Question

cos^2 2x-cos^2 6x=sin4x sin8x prove !

Answer 1

\[\cos^2(2x) - \cos^2(6x) = \sin(4x) \sin(8x)\]

Answer 2

thought i had that one.. rewrote it.. turns out i have no idea, im thinking sum to product identity but youve probly gathered that much on your own sorry

Answer 3

@nabsz.j its ok

Answer 4

You can write this as:
(cos2x+cos6x)(cos2x-cos6x)
Now,use cosA+cosB and cosA-cosB
You can solve now?

Answer 5

i did that and i used the next formula too .. tell me what to do after that !

Answer 6

okay so now..

Answer 7

You should get this:
(2 sin 4x cos 4x) (2 sin 2x cos 2x)

Answer 8

So,finally
sin 8x sin 4x is the answer :)

Answer 9

i got (2 cos 4x cos 2x ) (-2 sin 4x sin 2x )

Answer 10

how'd yu do it @DLS ?

Answer 11

which sin formua ?

Answer 12

sin2x=2sinxcosx

Answer 13

can you do it .. please , am gettin confused !

Answer 14

okay!

Answer 15

cos2(2x)−cos2(6x)⟹(cos(2x)−cos(6x))(cos(2x)+cos(6x))
So: Using:
cos(C)+cos(D)=2cos(C+D/2)cos(C−D)/2)
cos(C)−cos(D)=2sin(C+D/2)sin(D−C/2)
(2sin(4x)sin(2x))(2cos(4x)cos(2x))⟹(2sin(2x)cos(2x))⋅(2sin(4x)cos(4x))
⟹sin(4x)⋅sin(8x)

Answer 16

okay understood thanks DLS :)

Answer 17

One step answer :P
cos^2A−cos^2B=sin(A+B)sin(B−A)

Answer 18

how come (2sin(2x)cos(2x))⋅(2sin(4x)cos(4x))=sin(4x)⋅sin(8x) @DLS