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Math Help - Problem with finding theta

  1. #1
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    Exclamation Problem with finding theta

    A = (3cot2(θ/2)-1)/4
    I need to find theta, however I cannot seem to make it the subject of the equation. Can anyone help? the 2 on the cot2 is cot squared
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xlepher View Post
    A = (3cot2(θ/2)-1)/4
    I need to find theta, however I cannot seem to make it the subject of the equation. Can anyone help? the 2 on the cot2 is cot squared
    A = (3cot^2(@/2) - 1)/4
    => 4A = 3cot^2(@/2) - 1 ..............multiply both sides by 4
    => 4A + 1 = (3cot^2(@/2) ..............add 1 to both sides
    => (4A + 1)/3 = cot^2(@/2) .............divide both sides by 3
    => +/- sqrt[(4A + 1)/3] = cot(@/2) ....take the square root of both sides
    => cot^-1{+/- sqrt[(4A + 1)/3]} = @/2 ..take the inverse cotangent of both sides (you may call this arccot)

    => 2cot^-1{+/- sqrt[(4A + 1)/3]} = @
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  3. #3
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    Hello, xlepher!

    Hope I read your equation correctly . . .


    We have: . ¼[3·cot²(½θ) - 1] .= .A

    Multiply by 4: . 3·cot²(½θ) - 1 .= .4A

    Add 1: . 3·cot²(½θ) .= .4A + 1

    Divide by 3: . cot²(½θ) .= .(4A + 1)/3
    . . . . . . . . . . . . . . . . . . . . . . . . . _________
    Take square roots: . cot(½θ) .= .±√(4A + 1)/3
    . . . . . . . . . . . . . . . . . . . . . . . . . - - . . . _________
    Take inverse cotangent: . ½θ .= .arccot[±√(4A + 1)/3]
    . . . . . . . . . . . . . . . . . . . . . . . _________
    Multiply by 2: . θ .= .2·arccot[±√(4A + 1)/3]



    Edit: Too fast for me, Jhevon!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, xlepher!

    Hope I read your equation correctly . . .


    We have: . ¼[3·cot²(½θ) - 1] .= .A

    Multiply by 4: . 3·cot²(½θ) - 1 .= .4A

    Add 1: . 3·cot²(½θ) .= .4A + 1

    Divide by 3: . cot²(½θ) .= .(4A + 1)/3
    . . . . . . . . . . . . . . . . . . . . . . . . . _________
    Take square roots: . cot(½θ) .= .±√(4A + 1)/3
    . . . . . . . . . . . . . . . . . . . . . . . . . - - . . . _________
    Take inverse cotangent: . ½θ .= .arccot[±√(4A + 1)/3]
    . . . . . . . . . . . . . . . . . . . . . . . _________
    Multiply by 2: . θ .= .2·arccot[±√(4A + 1)/3]


    Edit: Too fast for me, Jhevon!
    Sorry again, Soroban. You must have started to suspect that I do this on purpose, I assure you I don't.
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  5. #5
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    Could arccot be simlified so that it could be used with a scientific calculator?
    So far i have:
    We have: . ¼[3·cot²(½θ) - 1] .= .1.29

    Multiply by 4: . 3·cot²(½θ) - 1 .= .4 * 1.29

    Add 1: . 3·cot²(½θ) .= .5.16 + 1

    Divide by 3: . cot²(½θ) .= .(6.16)/3
    . . . . . . . . . . . . . . . . . . . . . . . . . _________
    Take square roots: . cot(½θ) .= .√2.0533
    . . . . . . . . . . . . . . . . . . . . . . . . . - - . . .
    Take inverse cotangent: . ½θ .= .arccot 1.433
    . . . . . . . . . . . . . . . . . . . . . . .
    Multiply by 2: . θ .= .2·arccot1.433

    I do not know how to use arccot as i only have a simple scientific calculator that cot and arccot are not on there.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xlepher View Post
    Could arccot be simlified so that it could be used with a scientific calculator?
    So far i have:
    We have: . ¼[3·cot²(½θ) - 1] .= .1.29

    Multiply by 4: . 3·cot²(½θ) - 1 .= .4 * 1.29

    Add 1: . 3·cot²(½θ) .= .5.16 + 1

    Divide by 3: . cot²(½θ) .= .(6.16)/3
    . . . . . . . . . . . . . . . . . . . . . . . . . _________
    Take square roots: . cot(½θ) .= .√2.0533
    . . . . . . . . . . . . . . . . . . . . . . . . . - - . . .
    Take inverse cotangent: . ½θ .= .arccot 1.433
    . . . . . . . . . . . . . . . . . . . . . . .
    Multiply by 2: . θ .= .2·arccot1.433

    I do not know how to use arccot as i only have a simple scientific calculator that cot and arccot are not on there.
    well it depends on the calculator. but remember, cot(x) = 1/tan(x), so you can always do your computations in terms of tan(x). so i guess arccot would be 1/arctan, but you should verify this with some sort of test
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  7. #7
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    So would arccot(x) = 1/arccot(x) ?
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xlepher View Post
    So would arccot(x) = 1/arccot(x) ?
    i said arccot MAY BE 1/arctan(x)
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    Quote Originally Posted by Jhevon View Post
    A = (3cot^2(@/2) - 1)/4
    => 4A = 3cot^2(@/2) - 1 ..............multiply both sides by 4
    => 4A + 1 = (3cot^2(@/2) ..............add 1 to both sides
    => (4A + 1)/3 = cot^2(@/2) .............divide both sides by 3
    => +/- sqrt[(4A + 1)/3] = cot(@/2) ....take the square root of both sides
    Let's take Jhevon's post from here:
    cot(@/2) = (+/-) sqrt{(4A + 1)/3}

    Thus
    tan(@/2) = (+/-)1/sqrt{(4A + 1)/3}

    @/2 = atn[(+/-)1/sqrt{(4A + 1)/3}] = (+/-)atn[1/sqrt{(4A + 1)/3}]

    @ = (+/-)2*atn[1/sqrt{(4A + 1)/3}]

    Note:
    1/arccot(@) is NOT equal to arctan(@)!

    -Dan
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    Let's take Jhevon's post from here:
    cot(@/2) = (+/-) sqrt{(4A + 1)/3}

    Thus
    tan(@/2) = (+/-)1/sqrt{(4A + 1)/3}

    @/2 = atn[(+/-)1/sqrt{(4A + 1)/3}] = (+/-)atn[1/sqrt{(4A + 1)/3}]

    @ = (+/-)2*atn[1/sqrt{(4A + 1)/3}]

    Note:
    1/arccot(@) is NOT equal to arctan(@)!

    -Dan
    Fine, let's find out once and for all what arccot is in terms of tan.

    Let arccot(x) = y
    => x = cot(y)
    => x = 1/tan(y)
    => 1/x = tan(y)
    => arctan(1/x) = y

    thus, arccot(x) = arctan(1/x)

    so you can take the function that you want to find the arccot of, find it's inverse and use arctan on it in your calculator.

    Thanks for looking out again Dan.
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