Can you prove it?

Prove that,

* $\dfrac{d}{dx} \sin x=\cos x$

* $\dfrac{d}{dx} \cos x=-\sin x$

#Calculus

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Comments

We know that the definition of the derivative is the limit of the difference quotient, as such:

$\lim_{h\to0} \frac{f(x+h)-f(x)}{h}$

Let's replace sine in as our function.

$\lim_{h\to0} \frac{\sin(x+h)-\sin(x)}{h}$

Using Ptolemy's identity for sine, $\sin(\alpha + \beta) = \sin \alpha \cos\beta + \cos \alpha \sin \beta$ , we have

$\lim_{h\to0} \frac{\sin (x) \cos(h) + \cos(x) \sin (h)-\sin(x)}{h}$

Factor out the $\sin (x)$

$\lim_{h\to0} \frac{\sin (x) (\cos(h) -1)+ \cos(x) \sin (h)}{h}$

Split up the fraction

$\lim_{h\to0} \frac{\cos(h) -1}{h}\sin(x) + \lim_{h\to0}\frac{\sin(h)}{h}\cos(x)$

Using knowledge of the squeeze theorem, we know that $\lim_{x\to0} \frac{\sin(x)}{x} = 1$ and $\lim_{x\to0} \frac{\cos(x)-1} {x} = 0$ . Thus,

$\lim_{h\to0} 0\times \sin(x) + \lim_{h\to0} 1 \times \cos(x) \Longrightarrow \cos(x)$

Therefore,

$\frac{d}{dx} \sin(x) = \cos(x)$

For cosine, we have

$\lim_{h\to0} \frac{\cos(x+h)-\cos(x)}{h}$

Ptolemy's identity for cosine is $\cos(\alpha + \beta) = \cos \alpha \cos\beta - \sin \alpha \sin \beta$ , so

$\lim_{h\to0} \frac{\cos (x) \cos(h) - \sin(x) \sin (h)-\cos(x)}{h}$

Again, factor out the $\cos(x)$ and split up the fraction.

$\lim_{h\to0} \cos (x)\frac{( \cos(h) - 1)}{h} - \lim_{h\to0} \sin(x) \frac{ \sin (h)}{h}$

Using our sine and cosine equations from the squeeze theorem (above), we have

$\lim_{h\to0} \cos (x)\times 0 - \lim_{h\to0} \sin(x)\times 1 \Longrightarrow -\sin(x)$

Therefore,

$\frac{d}{dx} \cos(x) = -\sin(x)$

Lucas Chaves Meyles - 3 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$