Find the exact value of sin(x+y) given a. tan x = -4/3 and x lies in quadrant II b. cos y = 2/3 and y lies in

find the exact value of sin(x+y)
given
a. tan x = -4/3 and x lies in
quadrant II
b. cos y = 2/3 and y lies in
quadrant I

By the sine addition formula:
sin(x + y) = sin(x)cos(y) + cos(x)sin(y).

So, in order to find sin(x + y), we need to find sin(x), sin(y), cos(x), and sin(y).

Since x is in Quadrant II and y is in Quadrant I, we have:
cos(x) < 0, sin(x) > 0, cos(y) > 0, and sin(y) > 0.

By drawing a 3-4-5 triangle in Quadrant II such that |tan(x)| = opp/adj = 4/3, we have:
adj = 3, opp = 4, and hyp = 5 ==> |cos(x)| = adj/hyp = 3/5 and |sin(x)| = opp/hyp = 4/5.

Since cos(x) < 0 and sin(x) > 0, we have:
cos(x) = -3/5 and sin(x) = 4/5.

In a similar fashion, drawing a 2-√5-3 triangle in Quadrant I such that cos(y) = adj/hyp = 2/3 gives:
adj = 2, opp = √5, and hyp = 3 ==> |sin(y)| = opp/hyp = √5/3.

Since sin(y) > 0, we have:
sin(y) = √5/3.

Therefore:
sin(x + y) = sin(x)cos(y) + cos(x)sin(y)
= (4/5)(2/3) + (-3/5)(√5/3)
= 8/15 - 3√5/15
= (8 - 3√5)/15.

I hope this helps!

8/15 - 1/√5