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Math Help - Need help with trig identities

  1. #1
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    Need help with trig identities

    If you could go step by step it would be helpful!

    cotx+tanx = secxcscx



    sin^4x - cos^4x = 2sin^2x - 1



    cot^2x-1/1+cot^2x = 1-2sin^2x
    Thanks!
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  2. #2
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    Quote Originally Posted by Thatoneguy12345 View Post
    cot^2x-1/1+cot^2x = 1-2sin^2
    \frac{cot^2(x)-1}{cot^2(x)+1}
    \frac{cot^2(x)-1+1-1}{cot^2(x)+1}
    \frac{cot^2(x)+1-2}{cot^2(x)+1}
    \frac{cot^2(x)+1}{cot^2(x)+1}+\frac{-2}{cot^2(x)+1}
    1+\frac{-2}{cot^2(x)+1}
    1+\frac{-2}{csc^2(x)}
    1-2sin^2(x)
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  3. #3
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    Wait, how do you get the -1+1-1 on the 2nd step?
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  4. #4
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    Quote Originally Posted by Thatoneguy12345 View Post
    Wait, how do you get the -1+1-1 on the 2nd step?
    1-1 is zero right? If you add it to the equation it doesn't change the value of the equation. But you can use it to simplify the equation, like I did.

    For example if you have the equation

    5+2 = 7

    That's the same as

    5+2+(1-1)=7
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  5. #5
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    Oh whoa, thanks! Never knew you could do it that way!
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  6. #6
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    Q1
    Write LHS in terms of sin and cos
    Get a common denominator and add
    You should recognise and simplify the top
    Split up the product
    You should now have the RHS
    TRY IT!!

    Q2 Hint: a^4 - b^4 =(a^2 + b^2)(a^2 - b^2)
    The first bracket will be =1
    Look at the RHS (there is no cos in sight!)
    So, replace (cos x)^2 with 1 - (sin x)^2
    Too easy!

    Q3 Use the identity 1 + (cot x)^2 = (cosec x)^2 on the bottom
    Dividing by (cosec x)^2 is the same as multiplying by (sin x)^2
    So now you should have (sin x)^2 x ((cot x)^2 -1)
    Write cot^2 in terms of sin and cos
    Expand out bracket
    Do the same thing you did at the end of Q2 to get rid of cos^2
    Done
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  7. #7
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    Quote Originally Posted by Thatoneguy12345 View Post
    cotx+tanx = secxcscx
    cot(x)+tan(x)=sec(x)csc(x)
    \frac{cos(x)}{sin(x)}+\frac{sin(x)}{cos(x)}
    You choose the common denominator of cos(x)sin(x) and multiply through. You get:
    \frac{cos^2(x)+sin^2(x)}{sin(x)cos(x)}
    Use the identity sin^2(x)+cos^2(x)=1
    \frac{1}{sin(x)cos(x)}
    \frac{1}{sin(x)}*\frac{1}{cos(x)}
    csc(x)sec(x)

    Edit: If you still need me to show you the second one after Debsta explained it, post the request. Debsta did a good job of explaining it.
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  8. #8
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    My thoughts exactly!
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  9. #9
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    Thanks a lot you two!
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