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posted by  DeadBrain on 11/2/2009 8:09:00 PM  |  status: Closed  |  Earned Karma: 25

Problem with Verifying Pythogorean Identities

Course Textbook Chapter Problem Needs by
Trigonometry Advanced Mathematical Concepts 7.2 9 11/3/2009 at 8:00:00 AM
Question Details:
I simply can not solve this problem, I think I may be doing something wrong algebraically or something.  Could someone help me verify, simplifying the right side to look like the left?
(sinθ-cosθ)2 = 1-2sin2θ*cotθ
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Mentor
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posted by jbeede on 11/2/2009 8:33:56 PM  |  status: Live
Asker's Rating: Lifesaver   
Response Details:
First rewrite the right side, replacing cotθ with cosθ/sinθ to get:
This simplifes to 1-2sinθcosθ, do you see that?
sin2θ+cos2θ =1 so you can replace the "1' above to get: sin2θ+cos2θ-2sinθcosθ
Rearrange and you have sin2θ-2sinθcosθ+cos2θ
which factors to (sinθ-cosθ)2 and you are done!
 
Here's the same problem but approaching from the left side, I think it is easier.
First off you will want to FOIL the left side to get: sin2θ-2sinθcosθ+cos2θ
By re-grouping like this: [sin2θ+cos2θ]-2sinθcosθ it is easy to see that this simplifies to: 1-2sinθcosθ 
Now look at the right side:
since cotθ=cosθ/sinθ,  1-2sin2θ*cotθ can be re-written as:
which matches what we had for the left side above. This way does not require the double angle rule
Sage
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posted by Betel on 11/2/2009 8:36:33 PM  |  status: Live
Asker's Rating: Helpful   
Response Details:

Left side:

(sinx - cosx)2 = sin2x +cos2x -2cosxsinx

= 1 - 2cosxsinx

Right side:

1 - 2sin2xcotx = 1- 2sin2x (cosx/sinx) = 1 - 2sinxcosx


Left side matches right side.


Hope that helps!
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Scholar
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posted by eswarramesh on 11/3/2009 11:53:54 AM  |  status: Live
Asker's Rating: Helpful   
Response Details:
given          to prove  (sin θ-cos θ)2 = 1 - 2 sin2 θ * cot θ
       taking  left hand side = (sin θ - cos θ)2                      
                                       = sin2θ+cos2θ-2sin θcos θ        [(a-b)2 = a2+b2-2ab]
                                       =  1 - 2 sin2 θ cos θ / sinθ         [ sin2θ+cos2θ=1]                                                                   dividing numerator and denominator by sinθ in the second term of the expression
                                       =  1 - 2 sin2θ cot θ
                                hence proved
eswarramesh
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