diketahui sin A=3/5, tan B=5/12,

sinA = 3/5 = depan / miring, sehingga 'depan' = 3 dan 'miring' = 5

*samping = √ (miring^2 - depan^2) ; menggunakan rumus pytagoras
*samping = √ (5^2 - 3^2) = √(25 - 9) = √16 = 4, maka
cosA = 4/5

tanB = 5/12 = depan / samping. sehingga depan = 5, samping = 12 dengan pytagoras diperoleh miring = 13
sinB = 5/13
cosB = 12/13

a. sin(A+B) = sinA cosB + cosA sinB
sin(A+B = (3/5) (12/13) + (4/5) (5/13) = (36 + 20)/65 = 56 / 65

b. cos(A - B) = cosA cosB + sinA sinB
cos(A - B) = (4/5) (12/13) + (3/5) (5/13) = (48 + 15)/65 = 63 / 65