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Prove that cos2 2x – cos2 6x = sin 4x sin 8x
Let us start from the start..
\[cos^{2}2x -cos^{2}6x = sin4xsin8x\]?
have you tried cos^{2}2x = (1+cos4x)/2?
\[\cos^2 A - \cos^2 B = \sin(A+B)\sin(B-A)\] This is a standard formula. So, we need a proof right?
(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
@apoorvk yup proof needed..
@apoorvk it is cosA-cosBand not their squares for standard formula
\[\cos^2(2x) - \cos^2(6x) \implies (\cos(2x) - \cos(6x))(\cos(2x) + \cos(6x))\] So: Using: \[\cos(C) + \cos(D) = 2\cos(\frac{C+ D}{2})\cos(\frac{C-D)}{2})\] \[\cos(C) - \cos(D) = 2\sin(\frac{C+ D}{2})\sin(\frac{D-C}{2})\] \[(2\sin(4x) \sin(2x))(2\cos(4x)\cos(2x)) \implies (2\sin(2x)\cos(2x)) \cdot (2\sin(4x) \cos(4x))\] \[\implies \sin(4x) \cdot \sin(8x)\]
Using: \[2\sin(x)\cos(x) \implies \sin(2x)\]
I think above you by mistake written (B-A) @apoorvk
You are doing wrong I guess..
Rewrite for cos(A) - cos(B) = ??
@apoorvk u told me it is sin(a+b).sin(b-a) but u have written differ
yeah I have memory problems -.- Amnesia *sigh* *sniff*
\[\cos(A) - \cos(B) = 2\sin(\frac{A+B}{2}) \cdot \sin(\frac{B-A}{2})\]
And not sin(a-B) in the last..
@apoorvk if i use ur method we wont get that lol
NOOOOOOOO!!!!!!!!!!!!!!!!!! What have i been doing? /_\
Okay please tweak it and use it as needed, we know the procedure atleast. Or should I repost?
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
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!!!!!!!!!
Sir what you think is correct, if you want a confirmation. -.-
There must be condition for doing this formula : I am discussing with you @apoorvk
The above said formula will work if B > A..
\[\cos^2A−\cos^2B\] \[=(\cos A+\cos B)(\cos A−\cos B)\] \[=[2\cos(\frac{A+B}2)\cos(\frac{A−B}2)][2\sin(\frac{A+B}2)\sin(\frac{B-A}2)]\] \[=[2\sin(\frac{A+B}2)\cos(\frac{A+B}2)][2\sin(\frac{B-A}2\cos(\frac{B-A}2)]\] \[=\Large\boxed{\sin(A+B)\sin(B-A)}\] Happy nitpickers? --____--
Do you think It will work ??
Water in eyes, Fish in Water \(\large \rightarrow\) so Fish in eyes. -_-
I am asking you @apoorvk
If fish is eyes then why are looking upwards?? Is fish in your eyes??
I paarsonaaally bilib it's Phaathar bill wa-ark! -,- (dish phormula)
What is this?? @apoorvk
U know aandarishtand? oowhaat u know no inglis?? BAH HUMBUG!
Inglish pipul go, liebe children behind -_-
'Addition', man, addition ----> " sin (A '+' B) - Y U LOVE SUBTRACTION??
U inglish phail, I inglish oowhale. -____-
You failed in english? ok I got this.. Better luck next time.. @apoorvk
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?