Pre-Cal.-Proving Identities

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How can I prove this identity? I know the answer is "true," but I can't figure out exactly how to arrive at the answer.  

sinØ/(1-cosØ) - (1+cosØ)/sinØ = 0

Jan 18th, 2015

sinØ / (1 + cosØ) = (1 - cosØ) / sinØ 

I'll use a simpler, step by step approach, the same as when on solves an algebraic equation: 

multiply both sides by sinØ (this gets rid of the denominator on the right): 

sin^2Ø / (1 + cosØ) = (1 - cosØ) 

sin^2Ø simply means "sine squared of phi" 

it is the same as writing (sinØ)^2 

next step, multiply both sides by 1 + cosØ 

this gets rid of the denominator on the left 

sin^2Ø = (1 + cosØ)(1 - cosØ) 

sin^2Ø = 1 - cos^2Ø 

sin^2Ø + cos^2Ø = 1 

which is a well-known identity 

Using descriptions from plane geometry (right-angle triangle) 

(opp/hyp)^2 + (adj/hyp)^2 = 1 

(opposite^2 + adjacent^2)/hypotenuse^2 = 1 

Using Pythagorean theorem: 

hypotenuse^2 = opp.^2 + adj.^2 

we replace the top part 

hypotenuse^2 / hypotenuse^2 = 1 

which is true. 

You have 

sinØ-sinØcosØ/sinØsinØcosØ = cosØ/sinØsinØcosØ. 

I do not see your reasoning. 

At first, I thought you were going for a common denominator, but that would have been (for the denominator) 

(1 + cosØ)sinØ = sinØ + sinØcosØ 

(did you lose the + along the way?) 

Next, you would multiply each side by the missing part of the common denominator (which, in this case, is the denominator from the other side of the = sign) 

sinØsinØ / (sinØ + sinØcosØ) = (1-cosØ)(1+cosØ) / (sinØ + sinØcosØ) 

When you reach this stage, where each side (the whole side) is over the same common denominator, then you can ignore the denominator. 

simple example: let's say k is some fixed number and both sides have it as the denominator 

a/k = b/k 

then, obviously, the only way this can be true is if a=b (you can now ignore the k) 

This idea of dropping the common denominator leaves you with 

(sinØ)^2 = 1 - (cosØ)^2 

which becomes 

sin^2Ø + cos^2Ø = 1

Jan 18th, 2015

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