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.
The starting corner equals the product of its two nearest neighbors. For example, sin A = cos A ⋅ tan A {\\displaystyle \\sin A=\\cos A\\cdot \\tan A} The sum of the squares of the two items at the top of a triangle equals the square of the item at the bottom.
In this way, this trigonometric identity involving the tangent and the secant follows from the Pythagorean theorem. The angle opposite the leg of length 1 (this angle can be labeled φ = π/2 − θ ) has cotangent equal to the length of the other leg, and cosecant equal to the length of the hypotenuse.
Given a unit sphere, a "spherical triangle" on the surface of the sphere is defined by the great circles connecting three points u, v, and w on the sphere (shown at right). If the lengths of these three sides are a (from u to v ), b (from u to w ), and c (from v to w ), and the angle of the corner opposite c is C , then the (first) spherical ...
For example, the derivative of the sine function is written sin ′ (a) = cos(a), meaning that the rate of change of sin(x) at a particular angle x = a is given by the cosine of that angle. All derivatives of circular trigonometric functions can be found from those of sin( x ) and cos( x ) by means of the quotient rule applied to functions such ...
Using the usual notations for a triangle (see the figure at the upper right), where a, b, c are the lengths of the three sides, A, B, C are the vertices opposite those three respective sides, α, β, γ are the corresponding angles at those vertices, s is the semiperimeter, that is, s = a + b + c / 2 , and r is the radius of the inscribed circle, the law of cotangents states that
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 ...
Because PQ has length y 1, OQ length x 1, and OP has length 1 as a radius on the unit circle, sin(t) = y 1 and cos(t) = x 1. Having established these equivalences, take another radius OR from the origin to a point R(− x 1 , y 1 ) on the circle such that the same angle t is formed with the negative arm of the x -axis.