If cos^4(3x)-sin^4(3x)=cos(g(x)) then
g(x)=?
Use a^4-b^4=(a^2-b^2)(a^2+b^2)!
thanks in advance for any help!
cos^4(3x)-sin^4(3x)
= (cos^2 (3x) + sin^2 (3x)(cos^2 (3x ) - sin^2 (3x))
= (1)(cos (6x)
= cos 6x)
Besides using the difference of squares in the second line, I used:
sin^2 Ø + cos^2 Ø = 1 , and
cos 2Ø = cos^2 Ø - sin^2 Ø
Now I see how it works. Thanks for the help!
To find the value of g(x), we can use the given equation and the identity a^4 - b^4 = (a^2 - b^2)(a^2 + b^2).
In this case, a = cos^2(3x) and b = sin^2(3x).
Using the identity, we can rewrite the equation as:
cos^4(3x) - sin^4(3x) = cos^2(3x) - sin^2(3x) * (cos^2(3x) + sin^2(3x))
Simplifying further:
cos^4(3x) - sin^4(3x) = cos^2(3x) - sin^2(3x) * 1
cos^4(3x) - sin^4(3x) = cos^2(3x) - sin^2(3x)
Now, comparing this with the given equation cos(g(x)), we can equate:
cos^2(3x) - sin^2(3x) = cos(g(x))
Since cos^2(θ) - sin^2(θ) can be written as cos(2θ), we have:
cos(2 * 3x) = cos(g(x))
Therefore, g(x) = 2 * 3x
Simplifying, we get:
g(x) = 6x
So, g(x) = 6x is the value we found using the given equation and the identity.