Question

Verify: 2cosx2x/sin2x=cotx-tanx (more trig identities)

Answer 1

Oh! These are fun.
Quick question before I look, what do you mean by "2cosx2x"

Answer 2

@kaisan ?

Answer 3

oops sorry i typed it wrong its supposed to be: 2cos2x/sin2x=cotx-tanx

Answer 4

Okay. Thanks :D
\[2\cos2\theta/\sin2\theta=\cot \theta-\tan \theta\]
Let's first simplify the cotangent and tangent to their constituents.
\[2\cos2\theta/\sin2\theta=1/\tan \theta - \sin \theta/ \cos \theta\]
Then we can simplify 1/tangent to get

Answer 5

\[2\cos2\theta/\sin2\theta=\cos \theta/\sin \theta - \sin \theta/\cos \theta\]
Then we can further simplify the right hand side by subtracting and making common denominators. This gives:
\[2\cos2\theta/\sin2\theta=\cos^2 \theta-\sin^2\theta/\sin \theta \cos \theta\]

Answer 6

@kaisan , is it
\[2\cos2\theta\], or just \[\cos2\theta\]?

Answer 7

i can give u a pic of my homework

Answer 8

If the question is supposed to be \[\cos2\theta\]
then we apply or trig identities, which are:
\[\cos2\theta=\cos^2\theta-\sin^2\theta\] and \[\sin2\theta=2\sin \theta \cos \theta\]

Answer 9

So applying those to our problem, we get.
\[\cos^2\theta-\sin^2\theta/\sin \theta \cos \theta=\cos^2\theta-\sin^2\theta/\sin \theta \cos \theta\]

So this is true, so long as the question is "cos2x" not "2cos2x"

Answer 10

ok i'll post it the one i circled is the one i'm working on

Answer 11

@Argonx16 here it is

Answer 12

I feel like it's a typo (if we assume all are true), but I'm going to take a look at it again.

Answer 13

ok

Answer 14

So I checked on Wolfram Alpha (awesome resource fyi), and it appears that the give expression is true, so I think I did something like fraction subtraction wrong, so the right hand side should work out to be 2cos^2-2sin^2/sincos. Something like that. Definitely ask your teacher though. Sorry if this may have mislead you in any way!

Answer 15

its ok ^^ thx for trying