# trigonometry - Find the value of cos($\theta$) given cot($\theta$)

I am tasked with finding cos($\theta$) in quadrant I when given cot($\theta$) = 23. The book is showing the answer as:

23$\sqrt[]{530}$/530 but I keep getting $\cos(\theta)$ = 1/$\sqrt[]{530}$. Can someone please tell me what I’m doing wrong?

1 + $\cot^2(\theta)$ = 1/$\cos^2(\theta)$ // Pythagorean identity

1 + 529 = 1/$\cos^2(\theta)$ // Substitute $\cot^2(\theta)$

530 = 1/$\cos^2(\theta)$

530$\cos^2(\theta)$ = 1 // Divide both sides by $\cos^2(\theta)$

$\cos^2(\theta)$ = 1/530 // Divide both sides by 530

$\cos(\theta)$ = 1/$\sqrt[]{530}$ // Square root both sides

You made an error at the very beginning. The following is false: $$1 + \cot^2(\theta) = 1/\cos^2(\theta).$$

Instead note that $$1 + \cot^2(\theta) = \csc^2\theta,$$ so $$530\sin^2(\theta) = 1.$$ Hence $$\cos^2(\theta)=1-\sin^2(\theta)= 1-\frac{1}{530}= \frac{529}{530}.$$

• This is great as I see how you get cos^2($theta$) = 529/530 but the book is showing cos($\theta) = 23$\sqrt{529}$\530. – maybedave Dec 27 '17 at 1:04 • The formatting didn't work but I listed the book answer in my original question. Is the book wrong? – maybedave Dec 27 '17 at 1:05 • Ah...nevermind. I didn't remove the radical from the final answer. I got now.!! – maybedave Dec 27 '17 at 1:06 Your Pythagorean identity is wrong. To get the correct identity, we want to start with$\sin^2(\theta) + \cos^2(\theta) = 1$and divide by a term which gives us$\cot^2(\theta)$somewhere in there:$\cot^2(\theta) = \frac{\cos^2(\theta)}{\sin^2(\theta)}$, so we want to divide by$\sin^2(\theta)$, giving$1 + \cot^2(\theta) = \frac{1}{\sin^2(\theta)}$If$\cot\theta=23$, then adjacent side is$23$and opposite side$1$, so the hypotenuse becomes$\sqrt{23^2+1^2}=\sqrt{530}$. Thus$\cos\theta=\frac{23}{\sqrt{530}}=\frac{23\sqrt{530}}{530}\$.

HINTS

$$\cot \theta = \frac{\cos\theta}{\sin\theta}$$

$$\sin^2\theta + \cos^2\theta = 1$$

Hence

$$\cot\theta = 23 \to \cos\theta = 23\sin\theta$$

$$\cos\theta = 23(\sqrt{1 - \cos^2\theta})$$

Then you have to solve a second degree equations

$$\cos^2\theta = 23^2 - 23^2\cos^2\theta$$

$$(23^2 + 1)\cos^2\theta = 23^2$$

$$\cos\theta = \sqrt{\frac{23^2}{23^2+1}} = \sqrt{\frac{529}{530}} = 0.998113207$$