Let $A$ be in the fourth quadrant and let $\sec(A) = \dfrac{13}{5}$, Let $B$ be in the third quadrant and let $\csc(B)=\dfrac{-5}{3}$. Find $\sin(A + B)$ and determine in which quadrant $A + B$ terminates.

$\sec(A) = \dfrac{13}{5} \Rightarrow \cos(A) = \dfrac{5}{13}$
$\csc(B) = \dfrac{-5}{3} \Rightarrow \sin(B) = \dfrac{-3}{5}$
$\sin(A) = \dfrac{-12}{13}$ and $\cos(B) = \dfrac{-4}{5}$

$\sin(A + B) = \sin(A) \cos(B) + \sin(B) \cos(A)$

after performing the substitutions, I arrive at $\sin(A + B) = \dfrac{33}{65}$

$\cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B)$
= $\dfrac{5}{13}(\dfrac{-4}{5}) - \dfrac{-12}{13}(\dfrac{-3}{5}) = \dfrac{-56}{65}$

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    $\begingroup$ secant is the reciprocal of cosine. Cosecant is the reciprocal of sine $\endgroup$ – Will Jagy Jul 4 '18 at 23:51
  • $\begingroup$ Sorry, I just realized that a few moments ago. However, even with that mistake I guess my question is still valid. How do you determine in what quadrant a sum of angle lies in if the two angles are unknown? So I have to specifically solve for A and B? $\endgroup$ – Mutating Algorithm Jul 4 '18 at 23:59
  • $\begingroup$ fixed the mistake $\endgroup$ – Mutating Algorithm Jul 5 '18 at 0:03
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    $\begingroup$ if you find both $\cos(A+B)$ and $\sin(A+B)$ you will know the quadrant. It would be enough to know for sure what $\pm$ signs went with cosine, and which $\pm$ went with the sine. In your case, there is no difficulty finding the actual cosine and sine (for $A+B$) $\endgroup$ – Will Jagy Jul 5 '18 at 0:04
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    $\begingroup$ Should have a minus sign, $$ \cos(A+B) = \cos A \cos B - \sin A \sin B $$ $\endgroup$ – Will Jagy Jul 5 '18 at 0:40

Hint: Your calculations above are correct (I cleaned them up a little bit). If sine is positive but cosine is negative, what quadrant are you in? Does this give a hint to $A + B$?

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