Given sinx=3/5 and siny= 2/3 , where x and y are both in the first quadrant. Evaluate sin(x+y)

With both x and y in the first quadrant, we have each of sin(x), cos(x), sin(y), and cos(y) being positive.

Then from the Pythagorean identity, it follows that

cos(x) = √(1 - sin²(x)) = 4/5

cos(y) = √(1 - sin²(y)) = √5/3

Using the angle sum identity for sine,

sin(x + y) = sin(x) cos(y) + cos(x) sin(y)

… = (3/5) • (√5/3) + (4/5) • (2/3)

… = √5/5 + 8/15

… = (3√5)/15 + 8/15

… = (3√5 + 8)/15

hello from mrbilldoesmath!

answer:   x (x+6)^2, choice b

discussion:

factor x out of the equation:

x^3 + 12x^2 + 36x = x (x^2 + 12x + 36).

note that 6+6 = 12 and 6*6 = 36 so the above factors into:

x (x+6)^2

which is choice b

you,

mrb