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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

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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 $$ \cos^2(\theta)=1-\sin^2(\theta)= 1-\frac{1}{530}= \frac{529}{530}. $$

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  • $\begingroup$ 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. $\endgroup$ – maybedave Dec 27 '17 at 1:04
  • $\begingroup$ The formatting didn't work but I listed the book answer in my original question. Is the book wrong? $\endgroup$ – maybedave Dec 27 '17 at 1:05
  • $\begingroup$ Ah...nevermind. I didn't remove the radical from the final answer. I got now.!! $\endgroup$ – maybedave Dec 27 '17 at 1:06
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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

$1 + \cot^2(\theta) = \frac{1}{\sin^2(\theta)}$

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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 $\cos\theta=\frac{23}{\sqrt{530}}=\frac{23\sqrt{530}}{530}$.

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HINTS

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

In addition to

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

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

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

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

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

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