Question

which if the following are identities
sin3x/sinxcosx=4cosx-secx
(sinx+cosx)^2=1+sin2x
sin3x-sinx/cos3x+cosx=tanx
sin6x=2sin3xcos3x

Answer 1

The first is ambiguous and should be written more carefully.

Use your best known identities and prove them!

Answer 2

Well just a wild guess but the second one looks like one

Answer 3

Wild guesses are not acceptable. Prove it. Expand the left side and see.

Answer 4

I am not sure how that is done

Answer 5

The last is an eyeball question. Just look at it, recall the double angle formula for sine, and call it good.

Answer 6

Not sure how what is done? It's a binomial. Square it.

Answer 7

I see how the last one works though
since the double identity is
sin2x=2sinxcosx
and it says sin6x=2sin3xcos3x

Answer 8

I am still not seeing any other ones

Answer 9

Done.

Now, #2

(a+b)^2 = a^2 + 2ab + b^2

Apply this idea to the left side.

Answer 10

(sinx+cosx)^2=1+sin2x
sinx+cosx)^2=sin^2x+2sinxcosx+cos^2x

Answer 11

now what though?

Answer 12

You're done. Just look at it.

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

That should look familiar;

Just play with them. They won't break.