hcju/proofwikiqa huggingface dataset:
Let $K$ be an ellipse aligned in a cartesian plane in reduced form. Then from Equation of Ellipse in Reduced Form: :$\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$ Thus: :$y = \pm \dfrac b a \sqrt{a^2 - x^2}$ Consider a circle of radius $a$ whose center is at the origin. From Equation of Circle: Cartesian: Corollary 2, its equation is given by: :$x^2 + y^2 = a^2$ and so: :$y = \pm \sqrt{a^2 - x^2}$ The formulas show that each ordinate of the ellipse is $\dfrac b a$ the ordinate of the circle. Since the same thing is true of the vertical chords: {{begin-eqn}} {{eqn | l = \mathcal A_E       | r = \dfrac b a \mathcal A_C       | c = where $\mathcal A_E$ and $\mathcal A_C$ are the areas of the ellipse and the circle respectively }} {{eqn | r = \dfrac b a \paren {\pi a^2}       | c = Area of Circle }} {{eqn | r = \pi a b       | c =  }} {{end-eqn}} {{qed}}
:500px By definition, the foci $F_1$ and $F_2$ of $K$ are located at $\tuple {-c, 0}$ and $\tuple {c, 0}$ respectively. Let the vertices of $K$ be $V_1$ and $V_2$. By definition, these are located at $\tuple {-a, 0}$ and $\tuple {a, 0}$. Let the covertices of $K$ be $C_1$ and $C_2$. By definition, these are located at $\tuple {0, -b}$ and $\tuple {0, b}$. Let $P = \tuple {x, y}$ be an arbitrary point on the locus of $K$. From the equidistance property of $K$ we have that: :$F_1 P + F_2 P = d$ where $d$ is a constant for this particular ellipse. From Equidistance of Ellipse equals Major Axis: :$d = 2 a$ Also, from Focus of Ellipse from Major and Minor Axis: :$a^2 - c^2 = b^2$ Then: {{begin-eqn}} {{eqn | l = \sqrt {\paren {x - c}^2 + y^2} + \sqrt {\paren {x + c}^2 + y^2}       | r = d = 2 a       | c = Pythagoras's Theorem }} {{eqn | ll= \leadsto       | l = \sqrt {\paren {x + c}^2 + y^2}       | r = 2 a - \sqrt {\paren {x - c}^2 + y^2}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x + c}^2 + y^2       | r = \paren {2 a - \sqrt {\paren {x - c}^2 + y^2} }^2       | c = squaring both sides }} {{eqn | ll= \leadsto       | l = x^2 + 2 c x + c^2 + y^2       | r = 4 a^2 - 4 a \sqrt {\paren {x - c}^2 + y^2} + \paren {x - c}^2 + y^2       | c = expanding }} {{eqn | ll= \leadsto       | l = x^2 + 2 c x + c^2 + y^2       | r = 4 a^2 - 4 a \sqrt {\paren {x - c}^2 + y^2} + x^2 - 2 c x + c^2 + y^2       | c = further expanding }} {{eqn | ll= \leadsto       | l = a^2 - c x       | r = a \sqrt {\paren {x - c}^2 + y^2}       | c = gathering terms and simplifying }} {{eqn | ll= \leadsto       | l = \paren {a^2 - c x}^2       | r = a^2 \paren {\paren {x - c}^2 + y^2}^2       | c = squaring both sides }} {{eqn | ll= \leadsto       | l = c^2 x^2 - 2 c x a^2 + a^4       | r = a^2 x^2 - 2 c x a^2 + a^2 c^2 + a^2 y^2       | c = expanding }} {{eqn | ll= \leadsto       | l = c^2 x^2 + a^4       | r = a^2 x^2 + a^2 c^2 + a^2 y^2       | c = simplifying }} {{eqn | ll= \leadsto       | l = a^4 - a^2 c^2       | r = a^2 x^2 - c^2 x^2 + a^2 y^2       | c = gathering terms }} {{eqn | ll= \leadsto       | l = a^2 \paren {a^2 - c^2}       | r = \paren {a^2 - c^2} x^2 + a^2 y^2       | c = simplifying }} {{eqn | ll= \leadsto       | l = a^2 b^2       | r = b^2 x^2 + a^2 y^2       | c = substituting $a^2 - c^2 = b^2$ from $(2)$ }} {{eqn | ll= \leadsto       | l = 1       | r = \frac {x^2} {a^2} + \frac {y^2} {b^2}       | c = dividing by $a^2 b^2$ }} {{end-eqn}} {{qed}}
