Trigonometric and Hyperbolic Functions

Trigonometric Functions Hyperbolic Functions Inverse Trigonometric and Hyperbolic Functions

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Introduction

1. For real numbers θ we have eiθ = cos(θ)+ isin(θ).2. Replacing θ with −θ we obtain e−iθ = cos(θ)− isin(θ).

(Remember that the cosine is even and the sine is odd.)

3. Adding the two and dividing by 2 gives cos(θ) =eiθ + e−iθ

4. Subtracting the two and dividing by 2i gives sin(θ) =eiθ − e−iθ

5. The right sides above make sense for all complex numbers.

Introduction1. For real numbers θ we have eiθ = cos(θ)+ isin(θ).

2. Replacing θ with −θ we obtain e−iθ = cos(θ)− isin(θ).(Remember that the cosine is even and the sine is odd.)

Introduction1. For real numbers θ we have eiθ = cos(θ)+ isin(θ).2. Replacing θ with −θ we obtain e−iθ = cos(θ)− isin(θ).

Definition.

For any complex number z we define

cos(z) =eiz + e−iz

sin(z) =eiz− e−iz

Definition. For any complex number z we define

This Time, All Common Identities Carry Over

1. sin(z1 + z2) = sin(z1)cos(z2)+ cos(z1)sin(z2)2. cos(z1 + z2) = cos(z1)cos(z2)− sin(z1)sin(z2)3. sin2(z)+ cos2(z) = 14. sin(z+2π) = sin(z)5. cos(z+2π) = cos(z)

This Time, All Common Identities Carry Over1. sin(z1 + z2) = sin(z1)cos(z2)+ cos(z1)sin(z2)

2. cos(z1 + z2) = cos(z1)cos(z2)− sin(z1)sin(z2)3. sin2(z)+ cos2(z) = 14. sin(z+2π) = sin(z)5. cos(z+2π) = cos(z)

This Time, All Common Identities Carry Over1. sin(z1 + z2) = sin(z1)cos(z2)+ cos(z1)sin(z2)2. cos(z1 + z2) = cos(z1)cos(z2)− sin(z1)sin(z2)

3. sin2(z)+ cos2(z) = 14. sin(z+2π) = sin(z)5. cos(z+2π) = cos(z)

This Time, All Common Identities Carry Over1. sin(z1 + z2) = sin(z1)cos(z2)+ cos(z1)sin(z2)2. cos(z1 + z2) = cos(z1)cos(z2)− sin(z1)sin(z2)3. sin2(z)+ cos2(z) = 1

4. sin(z+2π) = sin(z)5. cos(z+2π) = cos(z)

This Time, All Common Identities Carry Over1. sin(z1 + z2) = sin(z1)cos(z2)+ cos(z1)sin(z2)2. cos(z1 + z2) = cos(z1)cos(z2)− sin(z1)sin(z2)3. sin2(z)+ cos2(z) = 14. sin(z+2π) = sin(z)

5. cos(z+2π) = cos(z)

This Time, All Common Identities Carry Over1. sin(z1 + z2) = sin(z1)cos(z2)+ cos(z1)sin(z2)2. cos(z1 + z2) = cos(z1)cos(z2)− sin(z1)sin(z2)3. sin2(z)+ cos2(z) = 14. sin(z+2π) = sin(z)5. cos(z+2π) = cos(z)

Some Proofs Actually Are Simpler Now

sin(z1)cos(z2)+ cos(z1)sin(z2)

=eiz1− e−iz1

2ieiz2 + e−iz2

eiz1 + e−iz1

2eiz2− e−iz2

(ei(z1+z2) + ei(z1−z2)− ei(−z1+z2)− e−i(z1+z2)

+ ei(z1+z2)− ei(z1−z2) + ei(−z1+z2)− e−i(z1+z2))

(2ei(z1+z2)−2e−i(z1+z2)

ei(z1+z2)− e−i(z1+z2)

2i= sin(z1 + z2)

This is why many people like to work with the complex definition ofthe trigonometric functions.

=eiz1− e−iz1

2ieiz2 + e−iz2

eiz1 + e−iz1

2eiz2− e−iz2

(2ei(z1+z2)−2e−i(z1+z2)

ei(z1+z2)− e−i(z1+z2)

2i= sin(z1 + z2)

=eiz1− e−iz1

2ieiz2 + e−iz2

eiz1 + e−iz1

2eiz2− e−iz2

(2ei(z1+z2)−2e−i(z1+z2)

ei(z1+z2)− e−i(z1+z2)

2i= sin(z1 + z2)

=eiz1− e−iz1

2ieiz2 + e−iz2

eiz1 + e−iz1

2eiz2− e−iz2

(2ei(z1+z2)−2e−i(z1+z2)

ei(z1+z2)− e−i(z1+z2)

2i= sin(z1 + z2)

