Answer to Question #184210 in Abstract Algebra for Rohit

Question

Use Cauchy’s mean value theorem to prove that:

{Cos(alpha)- cos(beta) }/{sin(alpha) -sin(beta) }=tan(theta)

Answer 1

Use Cauchy's Mean Value Theorem to prove the given result.

Let $f(x)=sinx,g(x)=cosx$ for all $x\in \left [ \alpha ,\beta \right ]\subset \left [ 0,\frac{\pi }{2} \right ]$

We know that $f(x)$ and $f(x)$ are derivable on $\left [ \alpha ,\beta \right ]\subset R$

And $f(x)=sinx,g(x)=cosx\Rightarrow f'(x)=cosx,g'(x)=-sinx\neq 0\forall x\in \left [ \alpha ,\beta \right ]$

By Cauchy's Mean Value theorem there exists $\theta \in \left ( \alpha ,\beta \right )$ so that

$\frac{f(\beta )-f(\alpha )}{g(\beta )-g(\alpha )}=\frac{f'(\theta )}{g'(\theta )}$

$\Rightarrow \frac{sin\beta -sin\alpha }{cos\beta -cos\alpha }=\frac{cos\theta }{-sin\theta }$

$\Rightarrow \frac{cos\beta -cos\alpha }{sin\beta -sin\alpha }=-\frac{sin\theta }{cos\theta }$

$\Rightarrow \frac{cos\beta -cos\alpha }{sin\beta -sin\alpha }=-tan\theta$

$\Rightarrow \frac{cos\alpha -cos\beta }{sin\alpha -sin\beta }=-tan\theta$

Therefore

$\Rightarrow \frac{cos\alpha -cos\beta }{sin\alpha -sin\beta }=-tan\theta$