you need to sketch right-angled triangles to get the missing sides
given: sinx = -1/4, y = -1, r = 4
x^2 + y^2 = r^2
x^2 + 1 = 16
x = -√15 in III
So sinx = -1/4 , cosx = -√15/4 , tanx = 1/√15
given: cosy = -4/5, x = -4, r = 5
then y = -3
siny = -3/5, cosy = -4/5 , tany = 3/4
Now we have all the parts, let's just use the definitions of ....
sin(x+y) = sinxcosy + cosxsiny
= (-1/4)(-4/5)+ (-√15/4)(-3/5)
= (4 + 3√15)/20
cos(x-y) = cosxcosy + sinxsinx
= .....
you try this one
tan(x+y) = (tanx + tany)/(1 - tanxtany)
= (1/√15 + 3/4)/(1 - (1/√15)(3/4))
= ( (4+3√15)/(4√15) )/( (4√15 - 3)/(4√15))
= (4 + 3√15)/(4√15 - 3)
Find sin(x+y), cos(x-y), tan(x+y), and the quadrant of (x+y) if sinx= -1/4, cosy= -4/5, with x and y in quadrant 3.
1 answer