Plane trigonometry
trigonometry in the Euclidean plane.
The elements of a triangle, its sides $ a, b, c $ and its angles $ A, B, C $( $ A $ opposite to $ a $, etc.), are related in various ways. In the Euclidean plane the most important relations are the angle sum formula
$$ A + B + C = \pi $$
(angles in radians), and the triangle inequalities
$$ a + b > c ,\ \ b + c > a ,\ \ c + a > b . $$
These inequalities are necessary and sufficient for three segments of positive length $ a, b, c $ to form the sides of a triangle.
Another relation is the cosine theorem:
$$ c ^ {2} = a ^ {2} + b ^ {2} - 2ab \cos C. $$
In particular, when $ C = \pi /2 $, the triangle is right-angled, and the cosine theorem becomes Pythagoras' theorem (cf. Pythagoras theorem)
$$ c ^ {2} = a ^ {2} + b ^ {2} . $$
In such a right-angled triangle,
$$ In a general triangle, further relations are provided by the [[Sine theorem|sine theorem]]: $$
\frac{a}{\sin A }
=
\frac{b}{\sin B }
= \
\frac{c}{\sin C }
= 2R ,
$$ where $ R $ is the radius of the circumcircle of the triangle (cf. [[Inscribed and circumscribed figures|Inscribed and circumscribed figures]]). A corollary of the sine theorem is the tangent formula $$ a- \frac{b}{a+}
b = \
\frac{ \mathop{\rm tan} [( A- B)/2] }{ \mathop{\rm tan} [( A+ B)/2] }
= \ \mathop{\rm tan} A-
\frac{B}{2}
\mathop{\rm cotan}
\frac{C}{2}
.
$$ With the notation $ s = ( a+ b+ c)/2 $ for the semi-perimeter of the triangle, the following half-angle formulas follow from the cosine theorem: $$ \cos ^ {2} \frac{A}{2}
= s( s-
\frac{a)}{bc}
,
$$ $$ \sin ^ {2} \frac{A}{2}
= ( s- b)( s-
\frac{c)}{bc}
,
$$ $$
\mathop{\rm tan} ^ {2}
\frac{A}{2}
= ( s- b)( s-
\frac{c)}{s(}
s- a) .
$$ =='"`UNIQ--h-0--QINU`"'Geometry of the triangle.== Among the many remarkable lines, points and circles connected with any triangle are the circumcircle with centre 0 and radius $ R $, the incircle and the three excircles with centres $ I $, $ I _ {a} $, $ I _ {b} $, $ I _ {c} $ and radii $ r $, $ r _ {a} $, $ r _ {b} $, $ r _ {c} $( cf. [[Inscribed and circumscribed figures|Inscribed and circumscribed figures]]), the medians (cf. [[Median (of a triangle)|Median (of a triangle)]]) $ m _ {a} $, $ m _ {b} $, $ m _ {c} $, with the centroid $ G $ as their common point, the inner bisectors $ AI $, $ BI $, $ CI $, and the outer bisectors $ I _ {b} I _ {c} $, $ I _ {c} I _ {a} $, $ I _ {a} I _ {b} $, the altitude lines $ h _ {a} $, $ h _ {b} $, $ h _ {c} $ with the [[Orthocentre|orthocentre]] $ H $ as their common point, the Euler line (cf. [[Euler straight line|Euler straight line]]) through $ O $, $ G $ and $ H $, and the [[Nine-point circle|nine-point circle]] through the midpoints of the sides, the feet of the altitude lines, and the midpoints of the segments connecting the vertices of the triangle to its orthocentre. The nine-point circle has radius $ R/2 $, its centre $ N $ is on the Euler line between $ G $ and $ H $ such that $ HN: NG : GO = 3: 1: 2 $, and the nine-point circle touches the incircle and the three excircles (Feuerbach's theorem). With the notation $ ( ABC) $ for the area of the triangle $ ABC $, the following relations are valid: $$ ( ABC) = \frac{1}{2}
ah _ {a} =
\frac{1}{2}
bc \sin A = \
\frac{abc}{4R\ }
=
$$ $$ = \ r \cdot s = r _ {a} ( s- a) = \sqrt {s( s- a)( s- b)( s- c) } . $$ It follows that, among others, $$ 4R = r _ {a} + r _ {b} + r _ {c} - r \ \textrm{ and } \ \
\frac{1}{r}
=
\frac{1}{r _ {a} }
+
\frac{1}{r _ {b} }
+
\frac{1}{r _ {c} }
.
$$ Very remarkable is Morley's theorem: The points of intersection of the adjacent trisectors of the angles of any triangle form the vertices of an equilateral triangle. Indeed, a direct calculation shows that the sides of Morley's triangle have length $$ 8R \sin \frac{A}{3}
\sin
\frac{B}{3}
\sin
\frac{C}{3}
,
$$ which is symmetric in $ A $, $ B $ and $ C $. =='"`UNIQ--h-1--QINU`"'The theorems of Ceva and Menelaus.== Let $ X, Y, Z $ be points on the (possibly extended) sides $ a, b, c $ of a triangle $ ABC $. Then, by the [[Ceva theorem|Ceva theorem]], $ AX $, $ BY $ and $ CZ $ are congruent if and only if $ ( BX: XC)( CY: YA)( AZ: ZB)= 1 $( signed distances) and by the [[Menelaus theorem|Menelaus theorem]], $ X $, $ Y $ and $ Z $ are collinear if and only if $ ( BX: XC)( CY: YA)( AZ: ZB) = - 1 $. =='"`UNIQ--h-2--QINU`"'Convex quadrangles.== Ptolemy's theorem (cf. [[Ptolemeus theorem|Ptolemeus theorem]]): For any point $ P $ in the plane of a triangle $ ABC $ the inequality $$ AB \cdot CP + BC \cdot AP \geq AC \cdot BP $$ holds, with equality if and only if $ P $ is on the arc $ CA $ of the circumcircle of $ ABC $( in this last case, $ ABCP $ is a circle quadrangle). Brahmagupta's formula states that for any convex cyclic quadrangle $ ABCD $ with area $ ( ABCD) $, sides $ a, b, c , d $ and semi-perimeter $ s = ( a+ b + c + d)/2 $, the relation $ ( ABCD) = \sqrt {( s- a)( s- b)( s- c)( s- d) } $ holds. In general, for any quadrangle $ ABCD $, the area $ ( ABCD) $ satisfies $$ ( ABCD) ^ {2} = ( s- a)( s- b)( s- c)( s- d) - abcd \cos ^ {2} A+ \frac{C}{2}
.
$$
It follows that among all quadrangles with given side lengths the inscribed quadrangles have maximum area (the cyclic order of the sides is immaterial).
Regular $ n $-gons.
A regular $ n $- gon inscribed in a circle with radius $ R $ has perimeter $ 2nR \sin ( \pi /n) $ and area $ ( n/2) R ^ {2} \sin ( 2 \pi /n) $; a regular $ n $- gon circumscribed about a circle with radius $ R $ has perimeter $ 2nR \mathop{\rm tan} ( \pi /n ) $ and area $ nR ^ {2} \mathop{\rm tan} ( \pi / n) $. See also Regular polyhedra.
References
[a1] | H.S.M. Coxeter, "Introduction to geometry" , Wiley (1969) pp. 3–23 |
[a2] | H.S.M. Coxeter, S.L. Greitzer, "Geometry revisited" , Random House (1967) |
[a3] | M. Berger, "Geometry" , II , Springer (1987) pp. Chapt. 10 |
Plane trigonometry. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Plane_trigonometry&oldid=49366