sin ( α ) = 3/8 and cos ( β ) = 6/7 Angle α terminates in Quadrant II, and Angle β terminates in Quadrant IV. What does sin ( α − β ) =? and What does cos ( α − β ) =?

Answer 1

Hi Brook,

when you have sin ( α ) = 3/8 it means you have cos(a) How?

sin (a ) = 3 / 8 = opp/ hyp --------> opp = 3 and hyp = 8 as you know hyp^2 = opp^2 + adj^2

64 = 9 + adj^2 both side minus 9

55 = adj^2 square root of both side√

±√55 = adj -----> a belong II quadrant so cos is negative as a result cos (a) = -√55/ 8

cos ( β ) = 6/7 = adj / hyp adj = 6 and hyp = 7

base of above information 49 = 36 + opp^2 -----> 13 = opp^2 ----> ±√13 = opp so sin(b) = opp/hyp = -√13/7

why -√13 ???? because sin at iv quadrant is negative

NOW sin ( a - b) = sina cosb - cosa sinb it is enough replace all needed information

sin ( a - b) = 3/8 * 6/7 - (-√55/8)(-√13/7) = ( 18 - √715)/56

cos (a - b) = cos a cos b + sin a sin b it is enough replace all needed information

cos (a - b) = (-√55/8) (6/7) + (3/8) (-√13/7) = ( - 6√55 - 3√13 ) / 56

I hope it is useful,

Minoo