identities.... i need help!

[candy]

Please show all the steps 
proving identities!!

 sin^2xcos^2x=1/8(1-cos4x)

[djammad]

solve the right hand side

[djammad]

i dont get the question write in words :S

[twilight]

i need help in this q too ...
how do u solve cos²2x

[twilight]

the q is
sin²x cos²x = 1/8(1- cos 4x)

[astarmathsandphysics]

Please show all the steps 
proving identities!!

 sin^2xcos^2x=1/8(1-cos4x)

 sin^2xcos^2x=(1/4)(2 sinxcosx)^2=1/4(sin2x)^2=

Now use cos4x=1-2sin^2(2x) so sin^2(2x)=1/2(1-cos4x)

[twilight]

how 2 solve sin^2(2x) ?

[astarmathsandphysics]

how 2 solve sin^2(2x) ?

you substitute this in

[twilight]

thx loads astar

[louis]

The above question can also be solved by the following method:

Remember to use the double angle formula for cos 2x

cos 2x = 2 cos² x  - 1                       cos ²x        =½ ( 1 +  cos 2x )
cos 2x = 1 - 2 sin²x                          sin² x         =½ ( 1  - cos 2x )

It follows that

cos 4x = cos 2 (2x)=  2 cos ²(2x)  - 1                       cos² (2x)       =½ ( 1 + cos 4x)
cos 4x = cos 2 (2x) = 1 - 2 sin² (2x)                         sin² (2x)        =½ ( 1 -  cos 4x )


sin² x     cos² x 

=½ ( 1 - cos 2x) * ½ ( 1 + cos 2x)                    From : (A-B)(A+B)=A² -  B²
=¼  ( 1 – cos² 2x)

=¼  { 1 - ½ ( 1+ cos 2(2x) ) }

=¼( 1 - ½ - ½ cos 4x )

=¼( ½  - ½ cos 4x)

=(¼) (½ )( 1 -  cos 4x)

=1/8 ( 1 -  cos 4x )

[candy]

thanxxx a ton louis and astar