Monday
December 9, 2013

# Posts by Hayden

Total # Posts: 108

trig!!
sin^4 x (cos^2 x)=(1/16)(1-cos 2x)(1-cos4x) work one 1 side only!

TRIG!
Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6 x=1 - (3/4)sin^2 2x work on one side only! Responses Trig please help! - Reiny, Monday, February 23, 2009 at 4:27pm LS looks like the sum of cubes sin^6 x + cos^6 x = (sin^2x)^3 + (cos^2x)^3 = (sin^2x+cos...

trig help again i cant get any of these
sin^4cos^2x=(1/16)(1-cos 2x)(1-cos4x) work one 1 side only!

(sin^2x+cos^2x)(sin^4x - (sin^2x)(cos^2x) + sin^4x) I DONT AGREE WITH THIS SHOULDNT IT BE (sin^2x+cos^2x)(sin^4x - (sin^2x)(cos^2x) + COS^4 X ??

sin^6 x + cos^6 x=1 - (3/4)sin^2 2x work on one side only!

precalc and trig
i posted a sorry. i posted this twice on that one. And i dont understand how RS=LS and thanks

precalc and trig
cos^2 (a+x) -sin^2 a=cos x * cos(2a+x)

Precalc and trig
how does cos^2x(cos^2a – sin^2a) - 2sinacosasinx = cos^2x(cos^2a – sin^2a) – 2cosxcosasinxsina

Precalc and trig
thanks! i accidentally posted this twice

Precalc and trig
cos^2 (x+a)-sin^2(a)= (cos x)(cos (2a+x)) prove using only side i worked on the left side...

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