Find the exact value of cos(x) if tan(x)=5 and is in quadrant I?

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Find the exact value of cos(x) if tan(x)=5 and is in quadrant I?

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Answer 1 · 2018-04-07T02:11:30

$cos (x) = \frac{1}{\sqrt{26}}$ $cos(x) = 1 / sqrt(26)$

Explanation:

This problem is solved using the Pythagorean Theorem and knowledge of "soh-cah-toa".

Recall that "soh-cah-toa" reminds us that $sin (x) = \frac{opposite}{hypotenuse}$ $sin(x) = "opposite"/"hypotenuse"$ , $cos (x) = \frac{adjacent}{hypotenuse}$ $cos(x) = "adjacent"/"hypotenuse"$ and $tan (x) = \frac{opposite}{adjacent}$ $tan(x) = "opposite"/"adjacent"$ , where 'opposite', 'adjacent' and 'hypotenuse' refer to sides of a right triangle.

Our equality $tan (x) = 5$ $tan(x) = 5$ refers to a right triangle where the side opposite from angle $x$ $x$ is 5 times larger than the side adjacent to $x$ $x$ . The following is one such triangle (not drawn to scale).

Sketchpad 5.1

Using the Pythagorean Theorem, we can find the length of the hypotenuse.

$a^{2} + b^{2} = c^{2}$ $a^2 + b^2 = c^2$ ,
$1^{2} + 5^{2} = c^{2}$ $1^2 + 5^2 = c^2$ ,
$26 = c^{2}$ $26 = c^2$ ,
$c = \sqrt{26}$ $c = sqrt(26)$ .

Then $cos (x)$ $cos(x)$ refers to our same triangle. See that $cos (x) = \frac{adjacent}{hypotenuse} = \frac{1}{\sqrt{26}}$ $cos(x) = "adjacent"/"hypotenuse" = 1 / sqrt(26)$ .

Note: This triangle extends up and to the right, into the first quadrant.

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Find the exact value of cos(x) if tan(x)=5 and is in quadrant I?

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