hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \frac {z_1} {z_2}       | r = z_1 \paren {z_2}^{-1}       | c = {{Defof|Complex Division}} }} {{eqn | r = \paren {a_1 + i b_1} \dfrac {a_2 - i b_2} {a_2^2 + b_2^2}       | c = Inverse for Complex Multiplication }} {{eqn | r = \frac {\paren {a_1 a_2 + b_1 b_2} + i \paren {a_2 b_1 - a_1 b_2} } {a_2^2 + b_2^2}       | c = {{Defof|Complex Multiplication}} }} {{eqn | r = \frac {a_1 a_2 + b_1 b_2} {a_2^2 + b_2^2} + i \frac {a_2 b_1 - a_1 b_2} {a_2^2 + b_2^2}       | c =  }} {{end-eqn}} {{qed}}
Proof by induction: For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition: :$\paren {r \paren {\cos x + i \sin x} }^n = r^n \paren {\map \cos {n x} + i \, \map \sin {n x} }$ $\map P 1$ is the case: :$\paren {r \paren {\cos x + i \sin x} }^1 = r^1 \paren {\map \cos {1 x} + i \, \map \sin {1 x} }$ which is trivially true. === Basis for the Induction === $\map P 2$ is the case: :$\paren {r \paren {\cos x + i \sin x} }^2 = r^2 \paren {\map \cos {n x} + i \, \map \sin {2 x} }$ From Product of Complex Numbers in Polar Form, we have: :$r_1 \paren {\cos x_1 + i \sin x_1 } r_2 \paren {\cos x_2 + i \sin x_2} = r_1 r_2 \paren {\map \cos {x_1 + x_2} + i \, \map \sin {x_1 + x_2} }$ Setting $r_1 = r_2 = r$ and $x_1 = x_2 = x$ gives the result. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\paren {r \paren {\cos x + i \sin x} }^k = r^k \paren {\map \cos {k x} + i \, \map \sin {k x} }$ Then we need to show: :$\paren {r \paren {\cos x + i \sin x} }^{k + 1} = r^{k + 1} \paren {\map \cos {\paren {k + 1} x} + i \, \map \sin {\paren {k + 1} x} }$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn| l = \paren {r \paren {\cos x + i \sin x} }^{k + 1}      | r = \paren {r \paren {\cos x + i \sin x} }^k \paren {r \paren {\cos x + i \sin x} } }} {{eqn| r = r^k \paren {\map \cos {k x} + i \, \map \sin {k x} } \paren {r \paren {\cos x + i \sin x} }      | c = Induction Hypothesis }} {{eqn| r = r^{k + 1} \paren {\map \cos {\paren {k + 1} x} + i \, \map \sin {\paren {k + 1} x} }      | c = Product of Complex Numbers in Polar Form }} {{end-eqn}} Hence, by induction, for all $n \in \Z_{> 0}$: :$\paren {r \paren {\cos x + i \sin x} }^n = r^n \paren {\map \cos {n x} + i \, \map \sin {n x} }$ {{qed}}
{{begin-eqn}} {{eqn | l = \frac {z_1} {z_2}       | r = \frac {r_1 \paren {\cos \theta_1 + i \sin \theta_1} } {r_2 \paren {\cos \theta_2 + i \sin \theta_2} }       | c = {{Defof|Polar Form of Complex Number}} }} {{eqn | r = \frac {\paren {r_1 \paren {\cos \theta_1 + i \sin \theta_1} } \paren {r_2 \paren {\cos \theta_2 - i \sin \theta_2} } } {\paren {r_2 \paren {\cos \theta_2 + i \sin \theta_2} } \paren {r_2 \paren {\cos \theta_2 - i \sin \theta_2} } }       | c = multiplying numerator and denominator by $r_2 \paren {\cos \theta_1 - i \sin \theta_1}$ }} {{eqn | r = \frac {r_1 r_2 \paren {\map \cos {\theta_1 - \theta_2} + i \map \sin {\theta_1 - \theta_2} } } {r_2^2 \paren {\map \cos {\theta_2 - \theta_2} + i \map \sin {\theta_2 - \theta_2} } }       | c = Product of Complex Numbers in Polar Form }} {{eqn | r = \frac {r_1 \paren {\map \cos {\theta_1 - \theta_2} + i \map \sin {\theta_1 - \theta_2} } } {r_2 \paren {\cos 0 + i \sin 0} }       | c =  }} {{eqn | r = \frac {r_1} {r_2} \paren {\map \cos {\theta_1 - \theta_2} + i \map \sin {\theta_1 - \theta_2} }       | c = Cosine of Zero is One and Sine of Zero is Zero }} {{end-eqn}} {{qed}}
