# Thread: Factor Formula

1. ## Factor Formula

When my textbook proving $\frac{d}{dx}(sin\theta)=cos\theta$, it mentions about "Factor Formula", but it never mention it detailly, so I don't understand it.
Can anyone tell me what is "Factor Formula"?

2. Originally Posted by SengNee
When my textbook proving $\frac{d}{dx}(sin\theta)=cos\theta$, it mentions about "Factor Formula", but it never mention it detailly, so I don't understand it.
Can anyone tell me what is "Factor Formula"?
No idea. Can you post the derivation? Perhaps it is different than the one I know.

-Dan

3. Originally Posted by topsquark
No idea. Can you post the derivation? Perhaps it is different than the one I know.

-Dan

To find $\lim_{\theta \rightarrow 0}{\frac{sin\theta}{\theta}}$

Area of $\bigtriangleup OAB$ $<$ Area of sector $OAB$ $<$ Area of $\bigtriangleup OAT$
$0.5(1)^2(sin\theta)<0.5(1^2)(\theta)<0.5(1)(tan\th eta)$
$sin\theta<\theta
$1<\frac{\theta}{sin\theta}<\frac{1}{cos\theta}$

As $\theta\rightarrow 0$,

$cos\theta\rightarrow 1$ and $\frac{1}{cos\theta}\rightarrow 1$.

Therefore, $\frac{\theta}{sin\theta}\rightarrow 1$
or
$\frac{sin\theta}{\theta}\rightarrow 1$
Hence
$\lim_{\theta \rightarrow 0}{\frac{sin\theta}{\theta}}=1$

Therefore, $\frac{\theta}{sin\theta}\rightarrow 1$
or
$\frac{sin\theta}{\theta}\rightarrow 1$
Hence
$\lim_{\theta \rightarrow 0}{\frac{sin\theta}{\theta}}=1$
Actually, I don't understand this part, because
$1<\frac{\theta}{sin\theta}<\frac{1}{cos\theta}$
$1<\frac{\theta}{sin\theta}<1$
What is this?
(1,1)?
Have range like this????
Why $\lim_{\theta \rightarrow 0}{\frac{sin\theta}{\theta}}=1$ ?
As $\theta \rightarrow 0$, the denominator is 0, it is undefined.

Derivative of $sinx$

Let $f(x)=sinx$ where $x$ is measured in radians.

$f(x)=sinx$
$f(x+\delta x)=sin(x+\delta x)$

$f'(x)=\lim_{\delta x \rightarrow 0}{\left[\frac{f(x+\delta x)-f(x)}{\delta x}\right]}$
$f'(x)=\lim_{\delta x \rightarrow 0}{\left[\frac{sin(x+\delta x)-sin x}{\delta x}\right]}$
$f'(x)=\lim_{\delta x \rightarrow 0}{\left[\frac{2cos(x+\frac{1}{2}\delta x)sin\frac{1}{2}\delta x}{\delta x}\right]}$ (factor formula)
$f'(x)=\lim_{\delta x \rightarrow 0}{\left[cos(x+\frac{1}{2}\delta x)\cdot \frac{sin\frac{1}{2}\delta x}{\frac{1}{2}\delta x}\right]}$
$f'(x)=cosx$

Since,
$\lim_{\delta x \rightarrow 0}{cos \left(x+\frac{1}{2}\delta x\right)}=cosx$
and
$\lim_{\delta x \rightarrow 0}{\left(\frac{sin\frac{1}{2}\delta x}{\frac{1}{2}\delta x}\right)}=1$ as $\left(\lim_{\delta x \rightarrow 0}{\frac{sin\theta}{\theta}}=1\right)$

So, what is factor formula???

4. Originally Posted by SengNee
Actually, I don't understand this part, because
$1<\frac{\theta}{sin\theta}<\frac{1}{cos\theta}$
$1<\frac{\theta}{sin\theta}<1$
What is this?
This winds up being an application of something called the "squeeze theorem." You take two expressions, one that is less than the expression you want the limit for and one that is greater. Then if, as you apply your limit, the limit of the lesser and greater expressions are the same then the limit of your expression, the one in the middle, must be the same as these.

