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There is a triangle in Quadrant II (where x= -3 and the hypoteneuse is 5) and a triangle in Quadrant III (where y=-1 and x=-2 in terms of any diagram. We know y is in Quad III because it is >180 but <270.
We can calculate the hypoteneuse (using pythag) in Quad III as we have the other 2 sides ( because `tan y=1/2)`
`r^2 = x^2 + y^2` and `therefore r^2 = (-2)^2 + (-1)^2`
To solve, also calculate the y value in Quad II. We know by calculation using Pythag `y^2= 5^2-(-3)^2`
`therefore y=sqrt16=4` and so `sin x = 4/5 and` `cos x=-3/5`
and we know `sin y = -1/sqrt5` and that `cos y=-2/sqrt5`
`cos(x+y) = cos x.cos y - sinx.siny`
`therefore =( -3)/5 times(-2)/sqrt5 - 4/5 times (-1)/sqrt5`
`therefore = 6/(5sqrt5) - (-4)/(5sqrt5)`
`therefore = 10/(5sqrt5)`
Now simplify = `2/sqrt5` and rationalize the denominator:
`therefore cos(x+y)= 2/sqrt5 times sqrt5/sqrt5`
`therefore = (2sqrt5)/5`