# Trigonometry: if $\tan(t)=-1/2$ is in Quadrant II | find $\sin(t)+\cos(t)$

Hello guys I have spent hours trying to solve this but I get stuck. I couldnt find help in the book or anywhere else. Thank you.

if $\tan(t)=-1/2$ is in Quadrant II

find $\sin(t)+\cos(t)$.

Hint

We know that $\tan(t) = \frac{\sin(t)}{\cos(t)} = -\frac12$

And we're in Quadrant II, meaning $t \in [\frac\pi2, \pi]$, which means that $\cos(t) \in [-1, 0]$ and $\sin(t) \in [0, 1]$

Since $\cos^2t+\sin^2t=1$, dividing both sides by $\cos^2 t$ we also have
$$1+\tan^2t=\frac 1{\cos^2t}$$
Also, in the second quadrant, $\cos t<0$ and $\sin t>0$.
Use the second equation and the restriction to find $\cos t$, then use the first equation and the restriction to find $\sin t$. Then add those for your final answer.
There are other ways, but this way is straightforward. The answer is easy to check: use a calculator to find the "reference angle" $u$ for which $u=\tan^{-1}(-1/2)$, then find the angle $t$ in the second quadrant with the same tangent, then use your calculator to find $\sin t+\cos t$. This should agree numerically with your answer. Unless you have a CAS calculator it will not give you an exact answer, which is irrational.