2cos^2a*cos^2b+2sin^asin^b-1=cos2acos2b how to prove it
You seem to have omitted exponents in the 2sin^asin^b term
2cos^2a*cos^2b+2sin^2asin^2b-1=cos2acos2b you are right... if you know how to prove it please help me
To prove the given equation, we need to manipulate the left side of the equation until it becomes equal to the right side. Let's start with the left side:
2cos^2(a)cos^2(b) + 2sin^a sin^b - 1
We can simplify this further using trigonometric identities. First, let's apply the double angle formulas for cosine and sine:
cos(2a) = cos^2(a) - sin^2(a)
cos(2b) = cos^2(b) - sin^2(b)
sin(2a) = 2sin(a)cos(a)
sin(2b) = 2sin(b)cos(b)
Now, let's substitute these identities into the left side of the equation:
2(cos^2(a)cos^2(b)) + 2(sin^a sin^b) - 1
= 2(cos^2(a)cos^2(b) - sin^2(a)sin^2(b)) + 2sin(a)sin(b) - 1
= 2(cos^2(a)cos^2(b) - (1 - cos^2(a))(1 - cos^2(b))) + 2sin(a)sin(b) - 1
= 2(cos^2(a)cos^2(b) + cos^2(a)cos^2(b) - cos^2(a) - cos^2(b) + 1) + 2sin(a)sin(b) - 1
= 4cos^2(a)cos^2(b) - 2(cos^2(a) + cos^2(b)) + 2sin(a)sin(b) - 1
= cos(2a)cos(2b) + 2sin(a)sin(b) - 1
Finally, we arrive at:
2cos^2(a)cos^2(b) + 2sin^a sin^b - 1 = cos(2a)cos(2b) + 2sin(a)sin(b) - 1
Thus, we have proven the given equation.