# Thread: Reducing a trig equation

1. ## Reducing a trig equation

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

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

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

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

I hope, you will now easier find the solution of trig. equ.

3. Try this:
$
sin(\theta) (cos(\theta) + csc(\theta)) = sin(\theta) (cos(\theta) + 1/sin(\theta)) = sin(\theta) cos(\theta) + 1 = 2cos^2(\theta)
$

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

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

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

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

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

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

$
\alpha = \frac 1 2 \pm \frac {\sqrt {5}}{10}
$

$
cos(\theta) = \pm \sqrt { \frac 1 2 \pm \frac {\sqrt {5}}{10} }
$

$
\theta = Arccos \left (\pm \sqrt { \frac 1 2 \pm \frac {\sqrt {5}}{10}} \right )
$

4. Brilliant
thanks