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Math Help - Trig Identities! Help Please!

  1. #1
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    Trig Identities! Help Please!

    I got several of these for homework. Several is >50....even after staying up laaaate last night I still cannot complete these 6. In all of them I have tried the breakdown into cosine and sine and work far into them. I think my problem is once I start working and I simplify certain things I become too commited to finishing what I already have done and don't want to scrap it and start over. I have tried for hours on these. I just can't get them out....please help.

    (cosecA-cotA)^2= 1-cosA/1+cosA


    (sinA+cosA)^2/(sinAcosA)= 2 + sexA coseA


    (sinA-cosA)(TanA+cotA)= secA- cosecA


    cotA-tanA/cotA+tanA= cos^2A-sin2^A


    (sinA-cosA)(TanA+cotA)=secA-cosecA


    (cotA-tanA)sinA=2cosA-secA
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  2. #2
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    Halifax, Nova Scotia
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    number 3 and 5 are the same questions.
    i have done 3 of these for you but rather than just copy my solution and easy way to see the "path" in solving these is to start with the left side by changing to sin and cos and doing any algebraic calculations (ie: squaring/dividing/simple factoring etc) then after 2 or 3 "obvious steps on the left, look back to the right side and try to deconstruct it ( chang to sin and cos, factor out, etc) I havent done these since precal in high school and im in my 4th year BSc(math), and it took me no more than 10 mins to get these 3 done. let me know if i can help out anymore.

    1)
    cotA-tanA/cotA+tanA= cos^2A-sin2^A

    cos/sin - sin/cos
    cos/sin + sin/cos

    cos^2 - sin^2 X sincos
    sincos cos^2 + sin^2

    cos^2 -sin^2
    cos^2 + sin^2

    = cos^2 -sin^2


    2) (cota - tanA)sinA = 2cosA- secA

    (cos/sin - sin/cos)sin

    cos - sin^2/cos

    cos - (1- cos^2)
    cos

    cos + cos - 1/cos

    = 2cosA - secA


    3)
    (sinA-cosA)(TanA+cotA)= secA- cosecA
    ** multiply through on left and get a common denominator you get...

    sin^3 + cos^2sin - sin^2cos - cos^3
    sincos

    sin(sin^2 + cos^2) - cos(sin^2 + cos^2)
    sincos sincos

    sin/sincos - cos/sincos

    1/cos - 1/sin

    = secA- cosecA

    Hope this helps
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  3. #3
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    Lexington, MA (USA)
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    Hello, Xelax!

    Here are the rest of them . . .


    4)\;\;\frac{\cot A-\tan A}{\cot A+\tan A} \:=\: \cos^2\!A-\sin^2\!A
    We have: . \frac{\dfrac{\cos A}{\sin A} - \dfrac{\sin A}{\cos A}} {\dfrac{\cos A}{\sin A} + \dfrac{\sin A}{\cos A}}

    Multiply by \frac{\sin A\cos A}{\sin A\cos A}\!:\;\;\frac{\sin A\cos A\left(\dfrac{\cos A}{\sin A} - \dfrac{\sin A}{\cos A}\right)} {\sin A\cos A\left(\dfrac{\cos A}{\sin A} + \dfrac{\sin }{\cos A}\right)}

    . . = \;\frac{\cos^2\!A -\sin^2\!A}{\underbrace{\cos^2\!A + \sin^2\!A}_{\text{This is 1}}} \;=\;\cos^2\!A - \sin^2\!A




    5)\;\;(\sin A-\cos A)(\tan A+\cot A)\:=\:\sec A-\csc A
    We have: . (\sin A - \cos A)\left(\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}\right) \;=\;(\sin A - \cos A)\cdot \frac{\overbrace{\sin^2\!A + \cos^2\!A}^{\text{This is 1}}}{\sin A\cos A}


    . . = \;\frac{\sin A - \cos A}{\sin A\cos A} \;=\;\frac{\sin A}{\sin A\cos A} - \frac{\cos A}{\sin A\cos A} \;=\; \frac{1}{\cos A} - \frac{1}{\sin A} \;=\;\sec A - \csc A




    6)\;\;(\cot A-\tan A)\sin A \:=\:2\cos A-\sec A
    We have: . \left(\frac{\cos A}{\sin A} - \frac{\sin A}{\cos A}\right)\sin A \;=\; \left(\frac{\cos^2\!A - \sin^2\!A}{\sin A\cos A}\right)\sin A \;=\;\frac{\cos^2\!A - \sin^2\!A}{\cos A}


    . . = \;\frac{\cos^2\!A - (1 - \cos^2\!A)}{\cos A} \;=\;\frac{2\cos^2\!A - 1}{\cos A} \;=\;\frac{2\cos^2\!A}{\cos A} - \frac{1}{\cos A} \;=\;2\cos A - \sec A

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