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derivatives of 
and 
Theorem $$\frac{d}{dx} (\sin x)=\cos x$$
Proof.
| $\displaystyle \frac{d}{dx} (\sin x)$ | $=\displaystyle \lim_{h \to 0} \frac{\sin (x+h)-\sin x}{h}$ |
| $=\displaystyle \lim_{h \to 0} \frac{\sin x \cos h+\cos x \sin h-\sin x}{h}$ by addition formula for sine | |
| $=\displaystyle \lim_{h \to 0} \frac{\sin x (\cos h-1)+\cos x \sin h}{h}$ | |
| $=\displaystyle \lim_{h \to 0} \left( \sin x \cdot \frac{\cos h-1}{h} +\cos x \cdot \frac{\sin h}{h} \right)$ | |
| $=\displaystyle \sin x \left( \lim_{h \to 0} \frac{\cos h-1}{h} \right) +\cos x \left( \lim_{h \to 0} \frac{\sin h}{h} \right)$ by this entry | |
| $=\displaystyle \sin x \cdot 0+\cos x \cdot 1$ by this theorem and its corollary | |
| $=\cos x$ |
Theorem $$\frac{d}{dx} (\cos x)=-\sin x$$
Proof.
| $\displaystyle \frac{d}{dx} (\cos x)$ | $=\displaystyle \lim_{h \to 0} \frac{\cos (x+h)-\cos x}{h}$ |
| $=\displaystyle \lim_{h \to 0} \frac{\cos x \cos h-\sin x \sin h-\cos x}{h}$ by addition formula for cosine | |
| $=\displaystyle \lim_{h \to 0} \frac{\cos x (\cos h-1)+\sin x \sin h}{h}$ | |
| $=\displaystyle \lim_{h \to 0} \left( \cos x \cdot \frac{\cos h-1}{h} -\sin x \cdot \frac{\sin h}{h} \right)$ | |
| $=\displaystyle \cos x \left( \lim_{h \to 0} \frac{\cos h-1}{h} \right) -\sin x \left( \lim_{h \to 0} \frac{\sin h}{h} \right)$ by this entry | |
| $=\displaystyle \cos x \cdot 0-\sin x \cdot 1$ by this theorem and its corollary | |
| $=-\sin x$ |
How to Cite This Entry
Warren Buck. "derivatives of and " (version 5). PlanetMath.org. Freely available at http://planetmath.org/DerivativesOfSinXAndCosX.html.
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