Search results
Results From The WOW.Com Content Network
Write the functions without "co" on the three left outer vertices (from top to bottom: sine, tangent, secant) Write the co-functions on the corresponding three right outer vertices (cosine, cotangent, cosecant) Starting at any vertex of the resulting hexagon: The starting vertex equals one over the opposite vertex.
The linear combination, or harmonic addition, of sine and cosine waves is equivalent to a single sine wave with a phase shift and scaled amplitude, [33] [34] a cos x + b sin x = c cos ( x + φ ) {\displaystyle a\cos x+b\sin x=c\cos(x+\varphi )}
Ebook version, in PDF format, full text presented. Trigonometry by Alfred Monroe Kenyon and Louis Ingold, The Macmillan Company, 1914. In images, full text presented. Trigonometry FAQ; Trigonometry on Mathwords.com index of trigonometry entries on Mathwords.com; Trigonometry on PlainMath.net Trigonometry Articles from PlainMath.Net
The cosine, cotangent, and cosecant are so named because they are respectively the sine, tangent, and secant of the complementary angle abbreviated to "co-". [32] With these functions, one can answer virtually all questions about arbitrary triangles by using the law of sines and the law of cosines. [33]
A geometric way of deriving the sine or cosine of 45° is by considering an isosceles right triangle with leg length 1. Since two of the angles in an isosceles triangle are equal, if the remaining angle is 90° for a right triangle, then the two equal angles are each 45°.
The sine and the cosine functions, for example, are used to describe simple harmonic motion, which models many natural phenomena, such as the movement of a mass attached to a spring and, for small angles, the pendular motion of a mass hanging by a string. The sine and cosine functions are one-dimensional projections of uniform circular motion.
3.2 Integrals involving hyperbolic sine and cosine functions. ... Download as PDF; Printable version; ... Integrals of hyperbolic tangent, cotangent, secant, cosecant ...
Illustration of the sine and tangent inequalities. 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. = = = =