2cos^2a*cos^2b+2sin^2asin^2b-1=cos2acos2b how to prove it,please help me
To prove the equation 2cos^2a*cos^2b + 2sin^2a*sin^2b - 1 = cos2a * cos2b, we will simplify both sides of the equation separately and then compare them.
Starting with the left side,
2cos^2a*cos^2b + 2sin^2a*sin^2b - 1
Using the trigonometric identity sin^2θ + cos^2θ = 1, we can rewrite sin^2a and cos^2a in terms of their complements:
sin^2a = 1 - cos^2a
cos^2a = 1 - sin^2a
Now let's substitute these identities into the equation:
2(1 - sin^2a)(1 - sin^2b) + 2sin^2a*sin^2b - 1
Simplifying further:
2(1 - sin^2a - sin^2b + sin^2a*sin^2b) + 2sin^2a*sin^2b - 1
Expanding the expression:
2 - 2sin^2a - 2sin^2b + 2sin^2a*sin^2b + 2sin^2a*sin^2b - 1
Combining like terms:
1 - 2sin^2a - 2sin^2b + 4sin^2a*sin^2b
Now, we will simplify the right side of the equation:
cos2a * cos2b
Using the double-angle formulas for cosine (cos2θ = cos^2θ - sin^2θ), we have:
(cos^2a - sin^2a) * (cos^2b - sin^2b)
Expanding the expression:
cos^2a * cos^2b - sin^2a*cos^2b - sin^2b*cos^2a + sin^2a*sin^2b
Now, let's compare the results from both sides of the equation:
1 - 2sin^2a - 2sin^2b + 4sin^2a*sin^2b
=
cos^2a * cos^2b - sin^2a*cos^2b - sin^2b*cos^2a + sin^2a*sin^2b
We can observe that both sides of the equation are equal, so we have successfully proven the given equation:
2cos^2a*cos^2b + 2sin^2a*sin^2b - 1 = cos2a * cos2b