If 180°<α<270°, cos ⁡α=−8/17, 270°<β<360°, and sin β=−4/5, what is sin⁡(α+β)?

Question

If 180°<α<270°, cos ⁡α=−8/17, 270°<β<360°, and sin β=−4/5, what is sin⁡(α+β)?

Answer 1

alpha in 3rd quadrant beta in 4th quadrant

SinB = -4/5 b = -53.13 use a calculator with inverse trig functions

what is sin(a+b)??

sin(a+b) = sinacosb + sinbcosa = sinacos53.13 + (-.8)cosa

= .6sina -.8cosa

180 < a < 270

sin180 < sina < sin270

0 < sina < -1 which is impossible a DNE does not exist, no such angle alpha

cos180 < cosa <cos270

-1< cosa < 0

.6sina -.8cosa

.6sina+.8cosa < .8sina -.8cosa < .6sina -0= .6sina