# Thread: Problem with finding theta

1. ## 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

2. Originally Posted by xlepher
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]} = @

3. Hello, xlepher!

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!

4. Originally Posted by Soroban
Hello, xlepher!

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.

5. 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.

6. Originally Posted by xlepher
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

7. So would arccot(x) = 1/arccot(x) ?

8. Originally Posted by xlepher
So would arccot(x) = 1/arccot(x) ?
i said arccot MAY BE 1/arctan(x)

9. Originally Posted by Jhevon
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

10. Originally Posted by topsquark
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.