# Thread: Trig Identities Question

1. ## Trig Identities Question

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

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

2. ## Re: Trig Identities Question

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

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

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

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

using the identity ...

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

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

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

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

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

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