Basic Trig Identity

saintv · #1 $Old$ November 8th, 2009, 11:34 AM

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!

e^(i*pi) · #2 $Old$ November 8th, 2009, 11:42 AM

Quote:

Originally Posted by saintv $View Post$

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 $\frac{1+cos(x)}{1+cos(x)}$ and thus get the difference of two squares.

LHS:

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

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

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

Note the numerator is the difference of two squares

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

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

$\frac{sin^2(x)}{sin(x)(1+cos(x))} = \frac{sin(x)}{1+cos(x)} = RHS$

saintv · #3 $Old$ November 8th, 2009, 11:52 AM

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?

e^(i*pi) · #4 $Old$ November 8th, 2009, 11:57 AM

Quote:

Originally Posted by saintv $View Post$

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 $1+cos(x)$ term on the RHS
The sin(x) on the RHS meant I'd have to use $sin^2(x)+cos^2(x)=1$ 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