=eiz1− e−iz1

2ieiz2 + e−iz2

eiz1 + e−iz1

2eiz2− e−iz2

(2ei(z1+z2)−2e−i(z1+z2)

ei(z1+z2)− e−i(z1+z2)

2i= sin(z1 + z2)

=eiz1− e−iz1

2ieiz2 + e−iz2

eiz1 + e−iz1

2eiz2− e−iz2

(2ei(z1+z2)−2e−i(z1+z2)

=ei(z1+z2)− e−i(z1+z2)

2i= sin(z1 + z2)

=eiz1− e−iz1

2ieiz2 + e−iz2

eiz1 + e−iz1

2eiz2− e−iz2

(2ei(z1+z2)−2e−i(z1+z2)

ei(z1+z2)− e−i(z1+z2)

= sin(z1 + z2)

=eiz1− e−iz1

2ieiz2 + e−iz2

eiz1 + e−iz1

2eiz2− e−iz2

(2ei(z1+z2)−2e−i(z1+z2)

ei(z1+z2)− e−i(z1+z2)

2i= sin(z1 + z2)

=eiz1− e−iz1

2ieiz2 + e−iz2

eiz1 + e−iz1

2eiz2− e−iz2

(2ei(z1+z2)−2e−i(z1+z2)

ei(z1+z2)− e−i(z1+z2)

2i= sin(z1 + z2)

Other Proofs Stay The Same

sin(z+2π) = sin(z)cos(2π)+ cos(z)sin(2π)= sin(z)

sin(z+2π)

= sin(z)cos(2π)+ cos(z)sin(2π)= sin(z)

sin(z+2π) = sin(z)cos(2π)+ cos(z)sin(2π)

= sin(z)

sin(z+2π) = sin(z)cos(2π)+ cos(z)sin(2π)= sin(z)

The Other Trigonometric Functions

1. tan(z) :=sin(z)cos(z)

2. cot(z) :=cos(z)sin(z)

3. sec(z) :=1

cos(z)

4. csc(z) :=1

sin(z)Note that secant and cosecant are not very common (around theworld).

The Other Trigonometric Functions1. tan(z) :=

sin(z)cos(z)

3. sec(z) :=1

cos(z)

4. csc(z) :=1

sin(z)cos(z)

3. sec(z) :=1

cos(z)

4. csc(z) :=1

sin(z)cos(z)

3. sec(z) :=1

cos(z)

4. csc(z) :=1

sin(z)cos(z)

3. sec(z) :=1

cos(z)

4. csc(z) :=1

sin(z)

Note that secant and cosecant are not very common (around theworld).

sin(z)cos(z)

3. sec(z) :=1

cos(z)

4. csc(z) :=1

The Derivatives

sin(z)=ddz

eiz−e−iz

ieiz+ie−iz

eiz+e−iz

2=cos(z)

cos(z) =−sin(z)

tan(z) =1

cos2(z)

cot(z) =− 1sin2(z)

sec(z) = sec(z) tan(z)

csc(z) =−csc(z)cot(z)

The Derivatives

sin(z)

eiz−e−iz

ieiz+ie−iz

eiz+e−iz

2=cos(z)

cos(z) =−sin(z)

tan(z) =1

cos2(z)

The Derivatives

sin(z)=ddz

eiz−e−iz

=ieiz+ie−iz

eiz+e−iz

2=cos(z)

cos(z) =−sin(z)

tan(z) =1

cos2(z)

The Derivatives

sin(z)=ddz

eiz−e−iz

ieiz+ie−iz

=eiz+e−iz

2=cos(z)

cos(z) =−sin(z)

tan(z) =1

cos2(z)

The Derivatives

sin(z)=ddz

eiz−e−iz

ieiz+ie−iz

eiz+e−iz

=cos(z)

cos(z) =−sin(z)

tan(z) =1

cos2(z)

The Derivatives

sin(z)=ddz

eiz−e−iz

ieiz+ie−iz

eiz+e−iz

2=cos(z)

cos(z) =−sin(z)

tan(z) =1

cos2(z)

The Derivatives

sin(z)=ddz

eiz−e−iz

ieiz+ie−iz

eiz+e−iz

2=cos(z)

cos(z) =−sin(z)

tan(z) =1

cos2(z)

The Derivatives

sin(z)=ddz

eiz−e−iz

ieiz+ie−iz

eiz+e−iz

2=cos(z)

cos(z) =−sin(z)

tan(z) =1

cos2(z)

The Derivatives

sin(z)=ddz

eiz−e−iz

ieiz+ie−iz

eiz+e−iz

2=cos(z)

cos(z) =−sin(z)

tan(z) =1

cos2(z)

The Derivatives

sin(z)=ddz

eiz−e−iz

ieiz+ie−iz

eiz+e−iz

2=cos(z)

cos(z) =−sin(z)

tan(z) =1

cos2(z)

The Derivatives

sin(z)=ddz

eiz−e−iz

ieiz+ie−iz

eiz+e−iz

2=cos(z)

cos(z) =−sin(z)

tan(z) =1

cos2(z)

The Derivatives

sin(z)=ddz

eiz−e−iz

ieiz+ie−iz

eiz+e−iz

2=cos(z)

cos(z) =−sin(z)

tan(z) =1

cos2(z)

Definition.

