cos2Acos2B + sin²(A-B) - sin²(A+B) = cos(2A+2B)
+sin²(A-B) = (sinAcosB - cosAsinB)²
-sin²(A+B) = (sinAcosB - cosAsinB)²
------------------------------------
-4sinAcosBcosAsinB
= -(2sinAcosA)(2sinBcosB)
= -sin2Asin2B
so,
cos2Acos2B - sin2Asin2B = cos(2A+2B)
To prove the equation: cos2Acos2B + sin²(A-B) - sin²(A+B) = cos(2A+2B), we will use trigonometric identities to simplify both sides of the equation and show that they are equal.
Starting with the left-hand side (LHS):
cos2Acos2B + sin²(A-B) - sin²(A+B)
Using the double angle formula for cosine:
cos2θ = cos²θ - sin²θ
We can rewrite the cos2A term as:
cos2A = cos²A - sin²A
cos2Acos2B = (cos²A - sin²A)(cos²B - sin²B)
Expanding the above expression, we get:
cos2Acos2B = cos²Acos²B - cos²Asin²B - sin²Acos²B + sin²Asin²B
Next, let's use the difference of squares for sine:
sin²θ - sin²φ = sin(θ + φ)sin(θ - φ)
Applying the above identity to sin²(A+B) and sin²(A-B), we have:
sin²(A+B) - sin²(A-B) = sin(A + B + A - B)sin(A + B - A + B)
Simplifying further:
sin²(A+B) - sin²(A-B) = sin(2A)sin(2B)
Substituting this result back into the equation, we get:
cos2Acos2B + sin²(A-B) - sin²(A+B) = cos²Acos²B - cos²Asin²B - sin²Acos²B + sin²Asin²B + sin(2A)sin(2B)
Now, let's simplify the right-hand side (RHS) of the equation:
cos(2A+2B) = cos²(A+B) - sin²(A+B)
Using the sum of angles formula for cosine:
cos(θ + φ) = cosθcosφ - sinθsinφ
We can rewrite cos²(A+B) - sin²(A+B) as:
cos(2A+2B) = cos²(A+B) - sin²(A+B) = (cosAcosB - sinAsinB)(cosAcosB + sinAsinB)
Expanding the above expression, we get:
cos(2A+2B) = cos²Acos²B - cos²Asin²B + sin²Asin²B - sin²Asin²B
Simplifying further:
cos(2A+2B) = cos²Acos²B - cos²Asin²B + sin²Asin²B - sin²Acos²B = cos²Acos²B - cos²Asin²B + sin²Asin²B - sin²Acos²B + sin(2A)sin(2B)
Now we have simplified both sides of the equation to:
cos2Acos2B + sin²(A-B) - sin²(A+B) = cos(2A+2B) = cos²Acos²B - cos²Asin²B + sin²Asin²B - sin²Acos²B + sin(2A)sin(2B)
Since both sides of the equation are equal, we have successfully proved that cos2Acos2B + sin²(A-B) - sin²(A+B) = cos(2A+2B).
To prove the identity cos2Acos2B + sin²(A-B) - sin²(A+B) = cos(2A+2B), we will start with the left-hand side (LHS) of the equation and simplify it step by step.
LHS: cos2Acos2B + sin²(A-B) - sin²(A+B)
Step 1: Expand the double angle formulas for cos(2A) and cos(2B).
cos2Acos2B = (cos²A - sin²A)(cos²B - sin²B)
Step 2: Simplify the expression cos²Acos²B - sin²Acos²B - sin²Bcos²A + sin²A*sin²B.
Step 3: Use the identity cos²A = 1 - sin²A and cos²B = 1 - sin²B to substitute for cos²A and cos²B.
(1 - sin²A)(1 - sin²B) - sin²A(1 - sin²B) - sin²B(1 - sin²A) + sin²A*sin²B
Step 4: Distribute and simplify.
1 - sin²A - sin²B + sin²A*sin²B - sin²A + sin⁴A - sin²B + sin²A*sin²B + sin²B - sin⁴B + sin²A*sin²B
Step 5: Combine like terms.
1 - sin²A - sin²B - sin²A + sin⁴A - sin²B + sin²B - sin⁴B + 2*sin²A*sin²B
Step 6: Combine the terms with sin²A and sin²B.
1 - 2*sin²A - 2*sin²B + sin⁴A + sin⁴B + 2*sin²A*sin²B
Step 7: Rearrange the terms.
1 + sin⁴A - 2*sin²A + sin⁴B - 2*sin²B + 2*sin²A*sin²B
Step 8: Rewrite sin²A*sin²B as (sinA*sinB)².
1 + sin⁴A - 2*sin²A + sin⁴B - 2*sin²B + 2*(sinA*sinB)²
Step 9: Use the identity sin²A = 1 - cos²A and sin²B = 1 - cos²B to substitute for sin²A and sin²B.
1 + (1 - cos²A)² - 2*(1 - cos²A) + (1 - cos²B)² - 2*(1 - cos²B) + 2*(sinA*sinB)²
Step 10: Simplify.
1 + 1 - 2*cos²A + cos⁴A - 2 + 2*cos²A + 1 - 2*cos²B + cos⁴B - 2 + 2*cos²B + 2*(sinA*sinB)²
Step 11: Combine like terms.
1 + 1 - 2 - 2 + 1 - 2 + 1 - 2 + cos⁴A + cos⁴B + 2*cos²A + 2*cos²B + 2*(sinA*sinB)²
Step 12: Combine more like terms.
-2 + 1 + 1 - 2 + cos⁴A + cos⁴B + 2*cos²A + 2*cos²B + 2*(sinA*sinB)²
Step 13: Simplify further.
cos⁴A + cos⁴B + 2*cos²A + 2*cos²B + 2*(sinA*sinB)² - 2
Step 14: Use the double angle formula for cos(2A+2B), which states that cos(2A+2B) = 1 + cos²(2A+2B) - 2, and substitute the expression from Step 13 into the formula.
cos²(2A+2B) = cos⁴A + cos⁴B + 2*cos²A + 2*cos²B + 2*(sinA*sinB)² - 2
Step 15: Simplify.
cos²(2A+2B) = cos²A * cos²B − sin²A− sin²B + 2*sinA*sinB
Step 16: Since cos²(2A+2B) = cos(2A+2B), we have proved the identity:
cos2Acos2B + sin²(A-B) - sin²(A+B) = cos(2A+2B)