Question

Verify Cauchy's mean value theorem for functions f(x) = sin x, g(x) = cos x in the interval 0,π/2​

Answer 1

We have to verify Cauchy's mean value theorem for the given functions

$f(x) = sinx \\g(x)= cosx$

for the interval x ∈ $(0,\frac{\pi }{2})$

• Cauchy's mean value theorem states that,

If two functions f(x) and g(x) are continous on closed interval [a.b] and     differentiable on open interval (a,b) and g(x) ≠ 0, ∀ x∈ (a,b) , Then there exist a point c ∈ (a,b) such that

               $\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)}$

• Now, verifying cauchy's mean value theorem for the given functions

           $f(x) = sinx \\g(x)= cosx$

• As sinx and cosx are continous and differential for all x ∈ R , therefore they are also continous for the interval x ∈ $[0,\frac{\pi }{2}]$ and differentiable for the interval x ∈ $(0,\frac{\pi }{2})$.

       Hence cauchy's mean value theorem is valid

• Now,

                  $\frac{f(\frac{\pi }{2} ) - f(0)}{g(\frac{\pi }{2} ) - g(0)} = \frac{f'(c)}{g'(c)}$

                  $\frac{sin\frac{\pi }{2} - sin0}{cos\frac{\pi }{2} - cos0} = \frac{cosc}{-sinc}$

                  $\frac{1-0}{0-1} = -\frac{cosc}{sinc}$

                  $\frac{cosc}{sinc} = 1\\cosc = sinc \\sin(\frac{\pi }{2} -c) = sinc\\\frac{\pi }{2} -c = c\\2c = \frac{\pi }{2}$

               ∴  $c = \frac{\pi }{4}$ ∈ $(0,\frac{\pi }{2})$

         Hence Cauchy's mean value theorem is verified.

Answer 2

Step-by-step explanation:

Given : f(x) = sin x, g(x) = cos x

To find: Verify Cauchy's mean value theorem for the function sin x and cos x in the interval [0,π/2]

Solution:

Cauchy's mean value theorem:if interval [a,b]

$\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)}$

Put the function and first order derivative in the formula

$\frac{sin \: \frac{\pi}{2} - sin \: 0}{cos \: \frac{\pi}{2} - cos \: 0} = \frac{cos \: c}{ - sin \: c} \\ \\ \frac{1 - 0}{0 - 1} = - \frac{cos \: c}{sin \: c} \\ \\ or \\ \\ -\frac{1}{1} = - cot \: c \: ...eq1$

Now,put the value of 1 in terms of angle of cotangent.

$- 1= - cot \: c$

$cot \:c = 1 \\\\ cot \:c=cot\:\left(\frac{\pi}{4}\right)\\\\ c=\frac{\pi}{4}$

c lies in the closed interval [0,π/2].Thus, Cauchy's mean value theorem holds.

Final answer:

c=π/4 lies in the closed interval [0,π/2].

Thus, Cauchy's mean value theorem holds and have been proved.

Hope it helps you.

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