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1 - cos2x

is

1 - cos2x equal to (sinx)^2

??????thanks:confused:
Reply 1
Avatar for qgujxj39
qgujxj39
17
Nearly. It's equal to 2(sinx)^2. Proof is from the double angle formula.
Reply 2
Avatar for rye225
rye225
13
No

Rule is

cos^2 x + sin^2 x =1
Reply 3
Avatar for sonic23
sonic23
OP
11
tommm
Nearly. It's equal to 2(sinx)^2. Proof is from the double angle formula.


thanks.

is that just one of the ones we have to remember?
Reply 4
Avatar for nuodai
nuodai
17
rye225
No

Rule is

cos^2 x + sin^2 x =1

You're thinking of the wrong rule; the rule you want is the double angle formula for cosine.

sonic23
thanks.

is that just one of the ones we have to remember?

You're given cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos (A+B) = \cos A \cos B - \sin A \sin B, so you can derive cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = \cos^2 x - \sin^2 x and play with it until you get it in the right form... but it's worth remembering these three, if nothing else:
cos⁡2x=cos⁡2x−sin⁡2x[br]=2cos⁡2x−1[br]=1−2sin⁡2x\newline \cos 2x = \cos^2 x - \sin^2 x\newline[br]= 2\cos^2 x - 1\newline[br]= 1 - 2\sin^2 x

Then you can rearrange it.
Reply 5
Avatar for qgujxj39
qgujxj39
17
sonic23
thanks.

is that just one of the ones we have to remember?


Pretty much, yes. As an experienced A-leveller, I can immediately recall cos⁡2x=2cos⁡2x−1=cos⁡2x−sin⁡2x=1−2sin⁡2x\cos 2x = 2 \cos^2 x - 1 = \cos^2 x - \sin^2 x = 1 - 2 \sin^2 x, but that's because I've used it so many times. Practice makes perfect.
Reply 6
Avatar for sonic23
sonic23
OP
11
nuodai
You're thinking of the wrong rule; the rule you want is the double angle formula for cosine.


You're given cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos (A+B) = \cos A \cos B - \sin A \sin B, so you can derive cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = \cos^2 x - \sin^2 x and play with it until you get it in the right form... but it's worth remembering these three, if nothing else:
cos⁡2x=cos⁡2x−sin⁡2x[br]=2cos⁡2x−1[br]=1−2sin⁡2x\newline \cos 2x = \cos^2 x - \sin^2 x\newline[br]= 2\cos^2 x - 1\newline[br]= 1 - 2\sin^2 x

Then you can rearrange it.



thanks
Reply 7
Avatar for qgujxj39
qgujxj39
17
Not a problem.

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