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?
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
Sorry but I meant to say LS = cos^2x-sin^2y
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
To prove or disprove the equation cos(x+y)cos(x-y) = cos^2(x) - sin^2(x), we can start by using the trigonometric identity for the cosine of a sum of angles:
cos(x + y) = cos(x)cos(y) - sin(x)sin(y).
Applying this identity to the left-hand side of the equation, we have:
cos(x + y)cos(x - y) = (cos(x)cos(y) - sin(x)sin(y))cos(x - y).
Next, we can simplify further by applying the trigonometric identity for the cosine of a difference of angles:
cos(x - y) = cos(x)cos(y) + sin(x)sin(y).
Substituting this identity into the equation, we get:
(cos(x)cos(y) - sin(x)sin(y))(cos(x)cos(y) + sin(x)sin(y)).
Now, we can simplify this expression by applying the distributive property:
= (cos(x)cos(y))^2 - (sin(x)sin(y))^2.
Since (cos(x)cos(y))^2 = cos^2(x)cos^2(y) and (sin(x)sin(y))^2 = sin^2(x)sin^2(y), we can rewrite the expression as:
= cos^2(x)cos^2(y) - sin^2(x)sin^2(y).
At this point, we may notice that the expression does not match the right-hand side of the original equation, which is cos^2(x) - sin^2(x). Therefore, we have disproven the equation:
cos(x + y)cos(x - y) ≠ cos^2(x) - sin^2(x).
So, the original statement is false. When distributing the cosines and simplifying the expression, we end up with different terms than those on the right-hand side of the equation.