Question

if tan^2(x) = sin^2(x)/cos^2(x) which by the half angle formulas = 1-cos(2x)/1+cos(2x)...the 1/2's cancel... does that meant that tan^2(x)=-1 or 0 or neither?

Answer 1

neither, how do you figure it should be -1 or zero?

Answer 2

\[1-\cos(2x)\div1+\cos(2x)\] take a -1 out of the top and you end up with \[-(-1+\cos(2x)\div1+\cos(2x)\]= -1? obviously this would be an identity if it were correct so i guess I'm just asking how it is wrong?

Answer 3

\[-\frac{-1+\cos(2x)}{1+\cos(2x)}\neq-1\]check the algebra...

Answer 4

\[\frac{1-\cos(2x)}{1+\cos(2x)}=-\frac{\cos(2x)-1}{1+\cos(2x)}\]but nothing is going to cancel here, so we just go round and round...

Answer 5

ah I see. I think my lack of parenthesis made me see something that was not truly there. thank you.

Answer 6

welcome :D