Let the point $\left({x, y}\right)$ satisfy the equations: : $x = a \cos \theta$ : $y = b \sin \theta$ Then: {{begin-eqn}} {{eqn | l = \frac {x^2} {a^2} + \frac {y^2} {b^2}       | r = \frac {\left({a \cos \theta}\right)^2} {a^2} + \frac {\left({b \sin \theta}\right)^2} {b^2}       | c =  }} {{eqn | r =  \frac {a^2} {a^2} \cos^2 \theta + \frac {b^2} {b^2} \sin^2 \theta       | c =  }} {{eqn | r = \cos^2 \theta + \sin^2 \theta       | c =  }} {{eqn | r = 1       | c = Sum of Squares of Sine and Cosine }} {{end-eqn}} {{qed}}
Let $K$ be aligned in a cartesian plane such that: :the major axis of $K$ is aligned with the $x$-axis :the minor axis of $K$ is aligned with the $y$-axis. Then from Equation of Ellipse in Reduced Form: parametric form: :$x = a \cos \theta, y = b \sin \theta$ Thus: {{begin-eqn}} {{eqn | l = \frac {\d x} {\d \theta}       | r = -a \sin \theta       | c = Derivative of Cosine Function }} {{eqn | l = \frac {\d y} {\d \theta}       | r = b \cos \theta       | c = Derivative of Sine Function }} {{end-eqn}} From Arc Length for Parametric Equations, the length of one quarter of the perimeter of $K$ is given by: {{begin-eqn}} {{eqn | l = \frac {\PP} 4       | r = \int_0^{\pi / 2} \sqrt {\paren {-a \sin \theta}^2 + \paren {b \cos \theta}^2} \rd \theta       | c =  }} {{eqn | r = \int_0^{\pi / 2} \sqrt {a^2 \paren {1 - \cos^2 \theta} + b^2 \cos^2 \theta} \rd \theta       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \int_0^{\pi / 2} \sqrt {a^2 - \paren {a^2 - b^2} \cos^2 \theta} \rd \theta       | c = simplifying }} {{eqn | r = a \int_0^{\pi / 2} \sqrt {1 - \paren {1 - \frac {b^2} {a^2} } \cos^2 \theta} \rd \theta       | c = extracting $a$ as a factor }} {{eqn | n = 1       | r = a \int_0^{\pi / 2} \sqrt {1 - k^2 \cos^2 \theta} \rd \theta       | c = setting $k^2 = 1 - \dfrac {b^2} {a^2} = \dfrac {a^2 - b^2} {a^2}$ }} {{end-eqn}} Since $\cos \theta = \map \sin {\dfrac \pi 2 - \theta}$ we can write for any real function $\map f x$: :$\ds \int_0^{\pi / 2} \map f {\cos \theta} \rd \theta = \int_0^{\pi / 2} \map f {\map \sin {\frac \pi 2 - \theta} } \rd \theta$ So substituting $t = \dfrac \pi 2 - \theta$ this can be converted to: {{begin-eqn}} {{eqn | l = \int_0^{\pi / 2} \map f {\cos \theta} \rd \theta       | r = -\int_{\pi / 2}^0 \map f {\sin t} \rd t       | c =  }} {{eqn | r = \int_0^{\pi / 2} \map f {\sin t} \rd t       | c =  }} {{end-eqn}} justifying the fact that $\cos$ can be replaced with $\sin$ in $(1)$ above, giving: :$\ds \PP = 4 a \int_0^{\pi / 2} \sqrt {1 - k^2 \sin^2 \theta} \rd \theta$ {{qed}}
From Rational Numbers under Multiplication form Monoid, $\struct {\Q, \times}$ is a monoid. Then we have:  === Commutativity === Rational Multiplication is Commutative. {{qed|lemma}} === Infinite === Rational Numbers are Countably Infinite. {{Qed}}