Let: : $w := r^{1 / n} \paren {\map \cos {\dfrac {\theta + 2 \pi k} n} + i \, \map \sin {\dfrac {\theta + 2 \pi k} n} }$ for $k \in \Z_{>0}$. Then: {{begin-eqn}} {{eqn | l = w^n       | r = \paren {r^{1 / n} \paren {\map \cos {\dfrac {\theta + 2 \pi k} n} + i \, \sin {\dfrac {\theta + 2 \pi k} n} } }^n       | c =  }} {{eqn | r = \paren {r^{1 / n} }^n \paren {\cos n \paren {\dfrac {\theta + 2 \pi k} n} + i \sin n \paren {\dfrac {\theta + 2 \pi k} n} }       | c = De Moivre's Formula }} {{eqn | r = r \paren {\map \cos {\theta + 2 \pi k} + i \, \map \sin {\theta + 2 \pi k} }       | c =  }} {{eqn | r = r \paren {\cos \theta + i \, \map \sin {\theta + 2 \pi k} }       | c = Cosine of Angle plus Multiple of Full Angle }} {{eqn | r = r \paren {\cos \theta + i \sin \theta}       | c = Sine of Angle plus Multiple of Full Angle }} {{end-eqn}} Now let $m = k + n$. Then: {{begin-eqn}} {{eqn | l = \frac {\theta + 2 m \pi} n        | r = \frac {\theta + 2 \paren {k + n} \pi} n       | c =  }} {{eqn | r = \frac {\theta + 2 k \pi} n + \frac {2 n \pi} n       | c =  }} {{eqn | r = \frac {\theta + 2 k \pi} n + 2 \pi       | c =  }} {{eqn | r = \frac {\theta + 2 k \pi} n + 2 \pi       | c =  }} {{eqn | ll= \leadsto       | l = \cos \frac {\theta + 2 m \pi} n + i \sin \frac {\theta + 2 m \pi} n       | r = \cos \frac {\theta + 2 k \pi} n + i \sin \frac {\theta + 2 k \pi} n       | c = from above }} {{end-eqn}} exploiting the fact that Sine and Cosine are Periodic on Reals. The result follows. {{qed}}
We are going to prove that $\map {f'} x = n x^{n - 1}$ holds for all real $n$. To do this, we compute the limit $\displaystyle \lim_{h \mathop \to 0} \frac {\paren {x + h}^n - x^n} h$: {{begin-eqn}} {{eqn | l = \frac {\paren {x + h}^n - x^n} h       | r = \frac {x^n} h \paren {\paren {1 + \frac h x}^n - 1}       | c =  }} {{eqn | r = \frac{x^n} h \paren {e^{n \map \ln {1 + \frac h x} } - 1}       | c =  }} {{eqn | r = x^n \cdot \frac {e^{n \map \ln {1 + \frac h x} } - 1} {n \map \ln {1 + \frac h x} } \cdot \frac {n \map \ln {1 + \frac h x} } {\frac h x} \cdot \frac 1 x       | c =  }} {{end-eqn}} Now we use the following results: :$\displaystyle \lim_{x \mathop \to 0} \frac {\exp x - 1} x = 1$ from Derivative of Exponential at Zero :$\displaystyle \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x = 1$ from Derivative of Logarithm at One to obtain: :$x^n \cdot \dfrac {e^{n \map \ln {1 + \frac h x} } - 1} {n \map \ln {1 + \dfrac h x} } \cdot \dfrac {n \map \ln {1 + \dfrac h x}} {\dfrac h x} \cdot \dfrac 1 x \to n x^{n - 1}$ as $h \to 0$ Hence the result. {{qed}}
