View Full Version : Free trigonomic identities solver
Jeff83
Apr 16, 2013, 10:28 PM
sin^2xcot^2x-1/sinxcos^2xcscx
ebaines
Apr 17, 2013, 05:51 AM
First, you ned to apply parentheses because it's not at all clear what the equation is. What you wrote is this:
\sin^2x\ cot^2x - \frac 1 {\sin x} \cos ^2x \ csc x
but I suspect that what you probably meant is this:
\frac {\sin^2x \ cot^2x - 1}{\sin x \ \cos ^2x \ csc x}
Second - what's your question?
Jeff83
Apr 17, 2013, 09:07 AM
Yes thank you that is what I meant , I'm trying to simplify . I'm just starting to learn .
ebaines
Apr 17, 2013, 09:19 AM
If you convert the cot(x) and csc(x) functions to their sine and cosine equivalents many terms cancel out and you can simplify this quite a bit. Also look for opportunitoes to use fundamental identities such as sin^2x + cos^2x = 1 (and its alternative arrangements such as 1- cos^2x = sin^2x). Post back with what you get for a simplified expression.