Question

If cos θ =-5/13 and sin θ < 0, what is tan θ?

Answer 1

You can find sin(theta) by using the identity \[(\cos(\theta))^{2}+(\sin(\theta))^{2}=1\]. Then you can find \[\tan(\theta)\] by the identity \[\tan(\theta)=\frac{ \sin(\theta) }{\cos(\theta) }\]

Answer 2

Well you can find theta using inverse cosine and a calculator.
arccos(-5/13) = theta
theta = 112.6198649

sin(112.6198649) = 0.923, that's less than 1.

Then tan(112.6198649) = -2.4

Answer 3

well.... hmmm so
recall your SOH CAH TOA
\(sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad
% cosine
{\color{brown}{ cos(\theta)=\cfrac{adjacent}{hypotenuse} }}

\\ \quad \\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\)

so, notice the cosine identity, that means \(cos(\theta)=-\cfrac{5}{13}\implies \cfrac{adjacent}{hypotenuse}\)

now, recall that, the hypotenuse is never negative, since it's just the radius
so on -5/13 it can't be 5/-13
so, that means the -5 is really the adjacent side
thus \(cos(\theta)=-\cfrac{5}{13}\implies \cfrac{adjacent}{hypotenuse}\implies \cfrac{adjacent=-5}{hypotenuse=13}\)