Question

Prove that cos2 2x – cos2 6x = sin 4x sin 8x

Answer 1

Let us start from the start..

Answer 2

it is cos^2...ok

Answer 3

\[cos^{2}2x -cos^{2}6x = sin4xsin8x\]?

Answer 4

yes

Answer 5

have you tried cos^{2}2x = (1+cos4x)/2?

Answer 6

no

Answer 7

\[\cos^2 A - \cos^2 B = \sin(A+B)\sin(B-A)\]

This is a standard formula.
So, we need a proof right?

Answer 8

(1+cos4x)/2-(1+cos12x)/2 = (1+cos 4x+cos12x+cos4xcos12x)/2
and then defactorize the last term...and factorize the second and third term

Answer 9

@apoorvk yup proof needed..

Answer 10

@apoorvk it is cosA-cosBand not their squares for standard formula

Answer 11

\[\cos^2(2x) - \cos^2(6x) \implies (\cos(2x) - \cos(6x))(\cos(2x) + \cos(6x))\]

So:

Using:

\[\cos(C) + \cos(D) = 2\cos(\frac{C+ D}{2})\cos(\frac{C-D)}{2})\]

\[\cos(C) - \cos(D) = 2\sin(\frac{C+ D}{2})\sin(\frac{D-C}{2})\]

\[(2\sin(4x) \sin(2x))(2\cos(4x)\cos(2x)) \implies (2\sin(2x)\cos(2x)) \cdot (2\sin(4x) \cos(4x))\]

\[\implies \sin(4x) \cdot \sin(8x)\]

Answer 12

Using:

\[2\sin(x)\cos(x) \implies \sin(2x)\]

Answer 13

I think above you by mistake written (B-A) @apoorvk

Answer 14

You are doing wrong I guess..

Answer 15

Rewrite for cos(A) - cos(B) = ??

Answer 16

@apoorvk u told me it is sin(a+b).sin(b-a) but u have written differ

Answer 17

yeah I have memory problems -.- Amnesia *sigh* *sniff*

Answer 18

\[\cos(A) - \cos(B) = 2\sin(\frac{A+B}{2}) \cdot \sin(\frac{B-A}{2})\]

Answer 19

And not sin(a-B) in the last..

Answer 20

@apoorvk if i use ur method we wont get that lol

Answer 21

NOOOOOOOO!!!!!!!!!!!!!!!!!! What have i been doing? /_\

Answer 22

Okay please tweak it and use it as needed, we know the procedure atleast.
Or should I repost?

Answer 23

See:

\[\cos^2(2x) - \cos^2(6x) = \sin(A+B)\sin(A-B) \implies \sin(2x - 6x)\sin(2x + 6x)\]

Don't you think we get negative ??? @apoorvk

Answer 24

yeah i got it ate already - unfortunately i can't view my raw LateX codes (some stupid bug :\), so I'll have to edit it manually. NOOOOOO!!!!!!!!!

Answer 25

Sir what you think is correct, if you want a confirmation. -.-

Answer 26

There must be condition for doing this formula :

I am discussing with you @apoorvk

Answer 27

The above said formula will work if B > A..

Answer 28

\[\cos^2A−\cos^2B\]
\[=(\cos A+\cos B)(\cos A−\cos B)\]
\[=[2\cos(\frac{A+B}2)\cos(\frac{A−B}2)][2\sin(\frac{A+B}2)\sin(\frac{B-A}2)]\]
\[=[2\sin(\frac{A+B}2)\cos(\frac{A+B}2)][2\sin(\frac{B-A}2\cos(\frac{B-A}2)]\]
\[=\Large\boxed{\sin(A+B)\sin(B-A)}\]




Happy nitpickers? --____--

Answer 29

Do you think It will work ??

Answer 30

Water in eyes, Fish in Water \(\large \rightarrow\) so Fish in eyes. -_-

Answer 31

I am asking you @apoorvk

Answer 32

If fish is eyes then why are looking upwards??

Is fish in your eyes??

Answer 33

I paarsonaaally bilib it's Phaathar bill wa-ark! -,- (dish phormula)

Answer 34

What is this?? @apoorvk

Answer 35

U know aandarishtand? oowhaat u know no inglis?? BAH HUMBUG!

Answer 36

Inglish pipul go, liebe children behind -_-

Answer 37

'Addition', man, addition ----> " sin (A '+' B) - Y U LOVE SUBTRACTION??

Answer 38

Okay my mistake..

Answer 39

U inglish phail, I inglish oowhale. -____-

Answer 40

You failed in english?

ok I got this..

Better luck next time.. @apoorvk

Answer 41

Hu shaid I phail in inglish?? U no si mai bandarfool inglish haan??? o.O
I gets the 105 out of 100 -__- u get how many?