1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $\sin(t) \in [0, 1]$

Does this help you along the way?

$\endgroup$
1
$\begingroup$

HINTS:

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.