Note this proof does not hold for $x = 0$. Let $y$ = $\map f x$. Then $y = x^n$. Then: {{begin-eqn}} {{eqn | l = y       | r = x^n }} {{eqn | ll= \leadsto       | l = \size y       | r = \size {x^n}       | c = taking the absolute value of both sides }} {{eqn | r = \size x^n       | c = Absolute Value of Power }} {{eqn | ll= \leadsto       | l = \ln \size y       | r = \map \ln {\size x^n}       | c = taking the natural logarithm of both sides }} {{eqn | r = n \ln \size x       | c = Logarithm of Power }} {{end-eqn}} Using: * Derivative of Composite Function * Derivative of Constant Multiple * Corollary to Primitive of Reciprocal and taking the derivative of both sides with respect to $x$ gives: :$\dfrac 1 y \dfrac {\d y} {\d x} = n \dfrac 1 x$ Multiplying both sides of the equation by $y$ yields: :$\dfrac {\d y} {\d x} = n \dfrac y x$ Substituting $x^n$ for $y$: :$\dfrac {\d y} {\d x} = n \dfrac {x^n} x$ From Quotient of Powers: :$\dfrac {\d y} {\d x} = n x^{n - 1}$ {{qed}}
{{begin-eqn}} {{eqn | l = a^0       | r = \map \exp {0 \ln a}       | c = {{Defof|Power to Real Number}} }} {{eqn | r = \map \exp 0       | c =  }} {{eqn | r = 1       | c = Exponential of Zero }} {{end-eqn}} {{qed}}
We have that the exponential function is the inverse of the natural logarithm function: : $\ln e = 1$ Hence the result. {{qed}}
We have that the exponential function is the inverse of the natural logarithm function: : $\ln 1 = 0$ Hence the result. {{qed}}
Using the definition of the exponential as a limit of a sequence: {{begin-eqn}} {{eqn | l = \exp 0       | r = \lim_{n \mathop \to \infty} \left({1 + \frac 0 n}\right)^n }} {{eqn | r = 1 }} {{end-eqn}} {{qed}}
This proof assumes the series definition of $\exp$. That is, let: :$\displaystyle \exp x = \sum_{k \mathop = 0}^{\infty} \frac {x^k} {k!}$ Then: {{begin-eqn}} {{eqn | l = \exp 0       | r = \sum_{k \mathop = 0}^{\infty} \frac {0^k} {k!} }} {{eqn | r = 1       | c = Definition of $0^k$ }} {{end-eqn}} {{qed}}
This proof assumes the Definition of $\exp x$ as the unique continuous extension of $e^x$. {{begin-eqn}} {{eqn | l = \exp 0       | r = e^0 }} {{eqn | r = 1       | c = {{Defof|Power (Algebra)|$x^0$}}, where $x \ne 0$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \exp {z + \paren {-z} }       | r = \exp z \, \map \exp {-z}       | c = Exponential of Sum }} {{eqn | ll= \leadsto       | l = \map \exp {z - z}       | r = \dfrac {\exp z} {\exp z}       | c =  }} {{eqn | ll= \leadsto       | l = \exp 0       | r = 1       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \log_b x - \log_b y       | r = \map {\log_b} {b^{\log_b x - \log_b y} }       | c = {{Defof|General Logarithm}} }} {{eqn | r = \map {\log_b} {\frac {\paren {b^{\log_b x} } } {\paren {b^{\log_b y} } } }       | c = Quotient of Powers }} {{eqn | r = \map {\log_b} {\frac x y}       | c = {{Defof|General Logarithm}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \log_b x - \log_b y       | r = \frac {\log_e x} {\log_e b} - \frac {\log_e y} {\log_e b}       | c = Change