Question

trig identity need help

Answer 1

need help with circled one

Answer 2

\[\cot( \alpha) -\tan(\alpha)=\frac{2 \cos(2 \alpha)}{\sin(2 \alpha)}\]

Do you remember the following identities:
\[\cos(2 \alpha)=\cos^2(\alpha)-\sin^2(\alpha)\]
\[\sin(2 \alpha)=2 \sin(\alpha)\cos(\alpha)\]

Answer 3

Try using these

Answer 4

\[\frac{2 (\cos^2(\alpha)-\sin^2(\alpha))}{2 \sin(\alpha) \cos(\alpha)}\]
We see that 2/2=1
So we have
\[\frac{ \cos^2(\alpha)-\sin^2(\alpha)}{ \sin(\alpha) \cos(\alpha)}\]

Answer 5

Now write as two separate fractions

Answer 6

And then you will be home free very soon

Answer 7

so after i separate then do i use the conjugate to get the same denominator?

Answer 8

No you don't want to combine the fractions
You want to separate them because you have two terms on the other side

Answer 9

oh! i see ^_^ so i just make the cot and tan look like the other side! ty so much

Answer 10

Well I was working with side that had double angles
And I was trying to show the side with double angles is cot(blah)-tan(blah)

Answer 11

Just so we are clear
This is what I'm saying to do
\[\frac{\cos^2(\alpha)-\sin^2(\alpha)}{\sin(\alpha) \cos(\alpha)}=\frac{\cos^2(\alpha)}{\sin(\alpha) \cos(\alpha)} -\frac{\sin^2(\alpha)}{\sin(\alpha)\cos(\alpha)}\]

Answer 12

Now do some canceling and some rewriting and then boom! :) You got it!

Answer 13

i'm confused >.<' \[\cos ^{2}\alpha \div \sin- \sin ^{2}\alpha \div \cos \alpha = \cos \alpha \div \sin \alpha -\sin \alpha \div \cos \alpha\]
this is where i get confused when we make the cot and tan look like sin amd cos i get confused on what to do after that

Answer 14

\[\frac{\cos(\alpha)}{\sin(\alpha)}=\cot(\alpha) \text{ and } \frac{\sin(\alpha)}{\cos(\alpha)}=\tan(\alpha)\]

Answer 15

I've tried to start from both sides. Here we go~

Answer 16

@Callisto what did you do on the RHS to make the numerator cos^2x-sin^2x

Answer 17

He found a common denominator so he could write as one fraction

Answer 18

She!!!!!!

Answer 19

I'm sorry callisto. I should had let you said that. :)

Answer 20

Nope, I'm glad that you help!! Your answer is the best!!! But I'm a she not a he...

Answer 21

@Callisto did u multiply by the reciprocal?