:300px By definition of the cartesian plane, the point $P$ is $x$ units from the $y$-axis and $y$ units from the $x$-axis. The $y$-axis and $x$-axis are perpendicular to each other, also by definition. Thus $x$, $y$ and $OP$ form a right-angled triangle. By Pythagoras' Theorem: :$OP^2 = x^2 + y^2$ Hence the result. {{qed}}
From Composite of Homomorphisms on Algebraic Structure is Homomorphism, $\psi \circ \phi$ is a homomorphism. From Composite of Injections is Injection, $\psi \circ \phi$ is an injection. A monomorphism is an injective homorphism. Hence $\psi \circ \phi$ is a monomorphism. {{qed}}
From Composite of Homomorphisms on Algebraic Structure is Homomorphism, $\psi \circ \phi$ is a homomorphism. From Composite of Surjections is Surjection, $\psi \circ \phi$ is a surjection. An epimorphism is a surjective homomorphism. Hence $\psi \circ \phi$ is an epimorphism. {{qed}}
From Composite of Homomorphisms on Algebraic Structure is Homomorphism, $\psi \circ \phi$ is a homomorphism. By the definition of a composite mapping, $\psi \circ \phi$ is a mapping from $S$ into $S$. Hence $\psi \circ \phi$ is an automorphism. {{qed}}
From Composite of Homomorphisms on Algebraic Structure is Homomorphism, $\psi \circ \phi$ is a homomorphism. From the definition of composition of mappings, $\psi \circ \phi$ is a mapping from $S$ into $S$. Hence $\psi \circ \phi$ is an endomorphism. {{qed}}
:500px Let a unit circle $C$ be drawn with its center at the origin $O$. Let $Q$ be the point on $C$ which intersects $OP$. $\angle OSP = \angle ORQ$, as both are right angles. Both $\triangle OSP$ and $\triangle ORQ$ share angle $\theta$. By Triangles with Two Equal Angles are Similar it follows that $\triangle OSP$ and $\triangle ORQ$ are similar. By definition of similarity: Then: {{begin-eqn}} {{eqn | l = \frac y r       | r = \frac {SP} {OP}       | c =  }} {{eqn | r = \frac {RQ} {OQ}       | c = {{Defof|Similar Triangles}} }} {{eqn | r = RQ       | c = $OP$ is Radius of Unit Circle }} {{eqn | r = \sin \theta       | c = {{Defof|Sine|subdef = Definition from Circle}} }} {{end-eqn}} When $\theta$ is obtuse, the same argument holds. When $\theta = \dfrac \pi 2$ we have that $x = 0$. Thus $y = r$ and $\sin \theta = 1 \dfrac y r$. Thus the relation holds for $\theta = \dfrac \pi 2$. When $\pi < \theta < 2 \pi$ the diagram can be reflected in the $x$-axis. In this case, both $\sin \theta$ and $y$ are negative. Thus the relation continues to hold. When $\theta = 0$ and $\theta = \pi$ we have that $y = 0$ and $\sin \theta = 0 = \dfrac y r$. Hence the result. {{qed}}
=== Definition from Triangle implies Definition from Circle === Let $\sin \theta$ be defined as $\dfrac {\text{Opposite}} {\text{Hypotenuse}}$ in a right triangle. Consider the triangle $\triangle OAP$. By construction, $\angle OAP$ is a right angle. Thus: {{begin-eqn}} {{eqn | l = \sin \theta       | r = \frac {AP} {OP}       | c =  }} {{eqn | r = \frac {AP} 1       | c = as $OP$ is the radius of the unit circle }} {{eqn | r = AP       | c =  }} {{end-eqn}} That is: : $\sin \theta = AP$ {{qed|lemma}} === Definition from Circle implies Definition from Triangle === Let $\sin \theta$ be defined as the length