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trigonometrically

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trigonometrically
ˌtrigonoˈmetrically, adv. [f. prec. + -ly2.] In a trigonometrical manner; by means, or by the method, of trigonometry.1666 Collins in Rigaud Corr. Sci. Men (1841) I. 115 Problems that seem easy trigonometrically, but not so analytically. 1751 J. Stuart in Lett. Lit. Men (Camden) 382 An exact Map of ... Oxford English Dictionary
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trigonometrically
trigonometrically/-klɪ; -klɪ/ adv. 牛津英汉双解词典
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Indicator function (complex analysis)
If , we simply say, that is trigonometrically convex. Such indicators have some special properties. If is -trigonometrically convex on , then , whenever . Notes References Complex analysis wikipedia.org
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trigonometrical
trigonometrical, a. (ˌtrɪgənəʊˈmɛtrɪkəl) [f. trigonometry or mod.L. trigonometria + -ic + -al1; after geometrical, etc.] Of, pertaining to, or performed by trigonometry. trigonometrical functions, those functions of an angle, or of an abstract quantity, used in trigonometry, viz. the sine, tangent, ... Oxford English Dictionary
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Bretschneider's formula
The trigonometric adjustment in Bretschneider's formula for non-cyclicality of the quadrilateral can be rewritten non-trigonometrically in terms of the wikipedia.org
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Is it possible to trigonometrically understand why $\cos(x)$ is even and $\sin(x)$ is odd? Is it possible to trigonometrically understand why $\cos(x)$ is even and $\sin(x)$ is odd? ![enter image description here](
Yes, it is: sending $\theta$ to $-\theta$ amounts to flipping the $y$-axis in your drawing, and then the cosine is unaffected since the $x$-axis is fixed, but the sine changes sign (since the $y$-axis is flipped).
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Gyroid
Curiously, like some other triply periodic minimal surfaces, the gyroid surface can be trigonometrically approximated by a short equation: The gyroid wikipedia.org
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List of representations of e
In trigonometry Trigonometrically, can be written in terms of the sum of two hyperbolic functions, at . wikipedia.org
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stasimetric
† stasiˈmetric, a. Surveying. Obs. [f. Gr. στάσι-ς standing, station (see stasis) + µέτρ-ον measure + -ic.] (See quots.)1774 M. Mackenzie Treat. Surveying (1819) Introd. 14 A Stasimetric Survey, is, when the mutual distances of three, or more, proper objects are carefully measured; and by means of t... Oxford English Dictionary
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Sulitjelma (mountain)
Using a mercury barometer, he determined the elevation of Vágŋatjåhkkå, and from that he trigonometrically calculated the elevations of Suliskongen, Vardetoppen wikipedia.org
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Property of bisectors of right triangle In triangle $ABC$ $\angle C=90^\circ$, $AA'$ and $BB'$ are angle bisectors intersecting at $I$ ($A'\in BC$, $B'\in AC$). What would be the easiest way to prove that projection o...
Draw the incircle (with center $I$ and radius $r$) of $\triangle ABC$, and let it meet $BC$ at $P$. Since $IP \perp BC$, we have $IP \parallel AC$. ( _Note: This is why the right angle at $C$ matters._ ) Also, $|IP|=r$. Let $Q$ be the point "near" $A^\prime$ on the incircle such that $IQ \parallel A...
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Glaswaldsee
The Glaswaldsee is one of the first lakes of the Black Forest to be trigonometrically surveyed and mapped. wikipedia.org
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Proof of $|\sin(x)| \leq 1$ and $|\cos(x)| \leq 1$ Using the series representations of $\sin(x)$ and $\cos(x)$, how does one show that both $|\sin(x)| \leq 1$ and $|\cos(x)| \leq 1$? I can do this easily algebraically...
To prove that $|\sin(x)|\le 1$ and $|\cos(x)|\le 1$ from the series definition, I would take a little "side step". From $$\sin(x)= \sum_{n= 0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$ we can differentiate "term by term" to show that $$(\sin(x))'= (-1)^n\sum_{n= 0}^\infty \frac{2n+1}{2n+ 1}!x^{2n}= \s...
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Casus irreducibilis
Trigonometric solution in terms of real quantities While casus irreducibilis cannot be solved in radicals in terms of real quantities, it can be solved trigonometrically wikipedia.org
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How to prove $\cos \frac{2\pi }{5}=\frac{-1+\sqrt{5}}{4}$? I would like to find the apothem of a regular pentagon. It follows from $$\cos \dfrac{2\pi }{5}=\dfrac{-1+\sqrt{5}}{4}.$$ But how can this be proved (geome...
Since $x := \cos \frac{2 \pi}{5} = \frac{z + z^{-1}}{2}$ where $z:=e^{\frac{2 i \pi}{5}}$, and $1+z+z^2+z^3+z^4=0$ (for $z^5=1$ and $z \neq 1$), $x^2+\frac{x}{2}-\frac{1}{4}=0$, and voilà.
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