

Step-by-step explanation:

To use Lagrange's Mean Value Theorem, we need to find a function that satisfies the conditions of the theorem. For this problem, we can consider the function f(x) = sin^(-1)(x), which is continuous and differentiable on the interval [0, 1].

Let's first show that π/6 + 1/8 > sin^(-1)(3/5).

By the Mean Value Theorem, there exists a number c in the interval (0, 3/5) such that:

sin^(-1)(3/5) - sin^(-1)(0) = f'(c)(3/5 - 0)

Taking the derivative of f(x) = sin^(-1)(x), we have:

f'(x) = 1 / √(1-x^2)

So, substituting the value of c, we get:

sin^(-1)(3/5) = f'(c)(3/5)

π/6 = f'(0)

Therefore,

sin^(-1)(3/5) - π/6 = f'(c)(3/5) - f'(0)(1/2)

= (1 / √(1-c^2)) (3/5) - (1/2)

To prove π/6 + 1/8 > sin^(-1)(3/5), it is sufficient to show that the right-hand side of the above inequality is less than 1/8.

Let's consider the function g(x) = (1 / √(1-x^2)) - 1/2 on the interval (0, 3/5). Taking the derivative of g(x), we get:

g'(x) = x / ((1-x^2)^(3/2))

Since x < √(3/5) < 1, we have g'(x) > 0 for all x in the interval (0, 3/5). Therefore, g(x) is increasing on the interval (0, 3/5).

Thus,

g(c) > g(0) = 0

which implies that

(1 / √(1-c^2)) - 1/2 > 0

Rearranging this inequality, we get:

(1 / √(1-c^2)) > 1/2

Squaring both sides, we get:

1 - c^2 > 1/4

c^2 < 3/4

c < √(3/4) = √3/2

Therefore,

sin^(-1)(3/5) - π/6 = (1 / √(1-c^2)) (3/5) - (1/2)

< (1 / √(1-(√3/2)^2)) (3/5) - (1/2)

= (1 / √(1/4)) (3/5) - (1/2)

= (2/5) - (1/2)

= -1/10

So,

π/6 + 1/8 > sin^(-1)(3/5)

Next, we need to show that sin^(-1)(3/5) > π/6 + 1/5√3.

By the Mean Value Theorem, there exists a number d in the interval (0, 3/5) such that:

sin^(-1)(3/5) - sin^(-1)(0) = f'(d)(3/5 - 0)

To know more about  lagrages mean value theorem refer:

https://brainly.in/question/34973122

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