of $AP$ in the triangle $\triangle OAP$. Compare $\triangle OAP$ with $\triangle ABC$ in the diagram above. We have that: :$\angle CAB = \angle POA = \theta$ :$\angle ABC = \angle OAP$ which is a right angle Therefore by Triangles with Two Equal Angles are Similar it follows that $\triangle OAP$ and $\triangle ABC$ are similar. By definition of similarity: {{begin-eqn}} {{eqn | l = \frac {\text{Opposite} } {\text{Hypotenuse} }       | r = \frac {BC} {AC}       | c = by definition }} {{eqn | r = \frac {AP} {OP}       | c = by definition of similarity }} {{eqn | r = AP       | c = as $OP$ is the radius of the unit circle }} {{eqn | r = \sin \theta       | c = by definition }} {{end-eqn}} That is: :$\dfrac {\text{Opposite} } {\text{Hypotenuse} } = \sin \theta$ {{qed}} === Definition from Circle equivalent to Definition from Real Numbers === Let, $\sin_G$ be the Geometric Sine from Definition:Sine/Definition from Circle. $\arcsin_G$ is the inverse of this function. Let $sin_A$ be the analytic sine function for real numbers, the one defined by Definition:Sine/Real Numbers. $\arcsin_A$ is the inverse of this function. We know from Arcsin as an Integral that $\arcsin_A$ and $\arcsin_G$ are the same function. $x=\sin_A\left({\theta}\right) \iff \arcsin_A\left(x\right)=\theta \iff \arcsin_G\left(x\right)=\theta \iff x=\sin_G\left({\theta}\right)$ {{qed}} Category:Sine Function 1scrax0c4uptod4hbu34yyiidb1j1v1
=== Definition from Triangle implies Definition from Circle === Let $\cos \theta$ be defined as $\dfrac {\text {Adjacent}} {\text {Hypotenuse}}$ in a right triangle. Consider the triangle $\triangle OAP$. By construction, $\angle OAP$ is a right angle. From Parallelism implies Equal Alternate Angles: :$\angle OPA = \theta$ Thus: {{begin-eqn}} {{eqn | l = \cos \theta       | r = \frac {AP} {OP}       | c =  }} {{eqn | r = \frac {AP} 1       | c = as $OP$ is the radius of the unit circle }} {{eqn | r = AP       | c =  }} {{end-eqn}} That is: :$\cos \theta = AP$ {{qed|lemma}} === Definition from Circle implies Definition from Triangle === Let $\cos \theta$ be defined as the length of $AP$ in the triangle $\triangle OAP$. Compare $\triangle OAP$ with $\triangle ABC$ in the diagram above. From Parallelism implies Equal Alternate Angles: :$\angle OPA = \theta$ We have that: :$\angle CAB = \angle OPA = \theta$ :$\angle ABC = \angle OAP$ which is a right angle Therefore by Triangles with Two Equal Angles are Similar it follows that $\triangle OAP$ and $\triangle ABC$ are similar. By definition of similarity: {{begin-eqn}} {{eqn | l = \frac {\text {Adjacent} } {\text {Hypotenuse} }       | r = \frac {AB} {AC}       | c = by definition }} {{eqn | r = \frac {AP} {OP}       | c = {{Defof|Similar Triangles}} }} {{eqn | r = AP       | c = as $OP$ is the radius of the unit circle }} {{eqn | r = \cos \theta       | c = by definition }} {{end-eqn}} That is: :$\dfrac {\text {Adjacent} } {\text {Hypotenuse} } = \cos \theta$ {{qed}} Category:Cosine Function ba20u20meju29n41namtd4sbpwlr2mq
