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How to solve (1+cosA)^2  (1cosA)^2  sin^2A = ctgA*sinA*cosAby makarov1901
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#1
Feb2711, 01:31 AM

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1. The problem statement, all variables and given/known data
(1+cosA)^2(1cosA)^2sin^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^2A1+2cosAcos^2A=1 When I cancel the left side I get: 4cosA=1 I'd be very grateful if anyone help me. 


#2
Feb2711, 01:49 AM

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Looks good so far. What's the problem?
(I assume when you wrote sin^2, you meant sin^{2} a.) 


#3
Feb2711, 01:55 AM

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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. 


#4
Feb2711, 02:04 AM

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How to solve (1+cosA)^2  (1cosA)^2  sin^2A = 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. 


#5
Feb2711, 02:13 AM

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My task is to transform both parts so that they're equal. I don't know why you say it's not an identity.



#6
Feb2711, 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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