Re: Trigonometric Identities

Well if the equation doesn't hold true for all values of $\displaystyle \displaystyle x$, then it's not an identity.

$\displaystyle \displaystyle \begin{align*} \frac{1 + \cos{x}}{\sin{x}} &= \frac{1 + \cos{\left(2\cdot \frac{x}{2}\right)}}{\sin{\left(2\cdot \frac{x}{2}\right)}} \\ &= \frac{1 + \cos^2{\frac{x}{2}} - \sin^2{\frac{x}{2}}}{2\sin{\frac{x}{2}}\cos{\frac{ x}{2}}} \\ &= \frac{2\cos^2{\frac{x}{2}}}{2\sin{\frac{x}{2}}\cos {\frac{x}{2}}} \\ &= \frac{\cos{\frac{x}{2}}}{\sin{\frac{x}{2}}} \\ &= \cot{\frac{x}{2}}\end{align*}$

It's true for all $\displaystyle \displaystyle x$.

Re: Trigonometric Identities

gah, I suck :(

Just a quick question though, since cot (x/2) = cos (x/2) / sin (x/2) how come you didnt use sin (x/2) as a restriction since it cannot ever be equal to zero?\

thanks in advance

Re: Trigonometric Identities

Quote:

Originally Posted by

**andrew2322** gah, I suck :(

Just a quick question though, since cot (x/2) = cos (x/2) / sin (x/2) how come you didnt use sin (x/2) as a restriction since it cannot ever be equal to zero?\

thanks in advance

This explains:

Quote:

Originally Posted by

**Prove It** Well if the equation doesn't hold true for all values of $\displaystyle \displaystyle x$, then it's not an identity.

Re: Trigonometric Identities

@mathgirl8:

You have to use [ tex ] ... [ /tex ], not [ tex ] ... [ \tex ]