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Last piece of study material

  1. Oct 27, 2013 #1
    these are a couple more math problems im having trouble with. if you can help me i would appreciate it. if you could just put the number of the problem your helping me with that would be great

    1)simplify -3sin^(5)x - 3sin^(3)x cos^(2)x

    2.)simplify 2cot^(2)x -3cotx - 9 / cot^(2)x - 9

    3.)simplify 5cos^(4)x - 5sin^(4)x and write in terms of cos x.

    4.)simplify tan^(2)x + (1 + sec x)^2 write in terms of sec x.
     
  2. jcsd
  3. Oct 27, 2013 #2

    Mentallic

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    Homework Helper

    What is the highest common factor in this expression?

    Also, notice that almost everything is in terms of sin but we have a [itex]\cos^2{x}[/itex] term. Remember this useful formula

    [tex]\sin^2{x}+\cos^2{x}=1[/tex]

    so

    [tex]\cos^2{x}=1-\sin^2{x}[/tex]

    if you apply this substitution, then you'll have everything in terms of sin and will probably be able to simplify even further.


    You should put brackets around the numerator and denominator as so

    (2cot^(2)x -3cotx - 9) / (cot^(2)x - 9)

    else it could be misinterpreted as being

    [tex]2\cot^2{x}-3\cot{x}-\frac{9}{\cot^2{x}}-9[/tex]

    Begin by letting [itex]y=\cot{x}[/itex] so you'll have a quadratic in y in the numerator and denominator, then factorize those quadratics and you should be able to cancel out a common factor, and then convert back to cot(x) at the end.

    Can you factorize [itex]x^4-y^4[/itex] as a difference of two squares?

    That important formula from earlier

    [tex]\sin^2{x}+\cos^2{x}=1[/tex]

    if we divide through by [itex]\cos^2{x}[/itex] then we get

    [tex]\frac{\sin^2{x}}{\cos^2{x}}+\frac{\cos^2{x}}{\cos^2{x}}=\frac{1}{\cos^2{x}}[/tex]

    [tex]\tan^2{x}+1=\sec^2{x}[/tex]

    So now you can use this to get rid of the tan(x) term in your expression.
     
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