Prove this: 1sin4xcos4x1sin6xcos6x=23?

1sin4xcos4x1sin6xcos6x=23

2 Answers
Mar 16, 2017

LHS=1sin4xcos4x1sin6xcos6x

=1((sin2x)2+(cos2x)2)1((sin2x)3+(cos2x)3)

=1((sin2x+cos2x)22sin2cos2x)1((sin2x+cos2x)33sin2xcos2x(sin2x+cos2x))

=1(sin2x+cos2x)2+2sin2cos2x1(sin2x+cos2x)3+3sin2xcos2x(sin2x+cos2x)

=112+2sin2cos2x113+3sin2xcos2x

=2sin2cos2x3sin2xcos2x=23=RHS

Proved

In step 3 the following formulae are used

a2+b2=(a+b)22ab

and

a3+b3=(a+b)33ab(a+b)

Mar 16, 2017

Please see the explanation. I confirmed each step of this proof using www.WolframAlpha.com

Explanation:

Multiply both sides by 3(1sin6(x)cos6(x))

33sin4(x)3cos4(x)=22sin6(x)2cos6(x)

Substitute 3(1cos2(x))2 for 3sin4(x)

33(1cos2(x))23cos4(x)=22sin6(x)2cos6(x)

Multiply the square:

33(12cos2(x)+cos4(x))3cos4(x)=22sin6(x)2cos6(x)

Distribute the -3:

33+6cos2(x)3cos4(x)3cos4(x)=22sin6(x)2cos6(x)

Combine like terms:

6cos2(x)6cos4(x)=22sin6(x)2cos6(x)

Divide both sides by 2:

3cos2(x)3cos4(x)=1sin6(x)cos6(x)

Substitute (1cos2(x))3 for sin6(x)

3cos2(x)3cos4(x)=1(1cos2(x))3cos6(x)

Expand the cube:

3cos2(x)3cos4(x)=1(13cos2(x)+3cos4(x)cos6(x))cos6(x)

Distribute the -1:

3cos2(x)3cos4(x)=11+3cos2(x)3cos4(x)+cos6(x)cos6(x)

Combine like terms:

3cos2(x)3cos4(x)=3cos2(x)3cos4(x)

The right is identical to the left. Q.E.D.