Trig Identities Question

• November 15th 2012, 11:27 AM
fogownz
Trig Identities Question
I'm trying to find $cos \alpha$. Earlier I determined using $tan \alpha = \frac{1}{cot \alpha}$ that $tan \alpha = \frac{1}{5}$. This is correct, according to my book.

Knowing this, can't I determine that because $tan \alpha = \frac{sin \alpha}{cos \alpha}$ that $cos \alpha = 5$. However, this logic seems incorrect. The answer my book is looking for is $\frac{5\sqrt{26}}{26}$. Can someone help me understand why my way doesn't work?
• November 15th 2012, 11:51 AM
skeeter
Re: Trig Identities Question
Quote:

Originally Posted by fogownz
I'm trying to find $cos \alpha$. Earlier I determined using $tan \alpha = \frac{1}{cot \alpha}$ that $tan \alpha = \frac{1}{5}$. This is correct, according to my book.

Knowing this, can't I determine that because $tan \alpha = \frac{sin \alpha}{cos \alpha}$ that $cos \alpha = 5$. However, this logic seems incorrect. The answer my book is looking for is $\frac{5\sqrt{26}}{26}$. Can someone help me understand why my way doesn't work?

because $\cos{\alpha} \ne 5$ ...

$-1 \le \cos(anything) \le 1$

using the identity ...

$\tan^2{\alpha} + 1 = \sec^2{\alpha}$

$\frac{1}{25} + 1 = \sec^2{\alpha}$

$\frac{26}{25} = \sec^2{\alpha}$

$\frac{25}{26} = \cos^2{\alpha}$

$\pm \frac{5}{\sqrt{26}} = \pm \frac{5\sqrt{26}}{26} = \cos{\alpha}$

you could also find $\cos{\alpha}$ by sketching a reference triangle in quads I and/or IV