Prove trigonometric equality: 1 - cosx = 2(sin^2)*(x/2)

[needingtoknow]

Homework Statement

It seems like a pretty straightforward equality but I when I tried to google it doesn't seem like it is known at all. All the paths I have tried have been dead ends. The question was initially:

Find the limit as x approaches 0 for the expression (1-cosx)/x^2

In the second step of the solution, the expression became (2(sin^2)*(x/2)) / x^2 and I didn't know how the numerator changed to that new expression.

Thank you for your help!

 

[LCKurtz]

Homework Statement

It seems like a pretty straightforward equality but I when I tried to google it doesn't seem like it is known at all. All the paths I have tried have been dead ends.


The question was initially:

Find the limit as x approaches 0 for the expression (1-cosx)/x^2

In the second step of the solution, the expression became (2(sin^2)*(x/2)) / x^2 and I didn't know how the numerator changed to that new expression.

Thank you for your help!

Do you know the identity $\cos(2x)= \cos^2x - \sin^2x = 1-2\sin^2x$? Solve for $1-\cos(2x)$ in terms of $\sin^2x$ and replace $x$ by $\frac\theta 2$ and you will have the identity you are looking for.

 

[needingtoknow]

Thank you that was exactly what I needed!

 

[chwala]

The OP did not show the end result, i think its necessary...
$\lim_{x→0}\left[ \dfrac{1-cos (x)}{x^2}\right]$=$\lim_{x→0}\left[\dfrac{2}{x^2}× \dfrac{sin^2(0.5x)}{0.5x^2}×\dfrac{x^2}{4}\right]$=$\dfrac{1}{2}$
since we know that, $\lim_{x→0}\left[\dfrac{sin^2(0.5x)}{0.5x^2}=1\right]$...

We could also use L'Hopital's rule here (or is it only used for indeterminate form $\frac {∞}{∞}?$) ...to end up with,
$\lim_{x→0}\left[ \dfrac{cos x}{2}=\dfrac{1}{2}\right]$

 

[Mark44]

The OP did not show the end result, i think its necessary...
$\lim_{x→0}\left[ \dfrac{1-cos (x)}{x^2}\right]$=$\lim_{x→0}\left[\dfrac{2}{x^2}× \dfrac{sin^2(0.5x)}{0.5x^2}×\dfrac{x^2}{4}\right]$=$\dfrac{1}{2}$
since we know that, $\lim_{x→0}\left[\dfrac{sin^2(0.5x)}{0.5x^2}=1\right]$

It's probably moot, since the OP hasn't been back for about 7 1/2 years.

But anyway, here is what I think is a simpler approach:
$\lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = \lim_{x \to 0} \frac{1 - \cos(x)}{x^2} \frac{1 + \cos(x)}{1 + \cos(x)}$
$= \lim_{x \to 0} \frac{1 - \cos^2(x)}{x^2(1 + \cos(x)} = \lim_{x \to 0} \frac{\sin^2(x)}{x^2(1 + \cos(x)}$
$= \lim_{x \to 0} \frac{\sin^2(x)}{x^2} \lim_{x \to 0}\frac 1 {1 + \cos(x)} = 1 \cdot \frac 1 2 = \frac 1 2$

We could also use L'Hopital's rule here (or is it only used for indeterminate form $\frac {∞}{∞}?$) ...to end up with,
$\lim_{x→0}\left[ \dfrac{cos x}{2}=\dfrac{1}{2}\right]$

L'Hopital's Rule can be used also for the indeterminate for $[\frac 0 0]$. However, as this was posted in the Precalc section, calculus techniquest wouldn't be appropriate.