of Base of Logarithm }} {{eqn | r = \frac {\log_e x - \log_e y} {\log_e b}       | c =  }} {{eqn | r = \frac {\log_e \left({\frac x y}\right)} {\log_e b}       | c = Difference of Logarithms: Proof for Natural Logarithm }} {{eqn | r = \log_b \left({\frac x y}\right)       | c = Change of Base of Logarithm }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {\log_b} {\frac x y} + \log_b y       | r = \map {\log_b} {\frac x y \times y}       | c = Sum of Logarithms }} {{eqn | r = \log_b x       | c =  }} {{eqn | ll= \leadsto       | l = \map {\log_b} {\frac x y}       | r = \log_b x - \log_b y       | c = subtracting $\log_b y$ from both sides }} {{end-eqn}} {{qed}}
Since $e$ is invertible, the power of $e$ is defined for all $n \in \Z$. We prove the case $n \ge 0$ by induction. === Basis for the Induction === By definition of power of monoid element: :$e^0 = e$ so the theorem holds for $n = 0$.  This is our basis for the induction. === Induction Hypothesis === Our induction hypothesis is that the theorem is true for $n = k$: :$e^k = e$ === Induction Step === In the induction step, we prove that the theorem is true for $n = k + 1$. We have: {{begin-eqn}} {{eqn | l = e^{k + 1}       | r = e^k \circ e       | c = {{Defof|Power of Element of Monoid}} }} {{eqn | r = e^k       | c = {{Defof|Identity Element}} }} {{eqn | r = e       | c = Induction Hypothesis }} {{end-eqn}} Therefore, by Principle of Mathematical Induction: :$\forall n \in \Z_{\ge 0} : e^n = e$ {{qed|lemma}} Now we prove the case $n < 0$. We have: {{begin-eqn}} {{eqn | l = e^n       | r = \paren {e^{-n} }^{-1}       | c = {{Defof|Power of Element of Monoid}} }} {{eqn | r = e^{-1}       | c = since $-n > 0$ }} {{eqn | r = e       | c = Inverse of Identity Element is Itself }} {{end-eqn}} Thus: :$\forall n \in \Z : e^n = e$ {{qed}}
{{begin-eqn}} {{eqn | l = \tan z       | r = \frac {\sin z} {\cos z}       | c = {{Defof|Complex Tangent Function}} }} {{eqn | r = \frac {\frac 1 2 i \paren {e^{-i z} - e^{i z} } }                   {\frac 1 2 \paren {e^{-i z} + e^{i z} } }       | c = Sine Exponential Formulation and Cosine Exponential Formulation }} {{eqn | r = i \frac {e^{-i z} - e^{i z} }                     {e^{-i z} + e^{i z} } }} {{eqn | r = i \frac {1 - e^{2 i z} }                     {1 + e^{2 i z} }       | c = multiplying numerator and denominator by $e^{i z}$ }} {{end-eqn}} {{qed}} Category:Tangent Exponential Formulation jupkkkg9vg2be4qs1g0evp04zf8wctk
{{begin-eqn}} {{eqn | l = \tan z       | r = \frac {\sin z} {\cos z}       | c = {{Defof|Complex Tangent Function}} }} {{eqn | r = \frac {e^{i z} - e^{-i z} } {2 i} / \frac {e^{i z} + e^{-i z} } 2       | c = Sine Exponential Formulation and Cosine Exponential Formulation }} {{eqn | r = \frac {e^{i z} - e^{-i z} } {i \paren {e^{i z} + e^{-i z} } }       | c = multiplying top and bottom by $2 i$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \tan z       | r = \frac {\sin z} {\cos z}       | c = {{Defof|Complex Tangent Function}} }} {{eqn | r = \frac {e^{i z} - e^{-i z} } {i \paren {e^{i z} + e^{-i z} } }       | c = Tangent Exponential Formulation: Formulation 2 }} {{eqn | r = -i \paren {\dfrac {e^{i z} - e^{-i z} } {e^{i z} + e^{-i z} } }       | c = multiplying numerator and denominator by $-i$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cot z       | r = \frac {\cos z} {\sin z}       | c = {{Defof|Complex Cotangent Function}} }} {{eqn | r = \frac {e^{i z} + e^{-i z} } 2 / \frac {e^{i z} - e^{-i z} } {2 i}       | c = Sine Exponential Formulation and Cosine Exponential Formulation }} {{eqn | r = i \frac {e^{i z} + e^{-i z} } {e^{i z} - e^{-i z} }       | c = multiplying numerator and denominator by $2 i$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cot z       | r = \frac 1 {\tan z}       | c = {{Defof|Complex Cotangent Function}} }} {{eqn | r = 1 / \dfrac {e^{i z} - e^{-i z} } {i \left({e^{i z} + e^{-i z} }\right)}       | c = Tangent Exponential Formulation/Formulation 2 }} {{eqn | r = i \frac {e^{i z} + e^{-i z} } {e^{i z} - e^{-i z} }       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \csc z       | r = \frac 1 {\sin z}       | c = {{Defof|Complex Cosecant Function}} }} {{eqn | r = 1 / \frac {e^{i z} - e^{-i z} } {2 i}       | c = Sine Exponential Formulation }} {{eqn | r = \frac {2 i} {e^{i z} - e^{-i z} }       | c = multiplying top and bottom by $2 i$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sec z       | r = \frac 1 {\cos z}       | c = {{Defof|Complex Secant Function}} }} {{eqn | r = 1 / \frac {e^{i z} + e^{-i z} } 2       | c = Sine Exponential Formulation and Cosine Exponential Formulation }} {{eqn | r = \frac 2 {e^{i z} + e^{-i z} }       | c = multiplying top and bottom by $2$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = e^{i \paren {\theta + 2 k \pi} }       | r = \map \cos {\theta + 2 k \pi} + i \, \map \sin {\theta + 2 k \pi}       | c = Euler's Formula }} {{eqn | r = \cos \theta + i \sin \theta       | c = Sine and Cosine are Periodic on Reals }} {{eqn | r = e^{i \theta}       | c = Euler's Formula }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = z_1 z_2       | r = r_1 e^{i \theta_1} r_2 e^{i \theta_2}       | c = {{Defof|Exponential Form of Complex Number}} }} {{eqn | r = r_1 \paren {\cos \theta_1 + i \sin \theta_1 } r_2 \paren {\cos \theta_2 + i \sin \theta_2}       | c = Euler's Formula }} {{eqn | r = r_1 r_2 \paren {\map \cos {\theta_1 + \theta_2} + i \map \sin {\theta_1 + \theta_2} }       | c = Product of Complex Numbers in Polar Form }} {{eqn | r = r_1 r_2 e^{i \paren {\theta_1 + \theta_2} }       | c = Euler's Formula }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac {z_1} {z_2}       | r = \frac {r_1 e^{i \theta_1} } {r_2 e^{i \theta_2} }       | c = {{Defof|Exponential Form of Complex Number}} }} {{eqn | r = \frac {r_1 \paren {\cos \theta_1 + i \sin \theta_1 } } {r_2 \paren {\cos \theta_2 + i \sin \theta_2} }       | c = Euler's Formula }} {{eqn | r = \frac {r_1} {r_2} \paren {\map \cos {\theta_1 - \theta_2} + i \map \sin {\theta_1 - \theta_2} }       | c = Division of Complex Numbers in Polar Form }} {{eqn | r = \frac {r_1} {r_2} e^{i \paren {\theta_1 - \theta_2} }       | c = Euler's Formula }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \paren {r e^{i \theta} }^\omega       | r = \paren {r \paren {\cos x + i \sin x} }^\omega       | c = {{Defof|Exponential Form of Complex Number}} }} {{eqn | r = r^\omega \paren {\cos \omega x + i \omega \sin x}       | c = De Moivre's Formula }} {{eqn | r = r^\omega e^{i \omega \theta}       | c = {{Defof|Exponential Form of Complex Number}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = z^{1 / n}       | r = \paren {r e^{i \theta} }^{1 / n}       | c = {{Defof|Exponential Form of Complex Number}} }} {{eqn | r = \paren {r \paren {\cos x + i \sin x} }^{1 / n}       | c = {{Defof|Polar Form of Complex Number}} }} {{eqn | r = \set {r^{1 / n} \paren {\cos \paren {\dfrac {\theta + 2 \pi k} n} + i \sin \paren {\dfrac {\theta + 2 \pi k} n} }: k \in \set {0, 1, 2, \ldots, n - 1} }       | c = Roots of Complex Number }} {{eqn | r = \set {r^{1 / n} e^{i \paren {\theta + 2 \pi k} / n}: k \in \set {0, 1, 2, \ldots, n - 1} }       | c = {{Defof|Exponential Form of Complex Number}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sech^2 x + \tanh^2 x       | r = \paren {\frac 2 {e^x + e^{-x} } }^2 + \tanh^2 x       | c = {{Defof|Hyperbolic Secant|index = 1}} }} {{eqn | r = \paren {\frac 2 {e^x + e^{-x} } }^2 + \paren {\frac {e^x - e^{-x} } {e^x + e^{-x} } }^2         | c = {{Defof|Hyperbolic Tangent|index = 1}} }} {{eqn | r = \frac {4 + e^{2 x} - 2 + e^{-2 x} } {e^{2 x} + 2 + e ^{-2 x} }       | c = Exponent Combination Laws }} {{eqn | r = \frac {e^{2 x} + 2 + e ^{-2 x} } {e^{2 x} + 2 + e ^{-2 x} } }} {{eqn | r = 1 }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \coth^2 x - \csch^2 x       | r = \paren {\frac {e^x + e^{-x} } {e^x - e^{-x} } }^2 - \csch^2 x       | c = {{Defof|Hyperbolic Cotangent|index = 1}} }} {{eqn | r = \paren {\frac {e^x + e^{-x} } {e^x - e^{-x} } }^2 - \paren {\frac 2 {e^x - e^{-x} } }^2       | c = {{Defof|Hyperbolic Cosecant|index = 1}} }} {{eqn | r = \frac {e^{2 x} + 2 + e^{-2 x} - 4} {e^{2 x} - 2 + e ^{-2 x} }       | c = Exponent Combination Laws }} {{eqn | r = \frac {e^{2 x} - 2 + e ^{-2 x} } {e^{2 x} - 2 + e ^{-2 x} } }} {{eqn | r = 1 }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \csch {-x}       | r = \frac 1 {\map \sinh {-x} }        | c = {{Defof|Hyperbolic Cosecant|index = 2}} }} {{eqn | r = \frac 1 {-\sinh x}       | c = Hyperbolic Sine Function is Odd }} {{eqn | r = -\csch x }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \coth {-x}       | r = \frac {\map \cosh {-x} } {\map \sinh {-x} }       | c = {{Defof|Hyperbolic Cotangent|index = 2}} }} {{eqn | r = \frac {\map \cosh {-x} }  {-\sinh x}       | c = Hyperbolic Sine Function is Odd }} {{eqn | r = \frac {\cosh x} {-\sinh x}       | c = Hyperbolic Cosine Function is Even }} {{eqn | r = -\coth x }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sinh a \cosh b + \cosh a \sinh b       | r = \frac {e^a - e^{-a} } 2 \frac {e^b + e^{-b} } 2 + \frac {e^a + e^{-a} } 2 \frac {e^b - e^{-b} } 2       | c = {{Defof|Hyperbolic Sine}} and {{Defof|Hyperbolic Cosine}} }} {{eqn | r = \frac {e^{a + b} - e^{-a + b} + e^{a - b} - e^{-a - b} } 4        | c = Exponential of Sum }} {{eqn | o =        | ro= +       | r = \frac {e^{a + b} + e^{-a + b} - e^{a - b} - e^{-a - b} } 4       | c =  }} {{eqn | r = \frac {2 e^{a + b} - 2 e^{-\paren {a + b} } } 4       | c =  }} {{eqn | r = \frac {e^{a + b} - e^{-\paren {a + b} } } 2       | c =  }} {{eqn | r = \map \sinh {a + b}       | c = {{Defof|Hyperbolic Sine}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | r = \cosh a \cosh b + \sinh a \sinh b       | o =        | c =  }} {{eqn | r = \frac {e^a + e^{-a} } 2 \frac {e^b + e^{-b} } 2 + \frac {e^a - e^{-a} } 2 \frac {e^b - e^{-b} } 2       | c = {{Defof|Hyperbolic Sine}} and {{Defof|Hyperbolic Cosine}} }} {{eqn | r = \frac {e^{a + b} + e^{-a + b} + e^{a + b} + e^{-a - b} } 4        | c = Exponential of Sum }} {{eqn | o =        | ro= +       | r = \frac {e^{a + b} - e^{-a + b} - e^{a - b} + e^{-a - b} } 4       | c =  }} {{eqn | r = \frac {2 e^{a + b} + 2 e^{-\paren {a + b} } } 4       | c =  }} {{eqn | r = \frac {e^{a + b} + e^{-\paren {a + b} } } 2       | c =  }} {{eqn | r = \map \cosh {a + b}       | c = {{Defof|Hyperbolic Cosine}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \tanh {a + b}       | r = \frac {\map \sinh {a + b} } {\map \cosh {a + b} }       | c = {{Defof|Hyperbolic Tangent|index = 2}} }} {{eqn | r = \frac {\sinh a \cosh b + \cosh a \sinh b} {\cosh a \cosh b + \sinh a \sinh b}       | c = Hyperbolic Sine of Sum and Hyperbolic Cosine of Sum }} {{eqn | r = \frac {\frac {\sinh a} {\cosh a} + \frac {\sinh b} {\cosh b} } {1 + \frac {\sinh a \sinh b} {\cosh a \cosh b} }       | c = dividing the top and bottom by $\cosh a \cosh b$ }} {{eqn | r = \frac {\tanh a + \tanh b} {1 + \tanh a \tanh b}       | c = {{Defof|Hyperbolic Tangent|index = 2}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \coth {a + b}       | r = \frac {\map \cosh {a + b} } {\map \sinh {a + b} }       | c = {{Defof|Hyperbolic Cotangent|index = 2}} }} {{eqn | r = \frac {\cosh a \cosh b + \sinh a \sinh b} {\sinh a \cosh b + \cosh a \sinh b}       | c = Hyperbolic Sine of Sum and Hyperbolic Cosine of Sum }} {{eqn | r = \frac {\frac {\cosh a \cosh b} {\sinh a \sinh b} + 1} {\frac {\cosh b} {\sinh b} + \frac {\cosh a} {\sinh a} }       | c = dividing the numerator and denominator by $\sinh a \sinh b$ }} {{eqn | r = \frac {\coth a \coth b + 1} {\coth b + \coth a}       | c = Definition of Hyperbolic Cotangent }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cosh 2 x       | r = \map \cosh {x + x} }} {{eqn | r = \cosh x \cosh x + \sinh x \sinh x       | c = Hyperbolic Cosine of Sum }} {{eqn | r = \cosh^2 x + \sinh^2 x }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \tanh 2 x       | r = \frac {\sinh 2 