d) cos(x + y) cos(x − y) = cos^2x + cos^2y − 1
We can start by using the trigonometric identity
cos(A + B)cos(A − B) = (cos^2A − sin^2B).
Applying this to the given equation, we have:
cos(x + y)cos(x − y) = cos^2(x) − sin^2(y).
Using the identity cos^2(x) = 1 − sin^2(x), we can simplify further:
cos^2(x) − sin^2(y) = (1 − sin^2(x)) − sin^2(y)
= 1 − sin^2(x) − sin^2(y)
= 1 − (sin^2(x) + sin^2(y))
= 1 − sin^2(x + y).
Now, using the identity sin^2(A) + cos^2(A) = 1, we can rewrite the equation as:
1 − sin^2(x + y) = 1 − cos^2(x + y).
Since both sides of the equation are equal to 1 − cos^2(x + y), we have:
1 − sin^2(x + y) = 1 − cos^2(x + y)
cos^2(x + y) = sin^2(x + y).
Finally, using the identity sin^2(A) = 1 − cos^2(A), we have:
cos^2(x + y) = 1 − cos^2(x + y).
Rearranging the terms, we get:
cos^2(x + y) + cos^2(x + y) = 1.
Simplifying further, we have:
2cos^2(x + y) = 1.
Dividing both sides by 2, we obtain:
cos^2(x + y) = 1/2.
Taking the square root of both sides, we have:
cos(x + y) = ±√(1/2).
Therefore, the given equation is not equal to cos^2(x) + cos^2(y) - 1.
There are always many ways to prove trig identities, but this one seems a bit more direct:
cos(x + y) cos(x − y)
= (cosx cosy - sinx siny)(cosx cosy + sinx siny)
= 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 + cos^2y - cos^2x cos^2y
= cos^2x + cos^2y - 1