prove that cos(6x)=32cos^6(x)-48cos^4(x)+18cos^2(x) -1
prove that cos(6x)=32cos^6(x)-48cos^4(x)+18cos^2(x) -1
Keshava Naidu , 10 Years ago
Grade 12th pass
Trigonometry> prove that cos(6x)=32cos^6(x)-48cos^4(x)+...
4 Answers
noogler
cos6x=cos[2(3x)]=2cos2(3x)-1 cos2x=2cos2x-1
=2[cos(3x)]2-1 cos3x=4cos3x-3cosx
=2[4cos3x-3cosx]2-1 (a-b)2=a2+b2-2ab
=2[16cos6x+9cos2x-24cos4x]-1
=32cos6x-48cos4x+18cos2x-1
Aman
cos6x=cos[2(3x)]=2cos2(3x)-1 cos2x=2cos2x-1 =2[cos(3x)]2-1 cos3x=4cos3x-3cosx =2[4cos3x-3cosx]2-1 (a-b)2=a2+b2-2ab =2[16cos6x+9cos2x-24cos4x]-1 =32cos6x-48cos4x+18cos2x-1
kirti
R.H.S. ==> cos(6x) = cos 2(3x) = 2cos2(3x) – 1
L.H.S. ==> 32cos6x – 48cos4x + 18cos2x – 1
=> 2(16cos6x – 24cos4x + 9cos2x) – 1
=> 2(4cos3x – 3cosx)2 – 1
=> 2cos2(3x) – 1
L.H.S. = R.H.S.
Hence, Proved
rishiraj
cos6x=cos3(2x) {cos3a=4cos3a-3cosa) (2x)=a
4cos3(2x)-3cos(2x)
4(cos2x)2-3cos(2x)
4(cos2x)3-3 6cos2x-1
[4(2cos2x-1)3-6cos2x+1]
[4(2cos2x-1)3-6cos2x+3]
4[8cos6x-2+3×2cos2x-3×4cos4x-6cos2x+3
32cos6x-4+24cos2x-48cos4x-6cos2x+3
32cos2x-48cos4x+18cos2x-1
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