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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.
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} .
This proof uses the power of a point theorem directly, without the auxiliary triangles obtained by constructing a tangent or a chord. Construct a circle with center B and radius a (see Figure 9), which intersects the secant through A and C in C and K. The power of the point A with respect to the circle is equal to both AB 2 − BC 2 and AC·AK.
which by the Pythagorean theorem is equal to 1. This definition is valid for all angles, due to the definition of defining x = cos θ and y sin θ for the unit circle and thus x = c cos θ and y = c sin θ for a circle of radius c and reflecting our triangle in the y-axis and setting a = x and b = y.
To extend the sine and cosine functions to functions whose domain is the whole real line, geometrical definitions using the standard unit circle (i.e., a circle with radius 1 unit) are often used; then the domain of the other functions is the real line with some isolated points removed.
As a special case, for C = π / 2 , then cos C = 0, and one obtains the spherical analogue of the Pythagorean theorem: cos c = cos a cos b {\displaystyle \cos c=\cos a\cos b\,} If the law of cosines is used to solve for c , the necessity of inverting the cosine magnifies rounding errors when c is small.
A quadratrix in the first quadrant (x, y) is a curve with y = ρ sin θ equal to the fraction of the quarter circle with radius r determined by the radius through the curve point. Since this fraction is 2 r θ π {\displaystyle {\frac {2r\theta }{\pi }}} , the curve is given by ρ ( θ ) = 2 r θ π sin θ {\displaystyle \rho (\theta ...
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.