Question

(cos^4x−sin^4x)/(1−tan^4x)=cos^4x
trig identity?

Answer 1

\[(\cos^4x−\sin^4x)/(1−\tan^4x)=\cos^4x\]

Answer 2

how do you prove this? :/

Answer 3

i would start by factoring the numerator, and then see that one of the factors is 1.
that should make the rest easier

Answer 4

you can factor Mr. Denominator as well

Answer 5

i've tried that. i've went
\[(\cos^2x+\sin^2x)(\cos^2x-\sin^2x)/(1-(\sin^4x/\cos^4x))\]

Answer 6

cos^2(x)+sin^2(x)=1

Answer 7

you don't need to get it all into sin and cos

Answer 8

cos^2(x)-sin^2(x)=cos(2x)

Answer 9

hmm.. when i try, numerator part and denominator part keep gets messed up :/

Answer 10

\[\frac{\cos(2x)}{1-\frac{\sin^4(x)}{\cos^4(x)}} \cdot \frac{\cos^4(x)}{\cos^4(x)}\]
\[\frac{\cos^4(x) \cos(2x)}{\cos^4(x)-\sin^4(x)}=\frac{\cos^4(x)\cos(2x)}{\cos(2x)}=\cos^4(x)\]

Answer 11

since we already showed \[\cos^4(x)-\sin^4(x)=\cos(2x) \]
earlier

Answer 12

great! thanks a lot.

Answer 13

you got it? :)

Answer 14

yup i've tried it out myself! thanks for the help.