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Ptolemy's theorem states that the sum of the products of the lengths of opposite sides is equal to the product of the lengths of the diagonals. When those side-lengths are expressed in terms of the sin and cos values shown in the figure above, this yields the angle sum trigonometric identity for sine: sin(α + β) = sin α cos β + cos α sin β.
There are a number of ways to demonstrate the validity of the small-angle approximations. The most direct method is to truncate the Maclaurin series for each of the trigonometric functions. Depending on the order of the approximation , cos θ {\displaystyle \textstyle \cos \theta } is approximated as either 1 {\displaystyle 1} or as 1 − 1 ...
In contrast, by the Lindemann–Weierstrass theorem, the sine or cosine of any non-zero algebraic number is always transcendental. [4] The real part of any root of unity is a trigonometric number. By Niven's theorem, the only rational trigonometric numbers are 0, 1, −1, 1/2, and −1/2. [5]
That cos nx is an n th-degree polynomial in cos x can be seen by observing that cos nx is the real part of one side of de Moivre's formula: + = ( + ). 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 .
A Fourier series (/ ˈ f ʊr i eɪ,-i ər / [1]) is an expansion of a periodic function into a sum of trigonometric functions.The Fourier series is an example of a trigonometric series. [2]
If vectors u and v have direction cosines (α u, β u, γ u) and (α v, β v, γ v) respectively, with an angle θ between them, their units vectors are ^ = + + (+ +) = + + ^ = + + (+ +) = + +. Taking the dot product of these two unit vectors yield, ^ ^ = + + = , where θ is the angle between the two unit vectors, and is also the angle between u and v.
Given the Euler angles α, β and γ, the quaternion representation should be (cos γ + k sin γ)(cos β + i sin β)(cos α + k sin α). I think. Anyway, the basic idea is just to take the quaternions for the fundamental rotations around the axes by the given angles and multiply those together in the proper order.
At this point we can either integrate directly, or we can first change the integrand to 2 cos 6x − 4 cos 4x + 2 cos 2x and continue from there. Either method gives Either method gives ∫ sin 2 x cos 4 x d x = − 1 24 sin 6 x + 1 8 sin 4 x − 1 8 sin 2 x + C . {\displaystyle \int \sin ^{2}x\cos 4x\,dx=-{\frac {1}{24 ...