(1+cosA)^2-(1-cosA)^2-sin^2=ctgA*sinA*cosA


by makarov1901
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makarov1901
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#1
Feb27-11, 01:31 AM
P: 3
1. The problem statement, all variables and given/known data
(1+cosA)^2-(1-cosA)^2-sin^2A=ctgA*sinA*cosA


2. Relevant equations



3. The attempt at a solution

I moved sin^2 to the right side, then expanded the left side and got to:
1+2cosA+cos^2A-1+2cosA-cos^2A=1
When I cancel the left side I get:
4cosA=1

I'd be very grateful if anyone help me.
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vela
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#2
Feb27-11, 01:49 AM
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Looks good so far. What's the problem?

(I assume when you wrote sin^2, you meant sin2 a.)
makarov1901
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#3
Feb27-11, 01:55 AM
P: 3
You assumed right.
The problem is that both parts of the identity should be equal.
As far as I know, 4cosA is not equal to 1.

vela
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#4
Feb27-11, 02:04 AM
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(1+cosA)^2-(1-cosA)^2-sin^2=ctgA*sinA*cosA


It's not an identity. For example, when a=pi/2, the lefthand side is -1 while the righthand side equals 0.

Perhaps there's a typo in the problem.
makarov1901
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#5
Feb27-11, 02:13 AM
P: 3
My task is to transform both parts so that they're equal. I don't know why you say it's not an identity.
vela
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#6
Feb27-11, 02:19 AM
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If it were an identity, it would hold for all values of a. It clearly doesn't; therefore, it's not an identity.


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