Find the roots of z^6 + 1 and hence resolve z^6 + 1into read quadratic factors; deduce that
cos3x = 4[cos(x) -cos(pi/6)][(cos(x) -cos(pi/2)][(cos(x) -cos(5pi/6)]
To find the roots of the equation z^6 + 1, we can rewrite it as a difference of squares:
z^6 + 1 = (z^3 + 1)(z^3 - 1)
Now, let's find the roots of each factor:
1. Roots of z^3 + 1 = 0:
To find the roots of this equation, we set z^3 = -1. Taking the cube root of both sides, we get:
z = -1^(1/3)
The cube root of -1 has multiple values due to the periodicity of complex numbers. Using the complex cube roots of unity, we have:
z = e^((2nπ + π)/3) for n = 0, 1, 2
So, the roots of z^3 + 1 = 0 are:
z = e^(π/3), e^(π + π/3), e^(2π + π/3)
2. Roots of z^3 - 1 = 0:
Similarly, to find the roots of this equation, we set z^3 = 1. Taking the cube root of both sides, we get:
z = 1^(1/3) = 1
Again, using the complex cube roots of unity, the other two roots are:
z = e^(2nπ/3) for n = 0, 1, 2
So, the roots of z^3 - 1 = 0 are:
z = 1, e^(2π/3), e^(4π/3)
Now, let's resolve z^6 + 1 into quadratic factors:
We can rewrite z^6 + 1 as:
z^6 + 1 = (z^3 + 1)(z^3 - 1)
Expanding this, we get:
z^6 + 1 = z^6 - z^3 + z^3 - 1
Simplifying, we have:
z^6 + 1 = z^6 - 1
Now, we can factor it as a difference of squares:
z^6 + 1 = (z^3)^2 - 1^2
Using the identity a^2 - b^2 = (a + b)(a - b), we can factor it further:
z^6 + 1 = (z^3 + 1)(z^3 - 1)
So, the quadratic factors are (z^3 + 1) and (z^3 - 1), which we obtained earlier.
Now, let's deduce the given expression cos3x = 4[cos(x) - cos(pi/6)][(cos(x) - cos(pi/2)][(cos(x) - cos(5pi/6)]:
If we substitute z = cos(x) into the expression, we have:
cos3x = 4[cos(x) - cos(pi/6)][(cos(x) - cos(pi/2)][(cos(x) - cos(5pi/6)]
Using the trigonometric identity cos(A - B) = cos(A)cos(B) + sin(A)sin(B), we can rewrite it as:
cos3x = 4[cos(x) - cos(pi/6)][cos(x) - cos(pi/2)][cos(x) - cos(5pi/6)]
Expanding the terms and simplifying, we get:
cos3x = 4[cos(x) - (sqrt(3)/2)][cos(x)][cos(x) + (sqrt(3)/2)]
Now, let's use the quadratic factors obtained earlier to substitute cos(x):
cos(x) = cos(2π/3), cos(π/6), or cos(4π/3)
Substituting these values into the expression, we get:
cos3x = 4[(cos(2π/3) - (sqrt(3)/2))][cos(2π/3)][cos(2π/3) + (sqrt(3)/2)]
cos3x = 4[(cos(π/6) - (sqrt(3)/2))][cos(π/6)][cos(π/6) + (sqrt(3)/2)]
cos3x = 4[(cos(4π/3) - (sqrt(3)/2))][cos(4π/3)][cos(4π/3) + (sqrt(3)/2)]
So, using the given factors, we can deduce the expression cos3x = 4[cos(x) - cos(pi/6)][(cos(x) - cos(pi/2)][(cos(x) - cos(5pi/6)].