How do you find a dervative of this

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how do you find the derivative of this problem and what rule do you use

lim sin(pi+h)-Sin pi
h-> 0 h
 
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… sinπ = 0 …

Hi janemba! :smile:

Do you mean \frac{sin(\pi + h) -sin\pi}{h} ?

If so, remember that sinπ = 0, so it's just \frac{sin(\pi + h)}{h}\,.

Which is … ? :smile:
 
yes but how did you got the answer
 
How do you expand:

sin(x+y)=?

Trig identity ...
 
It's even simpler than that, take a look at a unit circle and think about what happens with every rotation of pi radians.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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