someone solve this trig identity prob for me too
I typed it in a hurry and made a little mistake, The first 3 lines are correct and then in the fourth line substitute 1-cos^2x for sin^2x.
Here is the correct solution.
first use cos(x+y)=cosxcosy-sinxsiny.
cos(x+y)cos(x-y)=(cosxcosy-sinxsiny)(cosxcosy+sinxsiny)
=(cosxcosy)^2-(sinxsiny)^2 ; (use (a+b)(a-b)=a^2-b^2)
=cos^2xcos^2y-sin^2xsin^2y
=cos^2x(1-sin^2y)-(1-cos^2x)sin^2y
=cos^2x-cos^2xsin^2y-sin^2y+cos^2xsin^2y
=cos^2x-sin^2y.
Here is the correct solution.
first use cos(x+y)=cosxcosy-sinxsiny.
cos(x+y)cos(x-y)=(cosxcosy-sinxsiny)(cosxcosy+sinxsiny)
=(cosxcosy)^2-(sinxsiny)^2 ; (use (a+b)(a-b)=a^2-b^2)
=cos^2xcos^2y-sin^2xsin^2y
=cos^2x(1-sin^2y)-(1-cos^2x)sin^2y
=cos^2x-cos^2xsin^2y-sin^2y+cos^2xsin^2y
=cos^2x-sin^2y.
Originally Posted by gaus,Apr 16 2006, 09:03 PM
first use cos(x+y)=cosxcosy-sinxsiny.
cos(x+y)cos(x-y)=(cosxcosy-sinxsiny)(cosxcosy+sinxsiny)
=(cosxcosy)^2-(sinxsiny)^2 ; (use (a+b)(a-b)=a^2-b^2)
=cos^2xcos^2y-sin^2xsin^2y
=cos^2x(1-sin^2y)-(1-cos^2x)sin^2y
=cos^2x-cos^2xsin^2y-sin^2y+cos^2xsin^2y
=cos^2x-sin^2y.
cos(x+y)cos(x-y)=(cosxcosy-sinxsiny)(cosxcosy+sinxsiny)
=(cosxcosy)^2-(sinxsiny)^2 ; (use (a+b)(a-b)=a^2-b^2)
=cos^2xcos^2y-sin^2xsin^2y
=cos^2x(1-sin^2y)-(1-cos^2x)sin^2y
=cos^2x-cos^2xsin^2y-sin^2y+cos^2xsin^2y
=cos^2x-sin^2y.
hell yeah it is.....its the off topic and their aint no homework section so there we go lol
i dono if anyone would ever need to use this but all i know is that its in the course that im required to take for architecture
i dono if anyone would ever need to use this but all i know is that its in the course that im required to take for architecture
Originally Posted by dupmaronew,Apr 17 2006, 07:33 AM
. . . all i know is that its in the course that im required to take for architecture
Originally Posted by dupmaronew,Apr 16 2006, 06:39 PM
i meant cos square times x
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-Richard Feynman
You can replace physics with mathematics. This should give you an idea why some
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