Question

cos^2 2x−cos^2 6x=sin4x sin8x
please help me prove it

Answer 1

you can manipulate both sides, right ?

Answer 2

yes..but it would be prefered if we can begin with the left hand side

Answer 3

\(\normalsize\color{blue}{ \cos^2 2x−\cos^2 6x=\sin^4x \ sin^8x }\) like this ?

Answer 4

\[\cos ^{2} 2x - \cos ^{2} 6x = \sin4x \sin 8x \] like this..

Answer 5

Oh!! you know how to do !!! You figured out fast:)

Answer 6

\(\normalsize\color{blue}{ \cos^2 2x−\cos^2 6x=\sin(4x) \sin(8x) }\)

Answer 7

\(\normalsize\color{blue}{ \cos^2 2x−\cos^2 6x=\sin 4x \sin 8x }\)
\(\normalsize\color{blue}{ \cos^2 2x−\cos^2 (4x+2x)=\sin 4x \sin 8x }\)
\(\normalsize\color{blue}{ \cos^2 2x−(\cos^2 4x~~\sin^2 2x-\cos^2 2x~~\sin^2 4x)=\sin 4x \sin 8x }\)
\(\normalsize\color{blue}{ \cos^2 2x−\cos^2 4x~~\sin^2 2x+\cos^2 2x~~\sin^2 4x=\sin 4x \sin 8x }\)
\(\normalsize\color{blue}{ \cos^2 2x−(\cos^2 4x)~~\sin^2 2x+\cos^2 2x~~(\sin^2 4x)=\sin 4x \sin 8x }\)
\(\normalsize\color{blue}{ \cos^2 2x−(\cos^2 2x-\sin^2 2x)~~\sin^2 2x+\cos^2 2x~~(2\sin 2x)=\sin 4x \sin 8x }\)
\(\normalsize\color{blue}{ \cos^2 2x−\cos^2 2x+\sin^2 2x~~\sin^2 2x+\cos^2 2x~~(2\sin 2x)=\sin 4x \sin 8x }\) read this for now

Answer 8

ignore the last step I posted

Answer 9

Can I say something?

Answer 10

No I am lost... idk, sorry

Answer 11

OOPS go ahead

Answer 12

from the left hand side:
\[cos^2(2x) -cos^2(6x) = (cos (2x) -cos (6x))*(cos(2x)+cos(6x))\] so far you have 2 terms ,right?

Answer 13

how is \[\cos ^{2} 4x \] = \[\cos ^{2} 2x - \sin ^{2} 2x\] ??

Answer 14

I am sorry, the second term is
\[cos(2x) +cos (6x)= 2 cos(\dfrac{2x+6x}{2})cos(\dfrac{2x-6x}{2})= 2 cos (4x)cos(-2x)\]

Answer 15

ok..continue

Answer 16

the first term is
\[cos(2x) -cos (6x)= -2 sin (\dfrac{2x+6x}{2})sin(\dfrac{2x-6x}{2})= -2 sin (4x)sin(-2x)\]

Answer 17

Now, times them together, because originally they are the factor of the product above, right?
so at the end, you have the left hand side = 2cos(4x)cos(-2x)*(-2) sin(4x)sin (-2x)

Answer 18

Now I re-arrange it as
2cos (4x) sin(4x) *(-2) cos (-2x) sin(-2x) ok so far?

Answer 19

you can see \(2cos(4x) sin(4x) = sin (8x)\)

Answer 20

and \(2 cos (-2x) sin(-2x) = sin (-4x) = -sin (4x)\)
but in the front we have -2, not 2, so that it becomes sin(4x)
--> it turns to sin(8x)sin(4x) = the right hand side

Answer 21

hehehe... hopefully you can follow me. trigs is .... not hard but.... tedious!!

Answer 22

thank you so much!!