Question

Verify Cauchy's mean value theorem for the function sin x and cos x in the interval [a, b]

Answer 1

Step-by-step explanation:

Given : sin x and cos x

To find: Verify Cauchy's mean value theorem for the function sin x and cos x in the interval [a, b]

Solution:

Cauchy's mean value theorem:

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

Let f(x) = sin x and g(x)=cos x

Put the function and first order derivative in the formula

$\frac{sin \: b - sin \: a}{cos \: b - cos \: a} = \frac{cos \: c}{ - sin \: c} \\ \\ \frac{sin \: b - sin \: a}{cos \: b - cos \: a} = - \frac{cos \: c}{sin \: c} \\ \\ or \\ \\ \frac{sin \: b - sin \: a}{cos \: b - cos \: a} = - cot \: c \: ...eq1$

Put the formula for sin B-sin C and cos B-cos C as shown below

$sin B-sin C= 2cos\left( \frac{a + b}{2} \right)sin\left(\frac{b - a}{2} \right) \\ \\ cos B-cos C= 2sin\left(\frac{a + b}{2} \right)sin\left( \frac{a - b}{2} \right)$

Put these formulas in eq1

$\frac{ 2cos\left( \frac{a + b}{2} \right)sin\left( \frac{b - a}{2}\right ) }{2sin\left( \frac{a + b}{2}\right )sin\left( \frac{a - b}{2}\right )} = - cot \: c$

or

$-\frac{ 2cos\left( \frac{a + b}{2} \right)sin\left( \frac{a-b}{2} \right) }{2sin\left( \frac{a + b}{2}\right )sin\left( \frac{a - b}{2} \right)} = - cot \: c$

Cancel common terms from numerator and denominator

$- \frac{cos\left( \frac{a+b}{2}\right) }{sin\left( \frac{a+b}{2}\right) } = - cot \: c$$- cot \: \left( \frac{a+b}{2} \right) = - cot \: c \\ \\ c=\frac{a+b}{2} \in[a,b]$

c lies in the closed interval [a,b].Thus, Cauchy's mean value theorem holds.

Final answer:

c=(a+b)/2; which lies in the closed interval [a,b].

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

Hope it helps you.

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