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Algebraic Proof that

if f(x) = a sin(bx + c) and g(x) = a cos(bx + c),

then the PRODUCT of f and g is equivalent to

h(x) = a^2/2 cos(2(bx - (p/4 - c)))

(1) Work with f + g:

f(x) g(x) = a^2[(sin bx cos c + cos bx sin c)(cos bx cos c - sin bx sin c)]

= a^2[((cos c)^2 sin bx cos bx - (sin bx)^2 sin c cos c + (cos bx )^2 sin c cos c - (sin c)^2 sin bx cos bx)]

= a^2[((cos c)^2 - (sin c)^2)sin bx cos bx + ((cos bx )^2 - (sin bx)^2)sin c cos c]

= a^2(cos 2c sin bx cos bx + cos 2bx sin c cos c)


(2) Work with h:

h(x) = a^2/2 cos(2bx - 2(p/4 - c))

= a^2/2 [cos 2bx cos 2(p/4 - c) + sin 2bx sin 2(p/4 - c)]

= a^2/2 [cos 2bx cos(p/2 - 2c) + sin 2bx sin(p/2 - 2c)]

= a^2/2 [cos 2bx (cosp/2 cos 2c + sinp/2 sin 2c) + sin 2bx (sin p/2 cos 2c - cos p/2 sin 2c)]

= a^2/2 [cos 2bx sin 2c + sin 2bx cos 2c]

= a^2/2 [cos 2bx 2sin c cos c + 2sin bx cos bx cos 2c]

= (a^2/2)2 [cos 2bx sin c cos c + cos 2c sin bx cos bx]

= a^2 (cos 2c sin bx cos bx + cos 2bx sin c cos c) = f(x)g(x) above. QED.