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Camille Jordan showed [2] that any algebraic equation may be solved by use of modular functions. This was accomplished by Thomae in 1870. [ 3 ] Thomae generalized Hermite's approach by replacing the elliptic modular function with even more general Siegel modular forms and the elliptic integral by a hyperelliptic integral .
A formula for computing the trigonometric identities for the one-third angle exists, but it requires finding the zeroes of the cubic equation 4x 3 − 3x + d = 0, where is the value of the cosine function at the one-third angle and d is the known value of the cosine function at
A calculation confirms that z(0) = 1, and z is a constant so z = 1 for all x, so the Pythagorean identity is established. A similar proof can be completed using power series as above to establish that the sine has as its derivative the cosine, and the cosine has as its derivative the negative sine.
The angle between the horizontal line and the shown diagonal is 1 / 2 (a + b). This is a geometric way to prove the particular tangent half-angle formula that says tan 1 / 2 (a + b) = (sin a + sin b) / (cos a + cos b). The formulae sin 1 / 2 (a + b) and cos 1 / 2 (a + b) are the ratios of the actual distances to ...
The real part of the other side is a polynomial in cos x and sin x, in which all powers of sin x are even and thus replaceable through the identity cos 2 x + sin 2 x = 1. By the same reasoning, sin nx is the imaginary part of the polynomial, in which all powers of sin x are odd and thus, if one factor of sin x is factored out, the remaining ...
The sign of the square root needs to be chosen properly—note that if 2 π is added to θ, the quantities inside the square roots are unchanged, but the left-hand-sides of the equations change sign. Therefore, the correct sign to use depends on the value of θ.
r = | z | = √ x 2 + y 2 is the magnitude of z and; φ = arg z = atan2(y, x). φ is the argument of z, i.e., the angle between the x axis and the vector z measured counterclockwise in radians, which is defined up to addition of 2π. Many texts write φ = tan −1 y / x instead of φ = atan2(y, x), but the first equation needs ...
The third formula shown is the result of solving for a in the quadratic equation a 2 − 2ab cos γ + b 2 − c 2 = 0. This equation can have 2, 1, or 0 positive solutions corresponding to the number of possible triangles given the data.