Thread: trig identity

Hi smmmc

Originally Posted by smmmc

i subbed 1-tan(x)= sin(x)/cos(x)

You forgot the term "1"

yerh i did that.
Let me just show you my whole working out. sorry its not in latex, dont know how to use it

So,

cos(x)/(1- sin(x)/cos(x)) + sin(x)/(1-cos(x)/sin(x))

= cos(x)/(cos(x)-sin(x)/cos(x)) + sin(x)/(sin(x)-cos(x)/sin(x))

= cos(x) * cos(x)/(cos(x)-sin(x)) + sin(x) * (sin(x)/(sin(x)-cos(x))

= cos^2(x) + sin^2(x)/ (cos(x)-sin(x)*(sin(x)/(sin(x)-cos(x))

which equals

1/ (2cos(x)sin(x)-cos^2(x)-sin^2(x))

help please

Hi smmmc

Originally Posted by smmmc

cos^2(x) + sin^2(x)/ (cos(x)-sin(x)*(sin(x)/(sin(x)-cos(x))

What's this ??

cos(x) * cos(x)/(cos(x)-sin(x)) + sin(x) * (sin(x)/(sin(x)-cos(x))

It's good till this part. Then, you can manipulate it like this :





And a little suggestion from me : How about learning latex ? http://www.mathhelpforum.com/math-he...-tutorial.html

so it then becomes

cos^2(x)-sin^2(x)/ cox(x)-sin(x)

can you cross it out? simplify?

but that leads to cos(x)-sin(x).. and i dint know you could simplify like that?

help

Hi smmmc

doesn't simplify to