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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- March 19th 2007, 03:21 PMxlepherProblem 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 - March 19th 2007, 03:48 PMJhevon
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]} = @ - March 19th 2007, 04:05 PMSoroban
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! - March 19th 2007, 04:11 PMJhevon
- March 19th 2007, 04:13 PMxlepher
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.

- March 19th 2007, 04:21 PMJhevon
- March 19th 2007, 04:24 PMxlepher
So would arccot(x) = 1/arccot(x) ?

- March 19th 2007, 04:25 PMJhevon
- March 20th 2007, 04:53 AMtopsquark
- March 20th 2007, 08:12 AMJhevon
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.