# Trigonomic function question

• Jul 19th 2007, 05:32 AM
FireSoul
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.
• Jul 19th 2007, 06:25 AM
topsquark
Quote:

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
• Jul 19th 2007, 06:32 AM
topsquark
Quote:

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