Trig Identity  TOPIC_SOLVED

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.

Trig Identity

Postby j3nn4k on Fri Mar 21, 2014 2:16 am

So I'm working on my homework and cannot figure out how to verify this identity.

The instructions on the worksheet are simply to verify that the identity is true.

The equation is;

(sec(A)-tan(A))^2 = (1-sin(A))/(1+sin(A))

I can work with either the left or the right side to verify it, but neither seems to really be helping me out with this problem.
I've worked though several different starting points and keep hitting dead ends.

If someone could help me or at least help me get started in the right direction that would be wonderful.

Thanks in advance!
j3nn4k
 
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Re: Trig Identity  TOPIC_SOLVED

Postby buddy on Fri Mar 21, 2014 11:24 am

j3nn4k wrote:I've worked though several different starting points and keep hitting dead ends.

Lots of times that means that you have to work both sides down to some place where they're equal & then write it up down one side and then up the other.

j3nn4k wrote:(sec(A)-tan(A))^2 = (1-sin(A))/(1+sin(A))

Make the left all sines and cosines: (1/cos(A) - sin(A)/cos(A))^2 = ((1 - sin(A))/cos(A))^2

Make the right so the denom is in cosine: [(1 - sin(A))/(1 - sin(A))][(1 + sin(A))/(1 - sin(A))] = ((1 - sin(A))^2) / (1 - sin^2(A)) so the denom turns into cos^2(A)

So the left equals to the right. Do the proof by starting on one side, going down to where they're equal, and then going up the other side. (This is how they come up with those proofs in the book that make you think "where did THAT come from???")

Hope that helps.
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Re: Trig Identity

Postby j3nn4k on Fri Mar 21, 2014 3:44 pm

Okay, I think I can figure it out now. Thanks so much for the help!
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