Suppose 𝐴 is in Quadrant I and 𝐵 is in Quadrant IV. If cos(𝐴)=16/65 and sin(𝐵)=−40/41, then what is sin(𝐴+𝐵) and sin(A-B)

Answer 1

Use the sum and difference sine formulas

that gives an exact answer

as a check on those exact answers

use a calculator with an inverse trig function

sin(A+B) = -73/2665 = about -0.03

sin(A-B) = 1207/2665 = about 0.45

cosA = 16/65 A=about 75.75 degrees, using an inverse cosine function

sinB =-40/41 B= about -77.32 degrees using an inverse sine function

A+B = 75.75+ (-77.32) = -1.63 degrees

A-B = 75.7 - (-77.32) = 153.07 degrees

sin(A+B) = sin(-1.63) = -0.03

sin(A-B) sin153.07 = 0.45

cosA = 16/65 = adjacent side over hypotenuse

opposite side squared = hypotenuse squared minus adjacent side squared

= 65^2 - 16^2 = 4225-256 = 3969

opposite side = sqr3969 = 63

sinA = 63/65

sinB =-40/41

adjacent side squared = 41^2 -40^2 = 1681-1600 = 81

adjacent side = sqr81 = 9

cosB = 9/41

sin(A+B) = sinAcosB + sinBcosA = 63/65(9/41) - 40/41(16/65) = ((63)(9) - (40)16))/(65)(41) = -73/2665= about -0.03

sin(A-B) = sinAcosB - sinBcosA = (63/65)(9/41) - (-40/41)(16/65) = 1207/2665 =about 0.45