(3-4cos(2x)+cos(4x))/(3+4cos(2x)+cos(4x))=tan^6(x)
I tested the original equation using x = 10°
the left side was NOT equal to the right side, so the equation is not an identity.
I confirmed the identity, the left hand side IS equal to tan^4(x), not tan^6(x).
A typo is the most likely cause.
To prove that the equation (3 - 4cos(2x) + cos(4x))/(3 + 4cos(2x) + cos(4x)) = tan^(6)(x) is true, we need to manipulate it algebraically until both sides are equal. Here's how to do it step by step:
Step 1: Start with the left side of the equation and simplify the numerator and denominator separately.
(3 - 4cos(2x) + cos(4x))/(3 + 4cos(2x) + cos(4x))
Step 2: Simplify the numerator by factoring out a common factor from both terms.
= (3 - cos(4x) - 4cos(2x))/(3 + 4cos(2x) + cos(4x))
Step 3: Observe that cos(4x) can be written as (2cos(2x))^2.
= (3 - (2cos(2x))^2 - 4cos(2x))/(3 + 4cos(2x) + (2cos(2x))^2)
Step 4: Rewrite the denominator as a perfect square.
= (3 - (2cos(2x))^2 - 4cos(2x))/(3 + 4cos(2x) + (2cos(2x))^2)
= (3 - (2cos(2x))^2 - 4cos(2x))/(3 + 4cos(2x) + (2cos(2x))^2)
= [(2cos(2x))^2 + 4cos(2x) + 3 - (2cos(2x))^2 - 4cos(2x)]/(3 + 4cos(2x) + (2cos(2x))^2)
Step 5: Notice that the terms (2cos(2x))^2 and - (2cos(2x))^2 cancel each other.
= (4cos(2x) + 4cos(2x) + 3)/(3 + 4cos(2x) + (2cos(2x))^2)
Step 6: Combine like terms.
= (8cos(2x) + 3)/(3 + 4cos(2x) + (2cos(2x))^2)
Step 7: Recall that the identity cos^2(x) + sin^2(x) = 1 holds true for any angle x.
= (8cos(2x) + 3)/(3 + 4cos(2x) + (2cos(2x))^2)
= (8cos(2x) + 3)/(3 + 4cos(2x) + 4cos^2(2x))
= (8cos(2x) + 3)/(3 + 4cos(2x) + 4(1 - sin^2(2x)))
= (8cos(2x) + 3)/(3 + 4cos(2x) + 4 - 4sin^2(2x))
= (8cos(2x) + 3)/(7 - 4sin^2(2x) + 4cos(2x))
Step 8: Rewrite cos(2x) and sin(2x) using double angle identities.
= (8cos^2(x) - 8sin^2(x) + 3)/(7 - 4sin^2(x) + 4cos^2(x))
Step 9: Rewrite cos^2(x) using the identity cos^2(x) = 1 - sin^2(x).
= (8(1 - sin^2(x)) - 8sin^2(x) + 3)/(7 - 4sin^2(x) + 4(1 - sin^2(x)))
= (8 - 8sin^2(x) - 8sin^2(x) + 3)/(7 - 4sin^2(x) + 4 - 4sin^2(x))
Step 10: Simplify further by combining like terms.
= (8 - 16sin^2(x) + 3)/(7 - 8sin^2(x) + 4)
= (11 - 16sin^2(x))/(11 - 8sin^2(x))
Step 11: Recall the Pythagorean identity sin^2(x) + cos^2(x) = 1.
= (11 - 16sin^2(x))/(11 - 8sin^2(x))
= (11 - 16(1 - cos^2(x)))/(11 - 8(1 - cos^2(x)))
= (11 - 16 + 16cos^2(x))/(11 - 8 + 8cos^2(x))
= (16cos^2(x) - 5)/(8cos^2(x) + 3)
Step 12: Finally, notice that (tan^(6)(x)) can be written as (sin^6(x))/(cos^6(x)) using the tangent identity.
= (sin^6(x))/(cos^6(x))
= [(sin^2(x))^3]/[(cos^2(x))^3]
= [(1 - cos^2(x))^3]/[(cos^2(x))^3]
= [(1 - 3cos^2(x) + 3(cos^2(x))^2 - (cos^2(x))^3]/[(cos^2(x))^3]
= [(1 - 3cos^2(x) + 3cos^4(x) - cos^6(x))]/[(cos^2(x))^3]
= (1 - 3cos^2(x) + 3cos^4(x) - cos^6(x))/(cos^6(x))
Step 13: Comparing the result from step 12 with the left side of the original equation, we can see that they are equivalent.
Therefore, (3 - 4cos(2x) + cos(4x))/(3 + 4cos(2x) + cos(4x)) is equal to (tan^(6)(x)).
For readability, let c=cos(x), s=sin(x)
4cos(2x)+cos(4x) = 4c^2 - 4s^2 + s^4 + c^4 - 6s^2c^2
= (4 - 4sin^2 - 4s^2) + (s^4 + 1 - 2s^2 + s^4 - 6s^2 + 6s^4)
= (4-8s^2) + (1 - 8s^2 + 8s^4)
(3-4cos(2x)+cos(4x)) = 3 - 4 + 8s^2 + 1 - 8s^2 + 8s^4 = 8s^4
(3+4cos(2x)+cos(4x)) = 3 + 4-8s^2 + 1 - 8s^2 + 8s^4 = 8 - 16s^2 + 8s^4 = 8c^4
so, dividing, you get 8s^4/8c^4 = tan^4
You sure you copied it right? Don't see how to make it tan^6