sin(pi)' = -1

[jessebu]

The problem is to prove in detail that sin'(pi) = -1, without using the fact that cos (pi) = -1.

I've tried a number of approaches but none of them are working for me, unfortunately. Any advice on how to get started would be amazing.

 

[galactus]

The derivative is the slope of the line at that point.

See if a line with slope -1 is tangent to sin(x) at x=Pi.

 

[daon]

I circumvented the rule

\(\displaystyle -1 = -sin(\frac{\pi}{2}) = \lim_{h \to 0} \frac{cos(\frac{\pi}{2}+h) - cos(\frac{\pi}{2})}{h} = \lim_{h \to 0} \frac{cos(\frac{\pi}{2}+h)}{h}\)

From a know trig identity (or this "lemma"): \(\displaystyle cos(\pi/2+h) = cos(-\pi/2+\pi+h) = cos(-\pi/2)cos(\pi+h)-sin(-\pi/2)sin(\pi+h) = sin(\pi+h)\)

So,

\(\displaystyle -1 = \lim_{h \to 0} \frac{cos(\frac{\pi}{2}+h)}{h} = \lim_{h \to 0} \frac{sin(\pi+h)}{h}= \lim_{h \to 0} \frac{sin(\pi+h)-sin(\pi)}{h} = sin'(\pi)\)