Use a reference triangle to find the exact value of cos[arctan(-3/10)]

You should get:

\(\frac{10}{\sqrt{109}}\)


Didn't I answer this already?

https://biology-forums.com/index.php?topic=1900800.7#msg4904322

(reply back to the topic if you're still confuzzled)

may i know the step by step? thanks

Ok, I'll try my best to illustrate what's happening:

So we're told the following:

\(y=\arctan \left(\frac{1}{x}\right)\)

arctan is the same as \(\tan ^{-1}\left(\frac{1}{x}\right)\)

Also, to make things simple to understand, don't use y, use \(\theta\) instead, so we have:

\(\theta =\tan ^{-1}\left(\frac{1}{x}\right)\)

Take tangent of both sides, essentially reverse what's been done:

\(\tan \left(\theta \right)=\tan \left(\tan ^{-1}\left(\frac{1}{x}\right)\right)\)

The tan and the tan inverse technically cancel each other out on the right side:

\(\tan \left(\theta \right)=\frac{1}{x}\)

Now what do we know about tangent? We know it's \(\frac{opposite}{adjacent}\), that's the ratio, opposite over adjacent, so draw out a right triangle with theta as an angle, opposite of theta is 1 and adjacent to theta is x. Next, using the pythagorean theorem, you will get:

\(c^2=a^2+b^2\), where "a" is x and "b" is 1 and "c" is the hypotenuse that you need for cosine because cosine = \(\frac{adjacent}{hypotenuse}\)

So, finding "c", use the formula:

\(c^2=x^2+1^2\)
\(c=\sqrt{x^2+1}\rightarrow hypotenuse\)

Now:

\(\cos \left(\theta \right)=\frac{x}{\sqrt{x^2+1}}\)