Identity or not?

Question

a) I don't think this is an identity but what it the easiest way to prove it. cos6y= 2cosˆ2(3y)-1

and 2sinxcosx=sin(-2x) is an identity?

Answer 1

cos(6y)
= cos(3y + 3y)
= cos3ycos3y -- sin3ysin3y
= cos^2(3y) -- sin^2(3y)
= cos^2(3y) -- [1 -- cos^2(3y)]
= cos^2(3y) -- 1 + cos^2(3y)
= 2cos^2(3y) -- 1

sin(2x)
= sin(x + x)
= sinxcosx + cosxsinx
= 2sinxcosx ≠ sin(--2x)

Answer 2

2sinxcosx = sin(2x), not sin (-2x), as sin x does not = sin (-x)
-2sinxcosx = sin(-2x)

cos6y = 2cos^2(3y) - 1

Start with the right side.

Multiply everything by -1 / -1

= (-2cos^2 (3y) + 1) / -1
= (1 - 2cos^2(3y)) / -1
There is an identity for 1 - 2cos^2(x), it is cos(2x). In this case, x is 3y.
Therefore
=cos (2(3y)) / -1)
= cos (6y) / -1
= - cos (6y)

Because the negative of a cosine IS equal to the positive, this is an identity.

Answer 3

you know that cos2x = 2cos^2 x -1
so 2cos^2 3y -1 = cos2(3y) = cos6y is it's an identity

the 2nd one isn't an identity because
2sinxcosx = sin2x

Answer 4

cos 6y
= cos (3y + 3y) = cos 3y cos 3y - sin 3y sin 3y
= cos^2 3y - (1 - cos^2 3y)
= 2 cos^2 3y - 1

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

also sin -2x = -2 sin x cos x