=== Definition from Triangle implies Definition from Circle === Let $\tan \theta$ be defined as $\dfrac {\text{Opposite}} {\text{Adjacent}}$ in a right triangle. Consider the triangle $\triangle OAB$. By construction, $\angle OAB$ is a right angle. Thus: {{begin-eqn}} {{eqn | l = \tan \theta       | r = \frac {AB} {OA}       | c =  }} {{eqn | r = \frac {AB} 1       | c = as $OA$ is the radius of the unit circle }} {{eqn | r = AB       | c =  }} {{end-eqn}} That is: : $\tan \theta = AB$ {{qed|lemma}} === Definition from Circle implies Definition from Triangle === Let $\tan \theta$ be defined as the length of $AB$ in the triangle $\triangle OAB$. Compare $\triangle OAB$ with $\triangle ABC$ in the diagram above. We have that: :$\angle CAB = \angle BOA = \theta$ :$\angle ABC = \angle OAB$ which is a right angle Therefore by Triangles with Two Equal Angles are Similar it follows that $\triangle OAB$ and $\triangle ABC$ are similar. By definition of similarity: {{begin-eqn}} {{eqn | l = \frac {\text{Opposite} } {\text{Adjacent} }       | r = \frac {BC} {AB}       | c = by definition }} {{eqn | r = \frac {AB} {OA}       | c = by definition of similarity }} {{eqn | r = AB       | c = as $OA$ is the radius of the unit circle }} {{eqn | r = \tan \theta       | c = by definition }} {{end-eqn}} That is: :$\dfrac {\text{Opposite} } {\text{Adjacent} } = \tan \theta$ {{qed}} Category:Tangent Function 5ofok9i18yt87q3gyq6dlo0dmkh555x
Let $X \in \powerset S$. Since $X \subseteq S$, it follows that: :$X' \subseteq X \implies X' \in \powerset S$ because the Subset Relation is Transitive. From Cardinality of Power Set of Finite Set, it follows that for any $X \in \powerset S$: :$\set {X' \in \powerset S: X' \subseteq X}$ has $2^{\card X}$ elements. Therefore, the cardinality of $\subseteq$ is given by: :$\displaystyle \sum_{X \mathop \subseteq S} 2^{\card X}$ Let us split the sum over $\card X$: :$\displaystyle \sum_{X \mathop \subseteq S} 2^{\card X} = \sum_{k \mathop = 0}^n \sum_{\substack {X \mathop \subseteq S \\ \card X \mathop = n}} 2^{\card X}$ It now follows from Cardinality of Set of Subsets that: :$\displaystyle \card \subseteq = \sum_{k \mathop = 0}^n \binom n k 2^k$ From the Binomial Theorem: :$\displaystyle \sum_{k \mathop = 0}^n \binom n k 2^k = \paren {1 + 2}^n$ Hence: :$\card \subseteq = 3^n$ {{qed}}
From Quotient Structure is Well-Defined we have that $\circ_\RR$ is closed on $S / \RR$. Let $\eqclass x \RR, \eqclass y \RR, \eqclass z \RR \in S / \RR$. We shall prove that $\circ_\RR$ is associative: {{begin-eqn}} {{eqn | l = \paren {\eqclass x \RR \circ_{S / \RR} \eqclass y \RR} \circ_{S / \RR} \eqclass z \RR       | r = \eqclass {x \circ y} \RR \circ_{S / \RR} \eqclass z \RR       | c = {{Defof|Operation Induced on Quotient Set|Operation Induced on $S / \RR$ by $\circ$}} }} {{eqn | r = \eqclass {\paren {x \circ y} \circ z} \RR       | c = {{Defof|Operation Induced on Quotient Set|Operation Induced on $S / \RR$ by $\circ$}} }} {{eqn | r = \eqclass {x \circ \paren {y \circ z} } \RR       | c = $\circ$ is Associative }} {{eqn | r = \eqclass x \RR \circ_{S / \RR} \eqclass {y \circ z} \RR       | c = {{Defof|Operation Induced on Quotient Set|Operation Induced on $S / \RR$ by $\circ$}} }} {{eqn | r = \eqclass x \RR \circ_{S / \RR} \paren {\eqclass y \RR \circ_{S / \RR} \eqclass z \RR}       | c = {{Defof|Operation Induced on Quotient Set|Operation Induced on $S / \RR$ by $\circ$}} }} {{end-eqn}} Hence $\struct {S / \RR, \circ_\RR}$ is a semigroup. {{qed}}
