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

**fardeen_gen** · August 30th 2008, 05:23 AM

(sin θ + cos θ - 1)/(2 sin θ cos θ) = √(6) - 2

Find θ ?

**fardeen_gen** · August 30th 2008, 06:10 AM

I got a very messy arctan.

**ADARSH** · August 30th 2008, 06:14 AM

try to put it yaar

**Sean12345** · August 30th 2008, 06:17 AM

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;

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

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

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

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

$<br /> \implies \sin(2\theta)=\frac{1}{2}<br />$

Maybe there is an easier 'technique'

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