jika sin A = 6/10 tan B =9/40 . Tentukan cos (a-b) , cos (a+b) , sin (a-b) , tan (a-b) tolong beserta cara nya

sin A= depan/miring= 6/10
maka cos A= samping/ miring = x/10
memakai rumus phytagoras
x²=10²-6²= 100-36 = 64 maka x=8
cos A = 8/10
tan A= depan /samping = 6/8

tan B = depan / samping = 9/40
sin B = depan / miring = 9/x
x= √(9² + 40²) = √(81+1600) = 41
sin B = 9/41
cos B = samping / miring = 40/41

1. cos (a-b) = cos a cos b + sin a sin b = 8/10×40/41 + 6/10×9/41 = 207/205
2. cos (a+b) = cos a cos b - sin a sin b = 8/10×40/41 - 6/10×9/41= 163/205
3. sin (a-b) = sin a cos b - cos a sin b = 6/10×40/41 - 8/10×9/41 = 84/205
4. tan (a-b) = (tan a - tan b) / 1 +tan a tan b = (6/8 - 9/40)/ 1+ 6/8×9/40 = (21/40)/ 1+(27/160) = (21/40) / (160/160 + 27/160) =(21/40)/(187/160) = 84/187

angkanya besar2 hehe