# Find θ for given condition?(seems to be easy but I can't seem to proceed)?

• Aug 30th 2008, 06:23 AM
fardeen_gen
Find θ for given condition?(seems to be easy but I can't seem to proceed)?
(sin θ + cos θ - 1)/(2 sin θ cos θ) = √(6) - 2

Find θ ?
• Aug 30th 2008, 07:10 AM
fardeen_gen
I got a very messy arctan.(Worried)
• Aug 30th 2008, 07:14 AM
its 15
try to put it yaar
• Aug 30th 2008, 07:17 AM
Sean12345
Hi fardeen_gen,

Here is my very blunt approach:

Let's define $\sin(\theta)=s~,~\cos(\theta)=c$

We have $\frac{s+c-1}{2sc}=\sqrt{6}-2$ Call this equation $(*)$ Note that $2sc=\sin(2\theta)~\forall \theta \in \mathbb{Z}$ and thus $\frac{s+c-1}{\sin(2\theta)}=\sqrt{6}-2$ Now we can square both sides of the equation and do a bit of rearranging;

$\frac{(s+c-1)^2}{\sin^2(2\theta)}=(\sqrt{6}-2)^2$

$\implies \frac{(s^2+c^2)+2sc-2(s+c)+1}{\sin^2(2\theta)}=6-4\sqrt{6}+4$

$\implies \frac{2+\sin(2\theta)-2(s+c)}{\sin^2(2\theta)}=10-4\sqrt{6}$ Call this equation $(**)$

Note that from equation $(*)$ We have;

$s+c=\sin(2\theta)(\sqrt{6}-2)+1$

Now subbing this into equation $(**)$ gives;

$\frac{2+\sin(2\theta)-2[\sin(2\theta)(\sqrt{6}-2)+1]}{\sin^2(2\theta)}=10-4\sqrt{6}$

$\implies \frac{2+\sin(2\theta)-2\sin(2\theta)(\sqrt{6}-2)-2}{\sin^2(2\theta)}=10-4\sqrt{6}$

$\implies \frac{1-2(\sqrt{6}-2)}{\sin(2\theta)}=10-4\sqrt{6}$

$\implies \sin(2\theta)=\frac{5-2\sqrt{6}}{10-4\sqrt{6}}=\frac{5-2\sqrt{6}}{10-4\sqrt{6}}\cdot\frac{10+4\sqrt{6}}{10+4\sqrt{6}}$

$
\implies \sin(2\theta)=\frac{1}{2}
$

Maybe there is an easier 'technique' (Thinking)