How do you simplify $\frac{sin ((\frac{π}{2}) + h) - sin (\frac{π}{2})}{h}$ $[sin((pi/2)+h)-sin(pi/2)]/(h)$ ?

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How do you simplify $\frac{sin ((\frac{π}{2}) + h) - sin (\frac{π}{2})}{h}$ $[sin((pi/2)+h)-sin(pi/2)]/(h)$ ?

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Answer 1 · 2016-03-15T20:29:29

It is

$\frac{sin ((\frac{π}{2}) + h) - sin (\frac{π}{2})}{h} = \frac{(sin (\frac{π}{2}) cosh + cos (\frac{π}{2}) \cdot sinh) - sin (\frac{π}{2})}{h} = \frac{cosh - 1}{h}$ $[sin((pi/2)+h)-sin(pi/2)]/(h)=[(sin(pi/2)cosh+cos(pi/2)*sinh)-sin(pi/2)]/[h]=(cosh-1)/[h]$

Note that

1. $sin (a + b) = sin a \cdot cos b + cos a \cdot sin b$ $sin(a+b)=sina*cosb+cosa*sinb$
2. $sin (\frac{π}{2}) = 1$ $sin(pi/2)=1$ , $cos (\frac{π}{2}) = 0$ $cos(pi/2)=0$

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How do you simplify $\frac{sin ((\frac{π}{2}) + h) - sin (\frac{π}{2})}{h}$ $[sin((pi/2)+h)-sin(pi/2)]/(h)$ ?

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