Title: Use a reference triangle to find the exact value of cos[arctan(-3/10)] Post by: ccnnieeee on Jan 16, 2019 1. Use a reference triangle to find the exact value of cos[arctan(-3/10)]
2.y= arctan(1/x), Use a reference triangle to find cos y. Title: Re: Use a reference triangle to find the exact value of cos[arctan(-3/10)] Post by: duddy on Jan 16, 2019 1. Use a reference triangle to find the exact value of cos[arctan(-3/10)] You should get: \(\frac{10}{\sqrt{109}}\) Quote 2.y= arctan(1/x), Use a reference triangle to find cos y. 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) Title: Re: Use a reference triangle to find the exact value of cos[arctan(-3/10)] Post by: ccnnieeee on Jan 16, 2019 may i know the step by step? thanks
Title: Re: Use a reference triangle to find the exact value of cos[arctan(-3/10)] Post by: duddy on Jan 17, 2019 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: \(\cos \left(\theta \right)=\frac{x}{\sqrt{x^2+1}}\) |