a maths...(20PT)

1) cosx - cos3x = sinx - sin3x

cosx - 4(cos^3)x + 3cosx = sinx - 3sinx + 4(sin^3)x
-4(cos^3)x + 4cosx = 4(sin^3)x - 2sinx
2(sin^3)x - sinx + 2(cos^3)x- 2cosx = 0
2(sin^3)x - sinx + 2(cos^3)x- 2cosx = 0
sinx(2(sin^2)x - 1) + 2cosx(cos^2x - 1) = 0
sinx(2(sin^2)x - 1) - 2cosx(sin^2)x = 0
-sinx[1-2(sin^2)x + 2cosxsinx] = 0
-sinx(cos2x + sin2x) = 0
so sinx = 0 or cos2x+sin2x = 0
x = 0 or 180 or x = 67.5 or 157.5 for 0<= x<= 180

2) cos(x+15)cos(x-15) - sinx sin(x-30) = 165 <~ 有d 奇怪
我只好做左手邊only

{cos[(x+15)-(x-15)]+cos[(x+15)+(x-15)]}/2 - {cos[x-(x-30)]-cos[x+(x -30)]}/2
= [cos 30 + cos 2x - cos 30 + cos (2x - 30)]/2
= [cos 2x + cos(2x - 30)]/2
= cos[(2x+2x-30)/2]cos[(2x-2x+30)/2]
= cos (2x - 15)(cos 15)

it is impossible to be = 165!

prove the following identities.

1) sinA + sinB + sin(A + B) = 4cos A/2 cos B/2 sin(A+B)/2

LHS = sinA + sinB + sin(A + B)
= 2sin[(A+B)/2]cos[(A-B)/2] + sin(A+B)
= 2sin[(A+B)/2]cos[(A-B)/2] + sin[2(A+B)/2]
= 2sin[(A+B)/2]cos[(A-B)/2] + 2sin[(A+B)/2]cos[(A+B)/2]
= 2sin[(A+B)/2]{cos[(A-B)/2 + cos[(A+B)/2]}
= 2sin[(A+B)/2]{2cos[(A-B)/4 + (A+B)/4]cos[(A-B)/4 - (A+B)/4]
= 2sin[(A+B)/2][2cos(2A/4)cos(-2B/4)]
= 4sin[(A+B)/2]cos(A/2)cos(B/2), as cos(-x) = cos x
= R.H.S.

希望幫到你