Question

trig identity: cos^2x-sin^2x/1-tan^2x=cosx

Answer 1

Whats your question

Answer 2

For a proof?

Answer 3

just to prove it

Answer 4

I literally ddrw the whole thing out

Answer 5

and this dang site wont let me show it

Answer 6

Put sin 2(x + cos 2(x in place of 1

Answer 7

and cancel out stuff

Answer 8

so when i cancel everything ill be left with 1/ -tan^2 ?

Answer 9

I had -1 / -tan^2 and that came out to 1/tan^2

Answer 10

Give me a second to check if im right

Answer 11

k

Answer 12

Nope im wrong! sorry

Answer 13

There's a problem with the numerator. You'll never get sin^2 and cos^2 to have different signs in a Pythagorean identity.

Answer 14

its okay at least u tried

Answer 15

The numerator is = cos(2x), but that's not going to help.

Answer 16

?? it says in the paper cos^2x-sin^2x

Answer 17

is the denominator
and \(\huge \rm 1-tan^2x\)

Answer 18

yeah that is the denominator

Answer 19

we have \(\huge 1-tan^2x=\frac{cos^2x-sin^2x}{cos^2x}\)
you can see that \(\huge \rm cos^2x-sin^2x \) will cancel and you are left with
cos^2x

Answer 20

so u switched the denominator?

Answer 21

Nice! I was trying difference of squares and Pythagorean identities.

Answer 22

\(\huge \frac{\cancel{cos^2x-sin^2x}}{\frac{\cancel {cos^2x-sin^2x}}{cos^2x}}=\frac{1}{\frac{1}{cos^2x}}\)

Answer 23

flip it over and you get cos^2x

Answer 24

THANKS got it

Answer 25

oh Paythagoras will work but will take to a longer road

Answer 26

np!