How to show that \log(\frac{2a}{1-a^2}+\frac{2b}{1-b^2}+\frac{2c}{1-c^2})=\log\frac{2a}{1-a^2}+\log\frac{2b}{1-b^2}+\log\frac{2c}{1-c^2}



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How to show that

Answer & Explanation

Marcus Herman

Marcus Herman

Beginner2022-01-04Added 41 answers

Use loga+logb+logc=log(abc)
and then put a=tanA etc. to find
tanA=tanAA+B+C=nπ where n is any integer
and 2a1a2=2tanA1tan2A=tan2A
tan2A=tan2A  as  2A+2B+2C=2nπ
Now apply logarithm now


Beginner2022-01-05Added 36 answers

OK, heres


Expert2022-01-09Added 669 answers

The expression 2a1a2 should make you think of the tangent function, as in
Here I'm using capital A to refer to the angle and lower-case a to refer to its tangent: a=tanA.
log(a+b+c)=loga+logb+logc implies log(a+b+c)=log(abc), which then implies a+b+c=abc, and following the pattern in capital and lower-case letters above, we have tanA+tanB+tanC=tanAtanBtanC.
The identity you're trying to prove is equivalent to 2a1a2+2b1b2+2c1c2=2a1a22b1b22c1c2
(with lower-case a,b,c), so that's the same as
Now at this point I might not know how to proceed further if I hadn't seen the following at some point in the past. In the first place, the usual formula for the tangent of a sum implies with 30 seconds' more work that
and in the second place, if A+B+C=π radians or 180 or a half-circle then tan(A+B+C)=0. So the fraction is 0 and therefore the numerator is 0 and therefore tanA+tanB+tanC=tanAtanBtanC
So what you're being asked to prove is that if
tanA+tanB+tanC=tanAtanBtanC then
We've see that the fact that A+B+C=a half circle implies the first of these identities only because it implies that tan(A+B+C)=0. So you really just need to show that if tan(A+B+C)=0 then tan(2A+2B+2C)=0. It's not hard to show that that's the same as saying that if A+B+C is an integer multiple of a half-circle, then so is 2A+2B+2C.
Moral: Don't do this problem if you've forgotten your trigonometry.
However, since the problem as stated doesn't mention trigonometry, I'm wondering if there's another way to do it that avoids that. Probably there is. The tangent function and the identities we used involve a particular way of parametrizing the circle, as A(cosA,sinA). But the result doesn't seem to be one that should depend on such a choice of parametrization.

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