So 1 is always less than $\frac{\theta}{sin(\theta)}$ is always less than $sec(\theta)$. As you take the limit as $\theta \to 0$ we find that
$1 < \frac{\theta}{sin(\theta)} < 1$
so
$\lim_{\theta \to 0}\frac{\theta}{sin(\theta)} = 1$

Originally Posted by SengNee
$f'(x)=\lim_{\delta x \rightarrow 0}{\left[\frac{sin(x+\delta x)-sin x}{\delta x}\right]}$
$f'(x)=\lim_{\delta x \rightarrow 0}{\left[\frac{2cos(x+\frac{1}{2}\delta x)sin\frac{1}{2}\delta x}{\delta x}\right]}$ (factor formula)
What your instructor is calling the "factor formula" I call by the name of a "sum to product" formula. By using the sum of angles formulas for sine and cosine
$sin(A \pm B) = sin(A)~cos(B) \pm sin(B)~cos(A)$

$cos(A \pm B) = cos(A)~cos(B) \mp sin(A)~sin(B)$

we may construct the following formulas:
$sin(A) \pm sin(B) = 2~sin \left ( \frac{A \pm B}{2} \right )~cos \left ( \frac{A \mp B}{2} \right)$

$cos(A) + cos(B) = 2~cos \left ( \frac{A + B}{2} \right )~cos \left ( \frac{A - B}{2} \right )$

$cos(A) - cos(B) = -2~sin \left ( \frac{A + B}{2} \right )~sin \left ( \frac{A - B}{2} \right )$

For $sin(x + \delta x) - sin(x)$ we may use the top expression with $A = x + \delta x$ and $B = x$:
$sin(x + \delta x) - sin(x) = 2~sin \left ( \frac{(x + \delta x) - x)}{2} \right )~cos \left ( \frac{(x + \delta x) + x}{2} \right )$

$= 2~sin \left ( \frac{\delta x}{2} \right )~cos \left ( x + \frac{\delta x}{2} \right )$

-Dan

5. Originally Posted by topsquark
This winds up being an application of something called the "squeeze theorem." You take two expressions, one that is less than the expression you want the limit for and one that is greater. Then if, as you apply your limit, the limit of the lesser and greater expressions are the same then the limit of your expression, the one in the middle, must be the same as these.

So 1 is always less than $\frac{\theta}{sin(\theta)}$ is always less than $sec(\theta)$. As you take the limit as $\theta \to 0$ we find that
$1 < \frac{\theta}{sin(\theta)} < 1$
so
$\lim_{\theta \to 0}\frac{\theta}{sin(\theta)} = 1$

What your instructor is calling the "factor formula" I call by the name of a "sum to product" formula. By using the sum of angles formulas for sine and cosine
$sin(A \pm B) = sin(A)~cos(B) \pm sin(B)~cos(A)$

$cos(A \pm B) = cos(A)~cos(B) \mp sin(A)~sin(B)$

we may construct the following formulas:
$sin(A) \pm sin(B) = 2~sin \left ( \frac{A \pm B}{2} \right )~cos \left ( \frac{A \mp B}{2} \right)$

$cos(A) + cos(B) = 2~cos \left ( \frac{A + B}{2} \right )~cos \left ( \frac{A - B}{2} \right )$

$cos(A) - cos(B) = -2~sin \left ( \frac{A + B}{2} \right )~sin \left ( \frac{A - B}{2} \right )$

For $sin(x + \delta x) - sin(x)$ we may use the top expression with $A = x + \delta x$ and $B = x$:
$sin(x + \delta x) - sin(x) = 2~sin \left ( \frac{(x + \delta x) - x)}{2} \right )~cos \left ( \frac{(x + \delta x) + x}{2} \right )$

$= 2~sin \left ( \frac{\delta x}{2} \right )~cos \left ( x + \frac{\delta x}{2} \right )$

-Dan

How to prove them?

$\sin A\pm \sin B=2\sin\left(\frac{A\pm B}{2}\right)\cos\left(\frac{A\mp B}{2}\right)$
$\cos A + \cos B=2\cos\left(\frac{A + B}{2}\right)\cos\left(\frac{A - B}{2}\right)$
$\cos A - \cos B=-2\sin\left(\frac{A + B}{2}\right)\sin\left(\frac{A- B}{2}\right)$

$\cos A\cos B=\frac{1}{2}\left[\cos\left(A+B\right)+\cos\left(A-B\right)\right]$
$\sin A\sin B=-\frac{1}{2}\left[\cos\left(A+B\right)-\cos\left(A-B\right)\right]$
$\cos A\sin B=\frac{1}{2}\left[\sin\left(A+B\right)-\sin\left(A-B\right)\right]$
$\sin A\cos B=\frac{1}{2}\left[\sin\left(A+B\right)+\sin\left(A-B\right)\right]$