Rolle's theorem !!!!!!!!!!!!!!!!!!!!!?

Question

Show that there exists c, a<c<b such that

[sinb - sina ] divided by [cosa - cosb] equals cot c

Answer 1

Stop shouting! Most of us don&#039;t like noise.

As for your problem, it is a simple algebra problem which relies on the standard trig identities - no Rolle&#039;s theorem, no calculus:

http://en.wikipedia.org/wiki/List_of_tri...

[sin(b) - sin(a)] = 2 cos((b+a)/2) sin((b-a)/2)

[cos(a) - cos(b)] = - [cos(b) - cos(a)] = -[-2sin((b+a)/2) sin((b-a)/2)]

so [sin(b) - sin(a)] / [cos(a) - cos(b)] =

[2 cos((b+a)/2) sin((b-a)/2)] / -[-2sin((b+a)/2) sin((b-a)/2)]

= [cos((b+a)/2) sin((b-a)/2)] / [sin((b+a)/2) sin((b-a)/2)]

= cos((b+a)/2) / sin((b+a)/2) = cot ((b+a)/2)

Since (b+a)/2 is halfway between a and b, ...