2006-09-27 15:13:48 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
cos 6x = cos 3*2x = 4 cos^3(2x) - 3 cos 2x(using cos(3x) = cos^3(x) - 3cos(x))
= 4((2cos^2 x-1)^3 - 2(2cos^x -1)
= 4(8cos^6x - 12 cos^4 x + 6 cos^2 x -1) -3(2cos^x -1)
= 32 cos^6x - 48 cos^4 x + 18 cos^2 x - 1
2006-09-30 04:18:33 · answer #1 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
for such problems use the following relations:
cos(a+b) = cosa.cosb - sina.sinb
sin(a+b)=sina.cosb+cosa.sinb
so cos(6x) can be written as = cos(3x+3x) .... then 3x can be split further into x+2x etc
2006-09-28 12:56:17 · answer #2 · answered by m s 3 · 0⤊ 0⤋
cos(6x) = cos(2*3x) = cos^2(3x) - sin^2(3x)
cos(3x) = 4cos^3(x) - 3cos(x)
sin(3x) = 3sin(x) - 4sin^3(x)
cos(6x) = [4cos^3(x) - 3cos(x)]^2 - [3sin(x) - 4sin^3(x)]^2
2006-09-27 15:30:42 · answer #3 · answered by gp4rts 7 · 0⤊ 0⤋
-32(sinx)^2(cosx)^4
-16(cosx)^4
+18(cosx)^2-1
2006-09-27 23:14:09 · answer #4 · answered by Mamad 3 · 0⤊ 0⤋