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Taking the reciprocal of both sides of the Hill equation, rearranging, and inverting again yields: = [] = [] (). Taking the logarithm of both sides of the equation leads to an alternative formulation of the Hill-Langmuir equation:
The oldest and most elementary definitions are based on the geometry of right triangles and the ratio between their sides. The proofs given in this article use these definitions, and thus apply to non-negative angles not greater than a right angle. For greater and negative angles, see Trigonometric functions.
The reciprocal identities arise as ratios of sides in the triangles where this unit line is no longer the hypotenuse. The triangle shaded blue illustrates the identity 1 + cot 2 θ = csc 2 θ {\displaystyle 1+\cot ^{2}\theta =\csc ^{2}\theta } , and the red triangle shows that tan 2 θ + 1 = sec 2 θ {\displaystyle \tan ^{2 ...
To complete the square, form a squared binomial on the left-hand side of a quadratic equation, from which the solution can be found by taking the square root of both sides. The standard way to derive the quadratic formula is to apply the method of completing the square to the generic quadratic equation a x 2 + b x + c = 0 {\displaystyle ...
by Euler's criterion, but both sides of this congruence are numbers of the form , so they must be equal. Whether 2 {\displaystyle 2} is a quadratic residue can be concluded if we know the number of solutions of the equation x 2 + y 2 = 2 {\displaystyle x^{2}+y^{2}=2} with x , y ∈ Z p , {\displaystyle x,y\in \mathbb {Z} _{p},} which can be ...
They threaten the stability of industries on both sides of the border." ... Mexico’s retail sales were expected to take a huge hit, growing just 3.3% this year instead of the 4.5% growth ...
Joni Mitchell delivered a rare performance to benefit those affected by the Los Angeles fires. On Thursday, Jan. 30, the legendary folk singer-songwriter, 81, took the stage at the Kia Forum for ...
As an example of how Euler's criterion is used, we can use it to give a quick proof of the first supplemental case of determining for an odd prime p: By Euler's criterion () (), but since both sides of the equivalence are ±1 and p is odd, we can deduce that () = ().