Naram-Sin

(noun)

An Akkadian king who conquered Ebla, Armum, and Magan, and built a royal residence at Tell Brak.

Related Terms

Examples of Naram-Sin in the following topics:

  • The Akkadian Empire

    • The Akkadian Empire flourished in the 24th and 22nd centuries BCE, ruled by Sargon and Naram-Sin.
    • Manishtushu's son and successor, Naram-Sin (called, Beloved of Sin) (2254–2218 BCE), assumed the imperial title "King Naram-Sin, King of the Four Quarters."
    • Naram-Sin also conquered Magan and created garrisons to protect the main roads.
    • Bronze head of a king, most likely Sargon of Akkad but possibly Naram-Sin.
    • This stele commemorates Naram-Sin's victory against the Lullubi from Zagros in 2260 BCE.

  • Akkadian Government, Culture, and Economy

    • With Naram-Sin, Sargon's grandson, titular honors went even further than they did with Sargon.
    • Previously a ruler could, like the legendary Gilgamesh, become divine after death but the Akkadian kings, from Naram-Sin onward, were considered gods on earth in their lifetimes.
    • Both Sargon and Naram-Sin maintained control of the country by installing various members of their family in important positions around the empire.
    • Their daughters, Enheduanna and Emmenanna respectively, became high priestesses to Sin, the Akkadian version of the Sumerian moon deity, Nanna. at Ur, in the extreme south of Sumer.
    • Bassetki Statue from the reign of Naram-Sin with an inscription mentioning the construction of a temple in Akkad.

  • Sculpture in Mesopotamia

    • The Victory Stele of Naram Sin provides an example of the increasingly violent subject matter in Akkadian art, a result of the violent and oppressive climate of the empire.
    • In typical hieratic fashion, Naram Sin appears larger than his soldiers and his enemies.

  • Humans and Their Deities

    • In some cases, this included a fusion of animal form with the face of the king or queen (as in the Hatshepsut-Sphinx, ) or as the god-king victors of an epic battle, as in the stele of Naran-Sin of Akkad (the first Mesopotamian king to claim divinity for himself, ).
    • Naram-Sin, the first Mesopotamian king known to have claimed divinity for himself (worshiped as the Akkadian moon god), depicted on his victory stele

  • Trigonometric Limits

    • For $0 < x < \frac{ \pi}{2}$, $\sin x < x < \tan x.$
    • $\displaystyle{\lim_{x \to 0} \left ( \frac{x}{\sin x} \right ) = 1}$
    • $\displaystyle{\lim_{x \to 0} \left ( \frac{\sin x}{x} \right ) = 1}$
    • $\displaystyle{\frac{(1−\cos x)(1+\cos x)}{x(1+\cos x)}=\frac{(1−\cos^2x)}{x(1+\cos x)}=\frac{\sin^2x}{x(1+\cos x)}= \frac{\sin x}{x} \cdot \frac{\sin x}{1+\cos x}}$
    • $\displaystyle{\lim_{x \to 0}\left ( \frac{\sin x}{x} \frac{\sin x}{1 + \cos x} \right ) = \left (\lim_{x \to 0} \frac{\sin x}{x} \right ) \left ( \lim_{x \to 0} \frac{\sin x}{1 + \cos x} \right ) = \left (1 \right )\left (\frac{0}{2} \right )= 0}$

  • Double and Half Angle Formulae

    • Deriving the double-angle formula for sine begins with the sum formula that was introduced previously: $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$.
    • $\displaystyle{ \begin{aligned} \sin(\theta + \theta) &= \sin \theta \cos \theta + \cos \theta \sin \theta \\ \sin(2\theta) &= 2\sin \theta \cos \theta \end{aligned} }$
    • $\displaystyle{ \begin{aligned} \cos{\left(2\theta \right)} &= \cos^2 \theta - \sin^2 \theta \\ &= \left(1- \sin^2 \theta \right) - \sin^2 \theta \\ &= 1- 2\sin^2 \theta \end{aligned} }$
    • We will apply the double-angle formula for sine: $\sin(2\theta) = 2\sin \theta \cos \theta $.
    • $\displaystyle{ \begin{aligned} \sin{\left(60^{\circ}\right)} &= \sin{\left(30^{\circ} + 30^{\circ}\right)} \\ &= 2\sin (30^{\circ}) \cos (30^{\circ}) \end{aligned} }$

