# Basic Trig Identity

• Nov 8th 2009, 10:34 AM
saintv
Basic Trig Identity
I came across another in worksheet that I can't seem to figure out. I've been using cos^2x + sin^2x = 1 to help me out, except its only succeeded in confusing me more. I'd like to chalk it up to this not being equal, but really, its that fact I can't simplify anything for the life of me!

Here goes:

(1-cosx)/sinx = sinx/(1 + cosx)

Thank you!
• Nov 8th 2009, 10:42 AM
e^(i*pi)
Quote:

Originally Posted by saintv
I came across another in worksheet that I can't seem to figure out. I've been using cos^2x + sin^2x = 1 to help me out, except its only succeeded in confusing me more. I'd like to chalk it up to this not being equal, but really, its that fact I can't simplify anything for the life of me!

Here goes:

(1-cosx)/sinx = sinx/(1 + cosx)

Thank you!

The key to this question is seeing that you can multiply by $\displaystyle \frac{1+cos(x)}{1+cos(x)}$ and thus get the difference of two squares.

LHS:

Multiply by $\displaystyle \frac{1+cos(x)}{1+cos(x)}$

$\displaystyle \frac{1+cos(x)}{1+cos(x)} \times \frac{1-cos(x)}{sin(x)}$

$\displaystyle = \frac{(1-cos(x))(1+cos(x))}{sin(x)(1+cos(x))}$

Note the numerator is the difference of two squares

$\displaystyle \frac{1-cos^2(x)}{sin(x)(1+cos(x))}$

$\displaystyle 1-cos^2(x) = sin^2(x)$ and so a $\displaystyle sin(x)$ will cancel

$\displaystyle \frac{sin^2(x)}{sin(x)(1+cos(x))} = \frac{sin(x)}{1+cos(x)} = RHS$
• Nov 8th 2009, 10:52 AM
saintv
Thank you so much! I do have some quick questions for you:

What made you look at it and not choose to take the route of 1=cos^2x+sin^2x ? Was it just that you recognized the possible diffference of squares like that? Is it possible to solve using the other identity? And lastly, recognizing this just all comes with practice, right?
• Nov 8th 2009, 10:57 AM
e^(i*pi)
Quote:

Originally Posted by saintv
Thank you so much! I do have some quick questions for you:

What made you look at it and not choose to take the route of 1=cos^2x+sin^2x ? Was it just that you recognized the possible diffference of squares like that? Is it possible to solve using the other identity? And lastly, recognizing this just all comes with practice, right?

I did use that identity in the last step albeit in a different form.

Why I didn't use it

• There is no squared term in the initial equation
• There is a $\displaystyle 1+cos(x)$ term on the RHS
• The sin(x) on the RHS meant I'd have to use $\displaystyle sin^2(x)+cos^2(x)=1$ to get the numerator
• The best way to get a squared term from a binomial is the difference of two squares

Yeah, it does come with practice