Trig Identity Prove:
cos(x+y)cos(x-y)=cos^2(x)+cos^2(y)-1
Use the formulas for cos (x+y) and cos(x-y):
cos(x+y) = cosx cosy - sinx siny
cos(x-y) = cosx cosy + sinx siny
The product of the two is
cos^2x cos^2y - sin^2x sin^2y
= cos^2x cos^2y -(1-cos^2x)(1-cos^2y)
= cos^2x cos^2y -(1-cos^2x)(1-cos^2y)
You finish it off.
To prove the given trigonometric identity, we'll start with the left-hand side (LHS) and manipulate it to obtain the right-hand side (RHS).
LHS: cos(x+y)cos(x-y)
Now, let's use the trigonometric identity for the product of two cosines:
cos(A)cos(B) = (1/2)(cos(A+B) + cos(A-B))
By substituting A = (x+y) and B = (x-y), we get:
LHS = (1/2)(cos[(x+y)+(x-y)] + cos[(x+y)-(x-y)])
Simplifying further:
LHS = (1/2)(cos(2x) + cos(2y))
Using the identity for the cosine of a double angle:
cos(2θ) = cos^2(θ) - sin^2(θ)
We can express the LHS as:
LHS = (1/2)(2cos^2(x) - 1 + 2cos^2(y) - 1)
LHS = cos^2(x) + cos^2(y) - 1
This is the same as the RHS. Therefore, we have successfully proved the given trigonometric identity.