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anonymous
 3 years ago
Prove that cos2 2x – cos2 6x = sin 4x sin 8x
anonymous
 3 years ago
Prove that cos2 2x – cos2 6x = sin 4x sin 8x

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Let us start from the start..

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.1\[cos^{2}2x cos^{2}6x = sin4xsin8x\]?

Mimi_x3
 3 years ago
Best ResponseYou've already chosen the best response.1have you tried cos^{2}2x = (1+cos4x)/2?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\cos^2 A  \cos^2 B = \sin(A+B)\sin(BA)\] This is a standard formula. So, we need a proof right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0(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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@apoorvk yup proof needed..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@apoorvk it is cosAcosBand not their squares for standard formula

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\cos^2(2x)  \cos^2(6x) \implies (\cos(2x)  \cos(6x))(\cos(2x) + \cos(6x))\] So: Using: \[\cos(C) + \cos(D) = 2\cos(\frac{C+ D}{2})\cos(\frac{CD)}{2})\] \[\cos(C)  \cos(D) = 2\sin(\frac{C+ D}{2})\sin(\frac{DC}{2})\] \[(2\sin(4x) \sin(2x))(2\cos(4x)\cos(2x)) \implies (2\sin(2x)\cos(2x)) \cdot (2\sin(4x) \cos(4x))\] \[\implies \sin(4x) \cdot \sin(8x)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Using: \[2\sin(x)\cos(x) \implies \sin(2x)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think above you by mistake written (BA) @apoorvk

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You are doing wrong I guess..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Rewrite for cos(A)  cos(B) = ??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@apoorvk u told me it is sin(a+b).sin(ba) but u have written differ

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah I have memory problems . Amnesia *sigh* *sniff*

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\cos(A)  \cos(B) = 2\sin(\frac{A+B}{2}) \cdot \sin(\frac{BA}{2})\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And not sin(aB) in the last..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@apoorvk if i use ur method we wont get that lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0NOOOOOOOO!!!!!!!!!!!!!!!!!! What have i been doing? /_\

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay please tweak it and use it as needed, we know the procedure atleast. Or should I repost?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0See: \[\cos^2(2x)  \cos^2(6x) = \sin(A+B)\sin(AB) \implies \sin(2x  6x)\sin(2x + 6x)\] Don't you think we get negative ??? @apoorvk

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah 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!!!!!!!!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sir what you think is correct, if you want a confirmation. .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0There must be condition for doing this formula : I am discussing with you @apoorvk

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The above said formula will work if B > A..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\cos^2A−\cos^2B\] \[=(\cos A+\cos B)(\cos A−\cos B)\] \[=[2\cos(\frac{A+B}2)\cos(\frac{A−B}2)][2\sin(\frac{A+B}2)\sin(\frac{BA}2)]\] \[=[2\sin(\frac{A+B}2)\cos(\frac{A+B}2)][2\sin(\frac{BA}2\cos(\frac{BA}2)]\] \[=\Large\boxed{\sin(A+B)\sin(BA)}\] Happy nitpickers? ____

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Do you think It will work ??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Water in eyes, Fish in Water \(\large \rightarrow\) so Fish in eyes. _

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I am asking you @apoorvk

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If fish is eyes then why are looking upwards?? Is fish in your eyes??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I paarsonaaally bilib it's Phaathar bill waark! , (dish phormula)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What is this?? @apoorvk

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0U know aandarishtand? oowhaat u know no inglis?? BAH HUMBUG!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Inglish pipul go, liebe children behind _

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0'Addition', man, addition > " sin (A '+' B)  Y U LOVE SUBTRACTION??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0U inglish phail, I inglish oowhale. ____

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You failed in english? ok I got this.. Better luck next time.. @apoorvk

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hu 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?
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