The following proof follows on from the trig functions of adding or subtracting two angles. It is found useful in some circumstances to express the sum of two trig functions as a product. This is used to derive differential of sin and cos.
sin(A+B) = sinAcosB + cosAsinB ....(1)
sin(A-B) = sinAcosB - cosAsinB ....(2)
(1)+(2)
2sinAcosB = sin(A+B) + sin(A-B) ....(3)
(1)-(2)
2cosAsinB = sin(A+B) - sin(A-B) ....(4)
Now
cos(A+B) = cosAcosB - sinAsinB ....(5)
cos(A-B) = cosAcosB + sinAsinB ....(6)
(5)+(6)
2cosAcosB = cos(A+B) + cos(A-B) ....(7)
(5)-(6)
-2sinAsinB = cos(A+B) - cos(A-B) ...(8)
By writing
A+B=C and A-B=D adding
2A = C+D or A=(1/2)(C+D)
A+B=C and A-B=D subtracting
2B = C-D or B=(1/2)(C-D)
so we can rewrite (3) to (6)
sinC + sinD = 2sin((1/2)(C+D))cos((1/2)(C-D))
sinC - sinD = 2cos((1/2)(C+D))sin((1/2)(C-D))
cosC + cosD = 2cos((1/2)(C+D))cos((1/2)(C-D))
cosC - cosD = -2sin((1/2)(C+D))sin((1/2)(C-D))