If sin A = 2/3 and sin B = 1/3 , and A and B are in quadrant I, find sin (A + B).

Question

I need solutions and the steps. I cannot understand it even after researching for hours.

Answer 1

sinA = 2/3 cosA= sqr(9-4)/3= sqr5/3

sinB = 1/3 cosB = sqr(9-1)/3 = sqr8/3 = 2sqr2/3

sine is opposite side over hypotenuse, 3 is the hypotenuse

cosine is adjacent side over hypotenuse, and the square root of the hypotenuse squared minus opposite side squared

plug above values into the general formula for sine of sum of 2 angles

sin(A+B) = sinAcosB + sinBcosA

sin(A+B) = (2/3)(2sqr2)/3 + (1/3)sqr5/3 = (4/3)sqr2/3 + (1/3)sqr5/3 = about (4/3)(1.4)/3 + 2.2/3/3 = about 2.4/3= 0.8

that means A+B forms a right triangle with ratio of sides 3-4-5

an alternative method is just use a calculator with an inverse trig function and find A and B, add them, then use the calculator again to find the sine.

whatever you get, you know it has to be less than 1. cosine of anything is always less than or equal to 1 and greater than or equal to negative 1.