Diketahui sin A=1/2 dan sin B=1/3 dengan sudut A dan sudut

sin(A+B) = (2√2 + √3)/6, cos(A-B) = (2√6 + 1)/6, tan(A+B) = (9√3 + 8√2)/23

Pembahasan

Diketahui sin A = 1/2 dan sin B = 1/3, dengan sudut A dan B lancip. Kita perlu mencari nilai cos A, cos B, tan A, dan tan B terlebih dahulu.

Untuk sudut A:
Karena sin A = 1/2 dan A lancip, maka A = 30 derajat. 
cos A = sqrt(1 - sin^2 A) = sqrt(1 - (1/2)^2) = sqrt(1 - 1/4) = sqrt(3/4) = sqrt(3)/2.
tan A = sin A / cos A = (1/2) / (sqrt(3)/2) = 1/sqrt(3) = sqrt(3)/3.

Untuk sudut B:
cos B = sqrt(1 - sin^2 B) = sqrt(1 - (1/3)^2) = sqrt(1 - 1/9) = sqrt(8/9) = 2*sqrt(2)/3.
tan B = sin B / cos B = (1/3) / (2*sqrt(2)/3) = 1 / (2*sqrt(2)) = sqrt(2)/4.

a. sin(A + B) = sin A cos B + cos A sin B
= (1/2) * (2*sqrt(2)/3) + (sqrt(3)/2) * (1/3)
= 2*sqrt(2)/6 + sqrt(3)/6
= (2*sqrt(2) + sqrt(3))/6

b. cos(A - B) = cos A cos B + sin A sin B
= (sqrt(3)/2) * (2*sqrt(2)/3) + (1/2) * (1/3)
= 2*sqrt(6)/6 + 1/6
= (2*sqrt(6) + 1)/6

c. tan(A + B) = (tan A + tan B) / (1 - tan A tan B)
= (sqrt(3)/3 + sqrt(2)/4) / (1 - (sqrt(3)/3) * (sqrt(2)/4))
= ((4*sqrt(3) + 3*sqrt(2))/12) / (1 - sqrt(6)/12)
= ((4*sqrt(3) + 3*sqrt(2))/12) / ((12 - sqrt(6))/12)
= (4*sqrt(3) + 3*sqrt(2)) / (12 - sqrt(6))
Untuk merasionalkan penyebutnya:
= ((4*sqrt(3) + 3*sqrt(2)) * (12 + sqrt(6))) / ((12 - sqrt(6)) * (12 + sqrt(6)))
= (48*sqrt(3) + 4*sqrt(18) + 36*sqrt(2) + 3*sqrt(12)) / (144 - 6)
= (48*sqrt(3) + 12*sqrt(2) + 36*sqrt(2) + 6*sqrt(3)) / 138
= (54*sqrt(3) + 48*sqrt(2)) / 138
= (9*sqrt(3) + 8*sqrt(2)) / 23