should this factorise?

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February 13th 2010, 11:54 PM
200001

should this factorise?

http://img25.imageshack.us/img25/8563/trigs.jpg

Start of problem at the top
I believe i have done it right but it doesnt seem to factorise
Do i need to run it from the quadratic equation?
February 14th 2010, 12:03 AM
Prove It

Quote:

Originally Posted by 200001

http://img25.imageshack.us/img25/8563/trigs.jpg

Start of problem at the top
I believe i have done it right but it doesnt seem to factorise
Do i need to run it from the quadratic equation?

$\sin{\theta} + \cos^2{\theta} = \frac{1}{5}$

$\sin{\theta} + 1 - \sin^2{\theta} = \frac{1}{5}$

$0 = \sin^2{\theta} - \sin{\theta} - \frac{4}{5}$

Now Complete the Square:

$0 = \sin^2{\theta} - \sin{\theta} + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 - \frac{4}{5}$

$0 = \left(\sin{\theta} - \frac{1}{2}\right)^2 - \frac{1}{4} - \frac{4}{5}$

$0 = \left(\sin{\theta} - \frac{1}{2}\right)^2 - \frac{21}{20}$

$\frac{21}{20} = \left(\sin{\theta} - \frac{1}{2}\right)^2$

$\pm\sqrt{\frac{21}{20}} = \sin{\theta} - \frac{1}{2}$

$\pm \frac{\sqrt{21}}{2\sqrt{5}} = \sin{\theta} - \frac{1}{2}$

$\pm \frac{\sqrt{105}}{10} = \sin{\theta} - \frac{1}{2}$

$\frac{1}{2} \pm \frac{\sqrt{105}}{10} = \sin{\theta}$

$\frac{5 \pm \sqrt{105}}{10} = \sin{\theta}$

$\theta = \arcsin{\left(\frac{5 + \sqrt{105}}{10}\right)}$ or $\theta = \arcsin{\left(\frac{5 - \sqrt{105}}{10}\right)}$ .
February 14th 2010, 12:44 AM
200001

Yes, thats where i went with it but this is from a book that isnt driven at any equations that wont factorise and the level its picthed at would not go into that depth.
I had the same answer yet though I had gone wrong somewhere (Headbang)
February 14th 2010, 12:55 AM
Grandad

Hello 200001

Quote:

Originally Posted by 200001

http://img25.imageshack.us/img25/8563/trigs.jpg

Start of problem at the top
I believe i have done it right but it doesnt seem to factorise
Do i need to run it from the quadratic equation?

You have a sign wrong in the third line of your solution. It should be:
$\sin^2\phi-\sin\phi=\frac45$

$\Rightarrow 5\sin^2\phi-5\sin\phi-4=0$

$\Rightarrow \sin\phi = \frac{5\pm\sqrt{105}}{10}$

Now $-1\le\sin\phi\le 1$ . So the only valid root is:
$\sin\phi = \frac{5-\sqrt{105}}{10}$

$\Rightarrow \phi = n\pi +(-1)^n \arcsin\left(\frac{5-\sqrt{105}}{10}\right)$

Grandad
February 15th 2010, 10:14 PM
200001

Yes
that makes sense
Thanks!