EclipsedStar
  • EclipsedStar
Verify the given identity using double-angle formulas. Did I do this right?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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just_one_last_goodbye
  • just_one_last_goodbye
You got this one @AlexandervonHumboldt2 ? :)
AlexandervonHumboldt2
  • AlexandervonHumboldt2
you got wut to do JOLG?
just_one_last_goodbye
  • just_one_last_goodbye
Nah bruh ^_^ I'll go help someone else. ^_^ good luck

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More answers

triciaal
  • triciaal
can't see the question.
EclipsedStar
  • EclipsedStar
question: \[\frac{ cosx+\cos3x }{ 2\cos2x }=cosx\] work: \[\frac{ cosx+\cos(2x+x)}{ 2\cos2x }\] = \[\frac{ cosx+cos2xcosx-\sin^2x)-\sin^2xsinx}{ 2cos2x }\] = \[\frac{ cosx+(1-2\sin^2x)cosx-(2sinxcosx)sinx }{ 2\cos2x }\] =\[\frac{ cosx+cosx(1-2\sin) }{ 2\cos2x }\] =\[\frac{ 2cosx-4\sin^2xcosx }{ 2(1-2\sin^2x)}\] = cosx
EclipsedStar
  • EclipsedStar
The page refreshed on me ;-;
EclipsedStar
  • EclipsedStar
also I think I put in some typos...
triciaal
  • triciaal
|dw:1450032356999:dw|
triciaal
  • triciaal
|dw:1450032734832:dw|
triciaal
  • triciaal
something is off the original says cos x this gives me 1 please review carefully but the idea was to use the formula
EclipsedStar
  • EclipsedStar
I used this formula. |dw:1450033302575:dw|
triciaal
  • triciaal
|dw:1450033614662:dw|
triciaal
  • triciaal
|dw:1450033875756:dw|
EclipsedStar
  • EclipsedStar
What is the line over cosx for? XD
triciaal
  • triciaal
applying the formula n = 2 then 2-2 = 0 so cos (0x)
EclipsedStar
  • EclipsedStar
ohh
triciaal
  • triciaal
but cos of 0 = 1
EclipsedStar
  • EclipsedStar
Yeah I forgot that...haha
triciaal
  • triciaal
I have an error someplace but the principle is correct
EclipsedStar
  • EclipsedStar
I can't seem to find the error. XD I just learned this, so the concept is pretty confusing for me.
triciaal
  • triciaal
@sleepyjess can you find my error?
sleepyjess
  • sleepyjess
I haven't done this stuff in forever, and trying to review it is just confusing me :/
triciaal
  • triciaal
@sleepyjess thanks anyway especially considering you have more recent experience with this stuff
triciaal
  • triciaal
@EclipsedStar sorry if I confused you @TheSmartOne thank you for your input
EclipsedStar
  • EclipsedStar
It's fine, thanks though. :)
TheSmartOne
  • TheSmartOne
Let's start with the LHS \[\frac{ cosx+\cos3x }{ 2\cos2x }=cosx \] \[\frac{\cos(x) + 4 \cos^3 (x) - 3\cos x}{2\cos(2x)}\] \[\frac{4 \cos^3 (x) - 2\cos x}{2\cos(2x)}\] \[\frac{2( \cos^3 (x) - 1\cos x)}{2(\cos(2x))}\] \[\frac{\cos^3 (x) - 1\cos x}{\cos(2x)}\] \[\frac{\cos^3 (x) - 1\cos x}{\cos^2(x) -1}\] \[\frac{\cos(x)(\cos^2(x) - 1)}{\cos^2(x) -1}\] \[\cos(x)\] And, now, LHS = RHS
TheSmartOne
  • TheSmartOne
@triciaal Please check my work. Thanks :)
EclipsedStar
  • EclipsedStar
http://prntscr.com/9drqt4 my work...lol thanks tso
EclipsedStar
  • EclipsedStar
:)
TheSmartOne
  • TheSmartOne
|dw:1450036420041:dw|
EclipsedStar
  • EclipsedStar
lol...
TheSmartOne
  • TheSmartOne
|dw:1450036431193:dw|
TheSmartOne
  • TheSmartOne
there shouldn't be any + sign. It should be a multiplication sign :P
EclipsedStar
  • EclipsedStar
lol ok. XD
EclipsedStar
  • EclipsedStar
thanks :]
TheSmartOne
  • TheSmartOne
even though you later multiplied them xD
EclipsedStar
  • EclipsedStar
I'm just out of it today I guess :P
TheSmartOne
  • TheSmartOne
Everything else looks great. :) Great job :)
TheSmartOne
  • TheSmartOne
And I wonder why you say that you're bad at math xD Your like at my level in math. xD

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