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New Member
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May 3, 2007, 04:31 PM
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identities for sin2x and cos2x
How do I do a problem dealing with identities for sin2x and cos2x?
Must be proven algebraically. Sin^2x=1-cos2x/2
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Expert
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May 4, 2007, 11:08 AM
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Originally Posted by fountainviewkid
How do I do a problem dealing with identities for sin2x and cos2x?
Must be proven algebraically. sin^2x=1-cos2x/2
I think you meant sin^2x = (1/ - cos2x)/2, right?
Without doing the proof for you, here is a of hint: start with the identity for cos(2x) and combine it with sin^2x + cos^2x = 1
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Uber Member
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May 6, 2007, 03:43 PM
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Is your 2x meant to be 2 times x or x squared? Your original question sends mixed messages so it's impossible to give you an answer.
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Expert
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May 7, 2007, 12:55 PM
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I believe the convention is that sin^2x =
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Junior Member
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May 7, 2007, 02:32 PM
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*sigh* so hard to do it with words...
2 2
SIN X = COS X - 1.
There is no division. That's it.
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Junior Member
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May 8, 2007, 03:07 PM
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*sigh* I failed above post...
SIN^2 X = COS^2 X - 1
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Expert
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May 9, 2007, 12:10 PM
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Mr_Guy - True statement, but the original question was to show that:
sin^2x = (1-cos 2x)/2
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Junior Member
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May 9, 2007, 05:34 PM
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huh... well then, my bad.
AND this question stumped me for some time... ANDD it's going to be really hard without some mathematical things... *sigh*
--> SIN^2 X = (1 - COS 2X) / 2
[Given the identity (COS^2 X - SIN^2 X = COS 2X)... ]
--> SIN^2 X = (1 - (COS^2 X - SIN^2 X)) / 2
[The negative outside and inside the brackets turn positive... ]
--> SIN^2 X = (1 + (COS^2 X + SIN^2 X)) / 2
[Given the identity (COS^2 X + SIN^2 X = 1)... ]
--> SIN^2 X = (1+1) / 2
[Simple division]
--> SIN^2 X = 1
I don't know. We just finished this unit in math class, and we didn't really come across this question. If it's wrong, sorry.
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Expert
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May 10, 2007, 06:01 AM
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You made an error in signs in the 3rd equation. You should have:
sin^2x = (1 - cos^2x + sin^2x)/2 [note the negative sign in front of the cosine term]
Now add and subtract cos^2x in the parenthetical on the right, and you're back on track. You'll end up with sin^2x = 1 - cos^2x.
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Junior Member
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May 10, 2007, 02:49 PM
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... I get credit for being close, right? :P
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