angles (a) and (B) are obtuse angles in quadrant 2 (II).If csc a= 3 and tan B= -1/3,determine the exact value for cos (a+b)
From the sum formula for cosine:
cos(a + B) = cos a cos B - sin a sin B
Since csc a = 3 :
sin a = 1/3
Since sin a = 1/3 and cos2 a = 1 - sin2 a :
cos2a = 8/9 And since a is in quadrant II,
cos a = -√8 / 3
We can again use the Pythagorean identity and the fact that B is in quadrant II to get:
sin B = 1/ √10
cos B = -3 / √10
And so...
cos(a + B) = (-√8 / 3)(-3 / √10) - (1/3)(1/ √10)
Plug this into a calculator or simpify it to get a final answer.....I don't have time to write a better explanation right now, but if you have any questions about this please ask!