1. ## Trigonomic function question

Hey, I have been working a bit with the trig functions and came across this problem:

B = (A + sin(theta) )/ (sin(theta)*cos(theta))

when you separate the above fraction into the following:

C = A/(sin(theta)*cos(theta))

D = sin(theta) / (sin(theta)*cos(theta)) => 1/ cos(theta)

C+D != B if theta = 0 as sin(0) = 0 and cos(0) = 1

I came across this problem when trying to solve :

a - b*cos(theta) - b^2*sin(theta) = 0

Is there any method to make sure that this doesnt happen or do you just have to check the formula through for this?

Thanks for any replies.

2. Originally Posted by FireSoul
Hey, I have been working a bit with the trig functions and came across this problem:

B = (A + sin(theta) )/ (sin(theta)*cos(theta))

when you separate the above fraction into the following:

C = A/(sin(theta)*cos(theta))

D = sin(theta) / (sin(theta)*cos(theta)) => 1/ cos(theta)

C+D != B if theta = 0 as sin(0) = 0 and cos(0) = 1

I came across this problem when trying to solve :

a - b*cos(theta) - b^2*sin(theta) = 0

Is there any method to make sure that this doesnt happen or do you just have to check the formula through for this?

Thanks for any replies.
I would simply note that none of B, C, or D exist when $\theta = 0$, so there really is no problem.

-Dan

3. Originally Posted by FireSoul
I came across this problem when trying to solve :

a - b*cos(theta) - b^2*sin(theta) = 0
You are trying to solve for $\theta$? My recommendation:
$b^2sin(\theta) = a - b~cos(\theta)$<-- Square both sides

$b^4sin^2(\theta) = a^2 - 2ab~cos(\theta) + b^2 cos^2(\theta)$

Now, $sin^2(\theta) = 1 - cos^2(\theta)$, so
$b^4(1 - cos^2(\theta)) = a^2 - 2ab~cos(\theta) + b^2 cos^2(\theta)$

$b^4 - b^4cos^2(\theta) = a^2 - 2ab~cos(\theta) + b^2 cos^2(\theta)$

$(b^4 + b^2)cos^2(\theta) - 2ab~cos(\theta) + (a^2 - b^4) = 0$

Now let $y = cos(\theta)$. Then your equation is
$(b^4 + b^2)y^2 - 2aby + (a^2 - b^4) = 0$
which is a quadratic you can solve for y.

Note: When doing this you MUST check your solutions for $\theta$ in the original equation for $\theta$ as the method I am using here is likely to add spurious solutions.

-Dan