Asked by Greg
Prove or disprove cos(x+y)cos(x-y)=cos squared (x) - Sin squared (x)
I dstributed the cosines and attempted to cancel out terms but I can't get the signs right. Any help on what I am missing?
I dstributed the cosines and attempted to cancel out terms but I can't get the signs right. Any help on what I am missing?
Answers
Answered by
Reiny
LS
= [ cosxcosy - sinxsinx] [cosxcosy + sinxsiny]
= cos^2 x sin^2 x - sin^2 x sin^2 y
= (cosxcosy)^2 - (sinxsiny)^2
RS = cos^2 x - sin^2x = cos 2x
doesn't look like they are equal
test with values:
let x = 60, y=45
LS = cos(105) cos(15 = -0.25
RS = cos^2 60 - sin^2 60 = +.25 ≠ LS
or
let x = 81, y= 50
LS = cos131 cos31 = -.5623..
RS = cos^2 81 - sin^2 81 = -.95.. ≠ LS
not an identity
= [ cosxcosy - sinxsinx] [cosxcosy + sinxsiny]
= cos^2 x sin^2 x - sin^2 x sin^2 y
= (cosxcosy)^2 - (sinxsiny)^2
RS = cos^2 x - sin^2x = cos 2x
doesn't look like they are equal
test with values:
let x = 60, y=45
LS = cos(105) cos(15 = -0.25
RS = cos^2 60 - sin^2 60 = +.25 ≠ LS
or
let x = 81, y= 50
LS = cos131 cos31 = -.5623..
RS = cos^2 81 - sin^2 81 = -.95.. ≠ LS
not an identity
Answered by
Greg
Sorry but I meant to say LS = cos^2x-sin^2y
Answered by
Steve
cos(x+y)cos(x-y) = cos^2 x - sin^2 y
(cosx cosy - sinx siny)(cosx cosy + sinx siny)
(cosx cosy)^2 - (sinx siny)^2
cos^2 x cos^2 y - sin^2 x sin^2 y
cos^2 x (1-sin^2 y) - (1-cos^2 x) sin^2 y
cos^2x - cos^2x sin^2y - sin^2y + cos^2x sin^2y
cos^2x - sin^2y
(cosx cosy - sinx siny)(cosx cosy + sinx siny)
(cosx cosy)^2 - (sinx siny)^2
cos^2 x cos^2 y - sin^2 x sin^2 y
cos^2 x (1-sin^2 y) - (1-cos^2 x) sin^2 y
cos^2x - cos^2x sin^2y - sin^2y + cos^2x sin^2y
cos^2x - sin^2y
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