Post by: candy on May 21, 2009, 10:06:28 pm
proving identities!!
sin^2xcos^2x=1/8(1-cos4x)
Post by: djammad on May 21, 2009, 10:13:47 pm
Post by: djammad on May 21, 2009, 10:15:01 pm
Post by: twilight on May 21, 2009, 10:16:00 pm
how do u solve cos22x
Post by: twilight on May 21, 2009, 10:16:52 pm
sin2x cos2x = 1/8(1- cos 4x)
Post by: astarmathsandphysics on May 21, 2009, 10:18:42 pm
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)
Post by: twilight on May 21, 2009, 10:20:16 pm
Post by: astarmathsandphysics on May 21, 2009, 10:22:07 pm
how 2 solve sin^2(2x) ?
you substitute this in
Post by: twilight on May 21, 2009, 10:33:39 pm
Post by: louis on May 22, 2009, 05:15:39 am
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 )
Post by: candy on May 22, 2009, 10:02:18 am