# How do you find cos if tan=3/4?

May 13, 2018

$\tan \theta = \frac{3}{4}$ means an opposite of $3$, an adjacent of $4$, so a hypotenuse of $5$, because ${3}^{2} + {4}^{2} = {5}^{2}$.

$\cos \theta = \frac{\textrm{a \mathrm{dj} a c e n t}}{\textrm{h y p o t e \nu s e}} = \pm \frac{4}{5}$

#### Explanation:

In general the sign of sine, cosine or tangent is underdetermined when we know just one of the others. That's why we need the $\pm$.

For reference we should enumerate the forms like this that come up. They're pretty easy to write down when we think about the numerator and denominator as opposite, adjacent or hypotenuse as appropriate.

When $t = \tan \theta$ I like to write

$\theta = \arctan t$

I consider those two equations equivalent; i.e. $\arctan t$ is not (just) the principal value. I write the principal value as $\textrm{A r c} \textrm{\tan} \left(t\right)$ and get

$\arctan t = \textrm{A r c} \textrm{\tan} \left(t\right) + \pi k \quad$ integer $k$

That's all to say when I write the inverse trig functions below, they're just abbreviations for the question being asked here, e.g if $\tan \theta = \frac{a}{b}$ what's $\cos \theta$? We can be more sure about the signs if we stick to principal values, but I prefer this way.

$\cos \arctan \left(\frac{a}{b}\right) = \setminus \pm \frac{b}{\sqrt{{a}^{2} + {b}^{2}}}$

$\sin \arctan \left(\frac{a}{b}\right) = \setminus \pm \frac{a}{\sqrt{{a}^{2} + {b}^{2}}}$

sin arccos(a/b) = pm sqrt {b^2-a^2} /b

tan arccos(a/b) = pm sqrt {b^2-a^2} /a

cos arcsin(a/b) = pm sqrt {b^2-a^2} /b

tan arcsin(a/b) = pm a/ sqrt {b^2-a^2}

We note

$\sin \arccos \left(\frac{a}{b}\right) = \pm \cos \arcsin \left(\frac{a}{b}\right)$