x} {\cosh 2 x}       | c = {{Defof|Hyperbolic Tangent|index = 2}} }} {{eqn | r = \frac {2 \cosh x \sinh x} {\cosh^2 x + \sinh^2 x}       | c = Double Angle Formula for Hyperbolic Sine and Double Angle Formula for Hyperbolic Cosine }} {{eqn | r = \frac {\frac {2 \cosh x \sinh x} {\cosh^2 x} } {\frac {\cosh^2 x + \sinh^2 x} {\cosh^2 x} }       | c = dividing top and bottom by $\cosh^2 x$ }} {{eqn | r = \frac {2 \tanh x} {1 + \tanh^2 x}       | c = {{Defof|Hyperbolic Tangent|index = 2}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \tanh 2 x       | r = \tanh \left({x + x}\right) }} {{eqn | r = \frac {\tanh x + \tanh x} {1 + \tanh x \tanh x}       | c = Hyperbolic Tangent of Sum }} {{eqn | r = \frac {2 \tanh x} {1 + \tanh^2 x} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cosh x       | r = 1 + 2 \ \sinh^2 \frac x 2       | c = Double Angle Formula for Hyperbolic Cosine: Corollary 2 }} {{eqn | ll= \leadsto       | l = 2 \ \sinh^2 \frac x 2       | r = \cosh x - 1 }} {{eqn | ll= \leadsto       | l = \sinh \frac x2       | r = \pm \sqrt {\frac {\cosh x - 1} 2} }} {{end-eqn}} We also have that: :when $x \ge 0$, $\sinh x \ge 0$ :when $x \le 0$, $\sinh x \le 0$. {{qed}}
{{begin-eqn}} {{eqn | l = \cosh x       | r = 2 \ \cosh^2 \frac x 2 - 1       | c = Double Angle Formula for Hyperbolic Cosine: Corollary 1 }} {{eqn | ll= \leadsto       | l = 2 \ \cosh^2 \frac x 2       | r = \cosh x + 1 }} {{eqn | ll= \leadsto       | l = \cosh \frac x 2       | r = \pm \sqrt {\frac {\cosh x + 1} 2} }} {{end-eqn}} As $\forall x \in \R: \cosh x > 0$, the result follows. {{qed}}
{{begin-eqn}} {{eqn | l = \tanh \frac x 2       | r = \frac {\sinh \frac x 2} {\cosh \frac x 2}       | c = {{Defof|Hyperbolic Tangent|index = 2}} }} {{eqn | r = \frac {\pm \sqrt {\frac {\cosh x - 1} 2} } {\pm \sqrt {\frac {\cosh x + 1} 2} }       | c = Half Angle Formula for Hyperbolic Sine and Half Angle Formula for Hyperbolic Cosine }} {{eqn | r = \pm \sqrt {\frac {\cosh x - 1} {\cosh x + 1} } }} {{end-eqn}} We also have that: :when $x \ge 0$, $\tanh \dfrac x 2 \ge 0$ :when $x \le 0$, $\tanh \dfrac x 2 \le 0$. {{qed}}
{{begin-eqn}} {{eqn | l = \tanh \frac x 2       | r = \pm \sqrt {\frac {\cosh x - 1} {\cosh x + 1} }       | c = Half Angle Formula for Hyperbolic Tangent }} {{eqn | r = \pm \sqrt {\frac {\paren {\cosh x - 1} \paren {\cosh x + 1} } {\paren {\cosh x + 1}^2} }       | c = multiplying top and bottom by $\sqrt {\cosh x + 1}$ }} {{eqn | r = \pm \sqrt {\frac {\cosh x - 1} {\paren {\cosh x + 1}^2} }       | c = Difference of Two Squares }} {{eqn | r = \pm \sqrt {\frac {\sinh^2 x} {\paren {\cosh x + 1}^2} }       | c = Difference of Squares of Hyperbolic Cosine and Sine }} {{eqn | r = \pm \frac {\sinh x} {\cosh x + 1} }} {{end-eqn}} Since $\cosh x > 0$, it follows that $\cosh x + 1 > 0$. We also have that: :when $x \ge 0$, $\tanh \dfrac x 2 \ge 0$ and $\sinh x \ge 0$ :when $x \le 0$, $\tanh \dfrac x 2 \le 0$ and $\sinh x \le 0$. Thus: :$\tanh \dfrac x 2 = \dfrac {\sinh x} {\cosh x + 1}$ {{qed}}