Search results
Results from the WOW.Com Content Network
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.
Alternatively, the identities found at Trigonometric symmetry, shifts, and periodicity may be employed. By the periodicity identities we can say if the formula is true for −π < θ ≤ π then it is true for all real θ. Next we prove the identity in the range π/2 < θ ≤ π, to do this we let t = θ − π/2, t will now be in the range 0 ...
The figure at the right shows a sector of a circle with radius 1. The sector is θ/(2 π) of the whole circle, so its area is θ/2. We assume here that θ < π /2. = = = = The area of triangle OAD is AB/2, or sin(θ)/2.
The analog of the Pythagorean trigonometric identity holds: [2] + = If X is a diagonal matrix, sin X and cos X are also diagonal matrices with (sin X) nn = sin(X nn) and (cos X) nn = cos(X nn), that is, they can be calculated by simply taking the sines or cosines of the matrices's diagonal components.
The fixed point iteration x n+1 = cos(x n) with initial value x 0 = −1 converges to the Dottie number. Zero is the only real fixed point of the sine function; in other words the only intersection of the sine function and the identity function is sin ( 0 ) = 0 {\displaystyle \sin(0)=0} .
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 ...
Get AOL Mail for FREE! Manage your email like never before with travel, photo & document views. Personalize your inbox with themes & tabs. You've Got Mail!
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 ...