Prove cos2x-cos2y/cos2x+cos2y=tan(x+y)tan(y-x)
To prove the given equation:
cos2x - cos2y / cos2x + cos2y = tan(x+y)tan(y-x)
we will start by working on each side separately.
Left-hand side (LHS):
cos2x - cos2y / cos2x + cos2y
Using the identity cos2θ = cos²θ - sin²θ, we can rewrite the equation as:
(cos²x - sin²x) - (cos²y - sin²y) / (cos²x - sin²x) + (cos²y - sin²y)
Expanding the numerator and denominator, we get:
(cos²x - cos²y - sin²x + sin²y) / (cos²x - cos²y + sin²x - sin²y)
Rearranging terms, we have:
(cos²x - cos²y - sin²x + sin²y) / (cos²x + sin²x - cos²y - sin²y)
Using the identity sin²θ + cos²θ = 1, we can simplify further:
(1 - cos²x - 1 + cos²y) / (1 - cos²x + cos²y - 1)
Simplifying, we get:
(cos²y - cos²x) / (cos²y - cos²x)
Now, let's work on the right-hand side (RHS):
tan(x+y)tan(y-x)
Using the identity tan(A + B) = (tan A + tan B) / (1 - tan A tan B), we can rewrite the equation as:
(tan x + tan y) / (1 - tan x tan y) * (tan y - tan x)
Expanding, we get:
(tan x tan y + tan y² - tan x² - tan x tan y) / (1 - tan x tan y)
Simplifying, we have:
(tan y² - tan x²) / (1 - tan x tan y)
Now, we can see that both sides of the equation are equal:
(cos²y - cos²x) / (cos²y - cos²x) = (tan y² - tan x²) / (1 - tan x tan y)
Therefore, we have proven that cos2x - cos2y / cos2x + cos2y = tan(x+y)tan(y-x).
To prove the given identity:
cos(2x) - cos(2y) / cos(2x) + cos(2y) = tan(x+y) * tan(y-x)
We need to start with one side of the equation and manipulate it until it matches the other side. Let's start with the right-hand side (RHS) of the desired equation:
RHS = tan(x+y) * tan(y-x)
Now, let's express tan(x+y) and tan(y-x) in terms of sin and cos.
tan(x+y) = sin(x+y) / cos(x+y)
tan(y-x) = sin(y-x) / cos(y-x)
Multiplying these expressions together:
RHS = (sin(x+y) / cos(x+y)) * (sin(y-x) / cos(y-x))
= (sin(x)cos(y) + cos(x)sin(y)) / (cos(x)cos(y) - sin(x)sin(y))
Now, let's work on the left-hand side (LHS) of the equation:
LHS = cos(2x) - cos(2y) / cos(2x) + cos(2y)
Using the double-angle identity, we can express cos(2x) and cos(2y) in terms of sin and cos:
cos(2x) = cos^2(x) - sin^2(x)
cos(2y) = cos^2(y) - sin^2(y)
Substituting these values into the LHS:
LHS = (cos^2(x) - sin^2(x)) - (cos^2(y) - sin^2(y)) / (cos^2(x) - sin^2(x)) + (cos^2(y) - sin^2(y))
Now, let's simplify the numerator:
LHS = (cos^2(x) - sin^2(x) - cos^2(y) + sin^2(y)) / (cos^2(x) - sin^2(x) + cos^2(y) - sin^2(y))
Using the identity cos^2(x) - sin^2(x) = cos(2x) and cos^2(y) - sin^2(y) = cos(2y):
LHS = (cos(2x) - cos(2y)) / (cos(2x) + cos(2y))
Comparing this with the RHS, we see that the LHS matches the RHS. Thus, we have proven the given identity.