Here are some formula:
sinAcosB = (1/2)[sin(A+B)+sin(A-B)]
cosAsinB = (1/2)[sin(A+B)-sin(A-B)]
cosAcosB = (1/2)[cos(A+B)+cos(A-B)]
sinAsinB = (-1/2)[cos(A+B)-cos(A-B)]
sinA + sinB = 2 sin [(A+B)/2] cos [(A-B)/2]
sinA - sinB = 2 cos [(A+B)/2] sin [(A-B)/2]
cosA + cosB = 2 cos [(A+B)/2] cos [(A-B)/2]
cosA - cosB = -2 sin [(A+B)/2] sin [(A-B)/2]
更新1:
sin(A+B) = sinAcosB + cosAsinB sin(A - B) = sinAcosB - cosAsinB cos(A+B) = cosAcosB - sinAsinB cos(A-B) = cosAcosB + sinAsinB tan(A+B) = (tanA+tanB)/(1-tanAtanB) tan(A-B) = (tanA - tanB)/(1+tanAtanB)
更新2:
I think gladysbwlo1116 had given the best answer. His answer is simply as well as clear. For 納粹戰爭主義革命軍元首,I am sorry that I think you are better not to use the general solution for the sine function. You made it become more complex. Thanks for your answers.
更新3:
Brian Lui： 如果係你答先 我一定揀你o個個做best answer. 雖然amaths書ge答案都係in radian,不過我認為冇緊要gei...