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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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

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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

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.

I believe the convention is that sin^2x =

*sigh* so hard to do it with words...

2 2
SIN X = COS X - 1.

There is no division. That's it.

*sigh* I failed above post...

SIN^2 X = COS^2 X - 1

Mr_Guy - True statement, but the original question was to show that:

sin^2x = (1-cos 2x)/2

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.

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.

... I get credit for being close, right? :P