From Quotient Structure of Semigroup is Semigroup $\struct {S / \RR, \circ_\RR}$ is a semigroup. Let $\eqclass x {\RR} \in S / \RR$. Consider $\eqclass e \RR$: {{begin-eqn}} {{eqn | l = \eqclass x \RR \circ_{S / \RR} \eqclass e \RR       | r = \eqclass {x \circ e} \RR       | c = {{Defof|Operation Induced on Quotient Set|Operation Induced on $S / \RR$ by $\circ$}} }} {{eqn | r = \eqclass x \RR       | c = {{Defof|Identity Element}} }} {{end-eqn}} Furthermore: {{begin-eqn}} {{eqn | l = \eqclass e \RR \circ_{S / \RR} \eqclass x \RR       | r = \eqclass {e \circ x} \RR       | c = {{Defof|Operation Induced on Quotient Set|Operation Induced on $S / \RR$ by $\circ$}} }} {{eqn | r = \eqclass x \RR       | c = {{Defof|Identity Element}} }} {{end-eqn}} Hence $\eqclass e \RR$ is an identity. Hence $\struct {S / \RR, \circ_\RR}$ is a monoid. {{qed}}
From Quotient Structure of Monoid is Monoid $\struct {G / \RR, \circ_\RR}$ is a monoid with $\eqclass e \RR$ as its identity. Let $\eqclass x \RR \in S / \RR$. Consider $\eqclass {-x} \RR$ where $-x$ denotes the inverse of $x$ under $\circ$. {{begin-eqn}} {{eqn | l = \eqclass x \RR \circ_{S / \RR} \eqclass {-x} \RR       | r = \eqclass {x \circ -x} \RR       | c = {{Defof|Operation Induced on Quotient Set|Operation Induced on $S / \RR$ by $\circ$}} }} {{eqn | r = \eqclass e \RR       | c = {{Defof|Inverse Element}} }} {{end-eqn}} Furthermore: {{begin-eqn}} {{eqn | l = \eqclass {-x} \RR \circ_{S / \RR} \eqclass x \RR       | r = \eqclass {-x \circ x} \RR       | c = {{Defof|Operation Induced on Quotient Set|Operation Induced on $S / \RR$ by $\circ$}} }} {{eqn | r = \eqclass e \RR       | c = {{Defof|Inverse Element}} }} {{end-eqn}} Hence $\eqclass {-x} \RR$ is the inverse of $\eqclass x \RR$. Hence $\struct {G / \RR, \circ_\RR}$ is a group. {{qed}}
From Quotient Structure of Group is Group we have that $\struct {G / \RR, \circ_\RR}$ is a group. Let $\eqclass x \RR, \eqclass y \RR \in S / \RR$. {{begin-eqn}} {{eqn | l = \eqclass x \RR \circ_{S / \RR} \eqclass y \RR       | r = \eqclass {x \circ y} \RR       | c = {{Defof|Operation Induced on Quotient Set|Operation Induced on $S / \RR$ by $\circ$}} }} {{eqn | r = \eqclass {y \circ x} \RR       | c = $\circ$ is Commutative }} {{eqn | r = \eqclass y \RR \circ_{S / \RR} \eqclass x \RR       | c = {{Defof|Operation Induced on Quotient Set|Operation Induced on $S / \RR$ by $\circ$}} }} {{end-eqn}} Hence $\circ_{S / \RR}$ is commutative. Hence $\struct {G / \RR, \circ_\RR}$ is an abelian group. {{qed}}
=== Quotient Structure of Semigroup is Semigroup === {{:Quotient Structure of Semigroup is Semigroup}} === Quotient Structure of Monoid is Monoid === {{:Quotient Structure of Monoid is Monoid}} === Quotient Structure of Group is Group === {{:Quotient Structure of Group is Group}} === Quotient Structure of Abelian Group is Abelian Group === {{:Quotient Structure of Abelian Group is Abelian Group}} Category:Quotient Structures gk18p6202ns25er5kyotuajq3xk5pi4
