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

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 #text{Arc}text{tan}(t)# and get

#arctan t = text{Arc}text{tan}(t) + 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 = 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 (a/b) = \pm b/sqrt{a^2 + b^2} #

#sin arctan (a/b) = \pm 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(a/b) = pm cos arcsin(a/b) #