# Reducing a trig equation

• May 3rd 2010, 09:11 AM
200001
Reducing a trig equation
Hi
I am trying to solve this but cant get it in one/two trig ratios

http://img718.imageshack.us/img718/8164/coscosec.jpg
• May 3rd 2010, 11:45 AM
veljko
Quote:

Originally Posted by 200001
Hi
I am trying to solve this but cant get it in one/two trig ratios

http://img718.imageshack.us/img718/8164/coscosec.jpg

$\displaystyle \sin\phi(\cos \phi + \frac{1}{\sin\phi}) = \sin\phi(\frac{\cos\phi \sin\phi + 1}{\sin\phi}) =$ $\displaystyle \cos\phi \sin\phi + 1 = \cos\phi \sin\phi + \cos^2\phi + \sin^2\phi$

Than:
$\displaystyle \cos\phi \sin\phi + \cos^2\phi + \sin^2\phi = 2\cos^2\phi$
$\displaystyle \cos\phi \sin\phi + \sin^2\phi = \cos^2\phi$
$\displaystyle \cos\phi \sin\phi = \cos^2\phi - \sin^2\phi$
$\displaystyle \cos\phi \sin\phi = \cos2\phi$
$\displaystyle \frac{1}{2} \sin2\phi = \cos2\phi$

I hope, you will now easier find the solution of trig. equ.
• May 3rd 2010, 11:59 AM
ebaines
Try this:
$\displaystyle sin(\theta) (cos(\theta) + csc(\theta)) = sin(\theta) (cos(\theta) + 1/sin(\theta)) = sin(\theta) cos(\theta) + 1 = 2cos^2(\theta)$

$\displaystyle sin(\theta)cos(\theta) = 2 cos^2(\theta) -1$

Square both sides:
$\displaystyle sin^2(\theta) cos^2(\theta) = 4 cos^4(\theta) - 4 cos^2 (\theta) + 1$

Substitute $\displaystyle sin^2 \theta = 1 - cos^2 \theta$:

$\displaystyle (1 - cos^2(\theta))*cos^2 (\theta) = 4 cos^4(\theta) - 4 cos^2 (\theta) + 1$

$\displaystyle 5 cos^4(\theta) -5cos^2(\theta) +1 = 0$

Let $\displaystyle \alpha = cos^2(\theta)$:
$\displaystyle 5 \alpha^2 - 5 \alpha + 1 = 0$

Solve using the quadratic formula:
$\displaystyle \alpha = \frac 1 2 \pm \frac {\sqrt {5}}{10}$
$\displaystyle cos(\theta) = \pm \sqrt { \frac 1 2 \pm \frac {\sqrt {5}}{10} }$
$\displaystyle \theta = Arccos \left (\pm \sqrt { \frac 1 2 \pm \frac {\sqrt {5}}{10}} \right )$
• May 5th 2010, 10:24 AM
200001
Brilliant
thanks