  • Trigonometric Integrals

    • \\ \int\sin^2 {ax}\;\mathrm{d}x = \frac{x}{2} - \frac{1}{4a} \sin 2ax +C= \frac{x}{2} - \frac{1}{2a} \sin ax\cos ax +C\!
    • \\ \int x\sin^2 {ax}\;\mathrm{d}x = \frac{x^2}{4} - \frac{x}{4a} \sin 2ax - \frac{1}{8a^2} \cos 2ax +C\!
    • $\int\cos^2 {ax}\;\mathrm{d}x = \frac{x}{2} + \frac{1}{4a} \sin 2ax +C = \frac{x}{2} + \frac{1}{2a} \sin ax\cos ax +C$
    • Two simple examples of such integrals are $\int \sin^k x \cos x \; \mathrm d x$ and $\int \cos^k x \sin x\; \mathrm d x$ , which can be solved used the substitutions $u = \sin x$ and $u = \cos x$ , respectively.
    • If both $n$ and $m$ are even, then we can try to use a combination of the following three identities: $\cos^2 x = \frac{1}{2} (1 + \cos 2x)$, $\sin^2 x = \frac{1}{2} (1 - \sin 2x)$, and $\sin x \cos x = \frac{1}{2} \sin 2x$.

  • Triple Integrals in Spherical Coordinates

    • $f(x,y,z) \longrightarrow f(\rho \cos \theta \sin \varphi, \rho \sin \theta \sin \varphi, \rho \cos \varphi)$
    • $\displaystyle{\frac{\partial (x,y,z)}{\partial (\rho, \theta, \varphi)}} = \begin{vmatrix} \cos \theta \sin \varphi & - \rho \sin \theta \sin \varphi & \rho \cos \theta \cos \varphi \\ \sin \theta \sin \varphi & \rho \cos \theta \sin \varphi & \rho \sin \theta \cos \varphi \\ \cos \varphi & 0 & - \rho \sin \varphi \end{vmatrix} = \rho^2 \sin \varphi$
    • The $dx\, dy\, dz$ differentials therefore are transformed to $\rho^2 \sin \varphi \, d\rho \,d\varphi \,dz$.
    • $f(\rho \sin \varphi \cos \theta, \rho \sin \varphi \sin \theta, \rho \cos \varphi) = \rho^2$
    • $\begin{aligned} \iiint_D (x^2 + y^2 +z^2) \, dx\, dy\, dz &= \iiint_T \rho^2 \ \rho^2 \sin \theta \, d\rho\, d\theta\, d\phi, \\ &= \int_0^{\pi} \sin \phi \,d\phi \int_0^4 \rho^4 d \rho \int_0^{2 \pi} d\theta \\ &= 2 \pi \int_0^{\pi} \sin \phi \left[ \frac{\rho^5}{5} \right]_0^4 \, d \phi \\ &= 2 \pi \left[ \frac{\rho^5}{5} \right]_0^4 \left[- \cos \phi \right]_0^{\pi}= \frac{4096 \pi}{5} \end{aligned}$

  • The Law of Sines

    • $\displaystyle{\frac{ \sin \alpha}{a} = \frac{ \sin \beta}{b} = \frac{ \sin \gamma}{c}}$
    • $\displaystyle{ \begin{aligned} \frac{b \cdot \sin{\left(50^{\circ}\right)}}{10} &= \sin{\left(100^{\circ}\right)} \\ b \cdot \sin{\left(50^{\circ}\right)} &= 10 \cdot \sin{\left(100^{\circ}\right)} \\ b &= \frac{10 \cdot \sin{\left(100^{\circ}\right)}}{\sin{\left(50^{\circ}\right)}} \\ b &\approx 12.9 \end{aligned} }$

  • Derivatives of Trigonometric Functions

    • $\displaystyle{ \quad \quad = \lim_{h\rightarrow 0} \frac{\cos(x) \cdot \sin(h) + \cos(h) \cdot \sin(x) - \sin(x)}{h}}$
    • $\displaystyle{ \quad \quad = \lim_{h\rightarrow 0} \frac{\cos(x) \cdot \sin(h) + (\cos(h) - 1) \cdot \sin(x)}{h}}$
    • $\displaystyle{ \quad \quad = \lim_{h\rightarrow 0} \frac{\cos(x) \cdot \sin(h)}{h} + \lim_{h\rightarrow 0} \frac {(\cos(h) - 1) \cdot \sin(x)}{h}}$
    • $\displaystyle{ \quad \quad = \cos (x) (1) + \sin (x) (0)}$