![]() ![]() A Quick Proof of cos(pi/7)cos(2.pi/7)cos(3.A Trigonometric Observation in Right Triangle.Wonderful Trigonometry In Equilateral Triangle.The Law of Cosines and the Law of Sines Are Equivalent.Another Face and Proof of a Trigonometric Identity.The Concurrency of the Altitudes in a Triangle - Trigonometric Proof.Trigonometric Identities with Arctangents.ΔABC is right iff sin☪ + sin☫ + sin☬ = 2.A Trigonometric Solution to a Difficult Sangaku Problem.Addition and Subtraction Formulas for Sine and Cosine IV.Addition and Subtraction Formulas for Sine and Cosine III.Addition and Subtraction Formulas for Sine and Cosine II.Addition and Subtraction Formulas for Sine and Cosine.Maor, Trigonometric Delights, Princeton University Press, 1998 (There are additional simple proofs of those formulas.) References I don't think there's a sensible way to relate hyperbolic cosine to the Law of Cosines in Euclidean geometry. There is a Hyperbolic Law of Cosines, but it's not quite the same. \cos(\alpha + \beta ) &= \cos(\alpha ) \cos(\beta ) - \sin (\alpha ) \sin (\beta ),\\ In hyperbolic geometry, the definition of lines (and hence triangles) is different and the sum of the measures of the angles in a triangle is less than 180. $\displaystyle\frac - \alpha ) &= \sin \alpha.įrom these and the addition formulas for sine it is not difficult to derive the addition formulas for cosine: Repeating these steps with the other two angles $B$ and $C$ of $\Delta ABC$ we get the Law of Sines which in the standard notations appear as The resulting identity is, however, the same. This is Proposition III.21) On the diagram, $\angle BOC = 2\angle BAC (= 2A.)$ĭrop a perpendicular from $O$ on the side $BC.$ Assuming the radius of the circle is $R,$ $OB = OC = R.$ Also, $\angle BOP = \angle POC.$ In $\Delta BOP,$ $\sin (\angle BOP) = BP/OB = BC/2R.$ Therefore, $BC/\sin (\angle BOP) = 2R.$ When angle $A$ is obtuse, the center $O$ is located outside $\Delta ABC$ and the diagram looks differently. (As a corollary, from here it follows that all circumscribed angles subtending the same arc are equal irrespective of their position on the circle. The more common formulation asserts that an angle circumscribed in a circle is equal to half the central angle that subtends the same chord. In a circle the angle at the center is double of the angle at the circumference, when angles have the same circumference as base.
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