Prove that cos6x= 32cos⁶x-48cos⁴x+18 cos²x-1

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Solution

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

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Similar questions

cos 6 x = 32 {cos}^{6} x - 48 {cos}^{4} x + 18 {cos}^{2} x - 1

c o s 6 x = 32 c o s^{6} x - 48 c o s^{4} x + 18 c o s^{2} x - 1

cos 6x = 32 cos⁶ x – 48 cos⁴ x + 18 cos² x – 1

cos 6 x = 32 {cos}^{6} x - 48 {cos}^{4} x + 18 {cos}^{2} x - 1

cos 6 x = 32 {cos}^{6} x - 48 {cos}^{4} x + 18 {cos}^{2} x - m

m

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