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Mathematics Forum
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| Author | Message / Information |
| ryguy66 Quote | Reply | | proving trig identities posted on: 11/8/2005 12:32:49 PM I am having trouble proving these identities, please help me. 1. (csc^2y-cot^2y)/(sec^2y) = cos^2y 2. sinB + cosBcotB = cscB 3. (sinA)/(1-cosA) = cscA 4. (2tan13x)/(1+tan^2 13x) = sin26x 5. (secx -tanx)/(cscx-1) = tanx I tried these but they never seem to work, please help thanks |
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Euler
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proving trig identities
replied on: 6/14/2006 12:57:34 AM I doubt that you will even read this as it has been seven months; however: 1) change to: [(1/sin^2y)-(cos^2y/sin^2y)]/ (1/cos^2y) then get a common denominator on top and you have: [(1-cos^2y)/sin^2y]*cos^2y=cos^2y, the cos^2y drops out of both sides, then multiply both sides by sin^2y to get: 1-cos^2y=sin^2y, which can be rewritten as: cos^2y + sin^2y = 1, whiby changing the ch is true, thus proving the first one. 2) change cotB and you get: sinB + cosB(cosB/sinB)=cscB, distribute and get common denominator to get: (sin^2B + cos^2B)/ sinB, which equals: 1/sinB, which equals cscB, thus proving the second one. 3) by changing the cscA you get: sinA/(1-cosA)=1/sinA, then by multiplying both sides by sinA and 1-cosA you get: sin^2A=1-cosA, which means: sin^2A + cosA = 1, which is not true, so unless I made a mistake or you meant to have a cos^2A in the denominator, this one cannot be proven. 4) by changing tan13x in both the denominator and numerator you get: 2(sin13x/cos13x)/(1+sin^213x/cos^213x), then by getting a common denominator on the bottom you have: 2(sin13x/cos13x)/[(cos^213x+sin^213x)/cos^213x], and the top of the fraction in the denominator is equal to one, so then simplifying you get: 2sin13xcos13x=sin26x, which is true by the double-angle formula, thus proving the fourth one. 5) by changing everything into sin and cos, and getting a common denominator on top, you get: (1-sinx/cosx)/(1-sinx/sinx), and the quantity 1-sinx cancels leaving you with: sinx/cosx=tanx, which is true by the definition of tanx, thus proving he fifth one. Thank you. |
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