From Derivative of Secant Function: :$\map {D_x} {\sec x} = \dfrac {\sin x} {\cos^2 x}$ From Sine and Cosine are Periodic on Reals: Corollary: :$\forall x \in \openint {-\pi} \pi \setminus \set {-\dfrac \pi 2, \dfrac \pi 2}: \cos x \ne 0$ Thus, from Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive: :$\forall x \in \openint {-\pi} \pi \setminus \set {-\dfrac \pi 2, \dfrac \pi 2}: \cos^2 x > 0$ {{improve|Might need to find a less abstract-algebraic version of the above result}} From Sine and Cosine are Periodic on Reals: Corollary: :$\sin x > 0$ on the open interval $\openint 0 \pi$ It follows that: :$\forall x \in \openint 0 \pi \setminus \set {\dfrac \pi 2}: \dfrac {\sin x} {\cos^2 x} > 0$ From Sine and Cosine are Periodic on Reals: Corollary:: :$\sin x < 0$ on the open interval $\openint {-\pi} 0$ It follows that: :$\forall x \in \openint {-\pi} 0 \setminus \set {-\dfrac \pi 2}: \dfrac {\sin x} {\cos^2 x} < 0$ Thus, $(1)$ and $(2)$ follow from Derivative of Monotone Function and Differentiable Function is Continuous. From Zeroes of Sine and Cosine:: :$\cos - \dfrac \pi 2 = \cos \dfrac \pi 2 = \cos \dfrac {3 \pi} 2 = 0$ From Sine and Cosine are Periodic on Reals: Corollary: :$\cos x > 0$ on the open interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$ From the same source: :$\cos x < 0$ on the open interval $\openint {\dfrac \pi 2} {\dfrac {3 \pi} 2}$ Thus, $(3)$, $(4)$, $(5)$ and $(6)$ follow from Infinite Limit Theorem. === Graph of Secant Function === {{:Shape of Secant Function/Graph}} {{qed}}
From Derivative of Cosecant Function:: :$\map {D_x} {\csc x} = -\dfrac {\cos x} {\sin^2 x}$ From Sine and Cosine are Periodic on Reals: Corollary: :$\forall x \in \openint {-\dfrac \pi 2} {\dfrac {3 \pi} 2} \setminus \set {0, \pi}: \sin x \ne 0$ Thus, from Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive: :$\forall x \in \openint {-\dfrac \pi 2} {\dfrac {3 \pi} 2} \setminus \set {0, \pi}: \sin^2 x > 0$ {{improve|Might need to find a less abstract-algebraic version of the above result}} From Sine and Cosine are Periodic on Reals: Corollary: :$\cos x > 0$ on the open interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$ It follows that: :$\forall x \in \openint {-\dfrac \pi 2} {\dfrac \pi 2} \setminus \set 0: -\dfrac {\cos x} {\sin^2 x} < 0$ From Sine and Cosine are Periodic on Reals: Corollary: :$\cos x < 0$ on the open interval $\openint {\dfrac \pi 2} {\dfrac {3 \pi} 2}$ It follows that: :$\forall x \in \openint {\dfrac \pi 2} {\dfrac {3 \pi} 2} \setminus \set \pi: -\dfrac {\cos x} {\sin^2 x} > 0$ Thus, $(1)$ and $(2)$ follow from Derivative of Monotone Function. From Zeroes of Sine and Cosine: $\sin 0 = \sin \pi = \sin 2 \pi = 0$. From Sine and Cosine are Periodic on Reals: Corollary: :$\sin x > 0$ on the open interval $\openint 0 \pi$ From the same source: :$\sin x < 0$ on the open interval $\openint \pi {2 \pi}$ Thus, $(3)$, $(4)$, $(5)$ and $(6)$ follow from Infinite Limit Theorem. === Graph of Cosecant Function === {{:Shape of Cosecant Function/Graph}} {{qed}}