For any complex number z we define

cosh(z) =ez + e−z

sinh(z) =ez− e−z

sin(iz) =eiiz− e−iiz

e−z− ez

2i= ii

ez− e−z

2i= isinh(z)

cos(iz) = cosh(z)

cosh(z) =ez + e−z

e−z− ez

2i= ii

ez− e−z

2i= isinh(z)

cos(iz) = cosh(z)

cosh(z) =ez + e−z

e−z− ez

2i= ii

ez− e−z

2i= isinh(z)

cos(iz) = cosh(z)

cosh(z) =ez + e−z

e−z− ez

2i= ii

ez− e−z

2i= isinh(z)

cos(iz) = cosh(z)

cosh(z) =ez + e−z

sin(iz)

=eiiz− e−iiz

e−z− ez

2i= ii

ez− e−z

2i= isinh(z)

cos(iz) = cosh(z)

cosh(z) =ez + e−z

=e−z− ez

2i= ii

ez− e−z

2i= isinh(z)

cos(iz) = cosh(z)

cosh(z) =ez + e−z

e−z− ez

= iiez− e−z

2i= isinh(z)

cos(iz) = cosh(z)

cosh(z) =ez + e−z

e−z− ez

2i= ii

ez− e−z

= isinh(z)

cos(iz) = cosh(z)

cosh(z) =ez + e−z

e−z− ez

2i= ii

ez− e−z

2i= isinh(z)

cos(iz) = cosh(z)

cosh(z) =ez + e−z

e−z− ez

2i= ii

ez− e−z

2i= isinh(z)

cos(iz)

= cosh(z)

cosh(z) =ez + e−z

e−z− ez

2i= ii

ez− e−z

2i= isinh(z)

cos(iz) = cosh(z)

Another Connection Between Sine and Hyperbolic Sine

|sin(z)|2

= sin(x+ iy)sin(x+ iy)=

(sin(x)cos(iy)+ sin(iy)cos(x)

)sin(x+ iy)

sin(x)cosh(y)+ isinh(y)cos(x))×

sin(x)cosh(y)− isinh(y)cos(x))

= sin2(x)cosh2(y)+ sinh2(y)cos2(x)+sinh2(y)sin2(x)− sinh2(y)sin2(x)

= sin2(x)(cosh2(y)− sinh2(y)

)+ sinh2(y)

(cos2(x)+ sin2(x)

)= sin2(x)+ sinh2(y)

Another Connection Between Sine and Hyperbolic Sine|sin(z)|2

)sin(x+ iy)

)+ sinh2(y)

(cos2(x)+ sin2(x)

= sin(x+ iy)sin(x+ iy)

sin(x)cos(iy)+ sin(iy)cos(x))sin(x+ iy)

)+ sinh2(y)

(cos2(x)+ sin2(x)

)sin(x+ iy)

)+ sinh2(y)

(cos2(x)+ sin2(x)

)sin(x+ iy)

sin(x)cosh(y)+ isinh(y)cos(x))

)+ sinh2(y)

(cos2(x)+ sin2(x)

)sin(x+ iy)

)+ sinh2(y)

(cos2(x)+ sin2(x)

)sin(x+ iy)

= sin2(x)cosh2(y)+ sinh2(y)cos2(x)

+sinh2(y)sin2(x)− sinh2(y)sin2(x)= sin2(x)

(cosh2(y)− sinh2(y)

)+ sinh2(y)

(cos2(x)+ sin2(x)

)sin(x+ iy)

)+ sinh2(y)

(cos2(x)+ sin2(x)

)sin(x+ iy)

)+ sinh2(y)

(cos2(x)+ sin2(x)

= sin2(x)+ sinh2(y)

)sin(x+ iy)

)+ sinh2(y)

(cos2(x)+ sin2(x)

Theorem.

The only zeros of the complex sine function are thenumbers nπ where n is an integer.

Proof. The only solutions of

0 =∣∣sin(z)

∣∣2 =∣∣sin(x+ iy)

∣∣2 = sin2(x)+ sinh2(y)

are y = 0 and x = nπ . So z = nπ where n is an integer are the zeros ofthe sine function.

Theorem. The only zeros of the complex sine function are thenumbers nπ where n is an integer.

0 =∣∣sin(z)

∣∣2 =∣∣sin(x+ iy)

∣∣2 = sin2(x)+ sinh2(y)

Proof.

The only solutions of

0 =∣∣sin(z)

∣∣2 =∣∣sin(x+ iy)

∣∣2 = sin2(x)+ sinh2(y)

0 =∣∣sin(z)

∣∣2

=∣∣sin(x+ iy)

∣∣2 = sin2(x)+ sinh2(y)

0 =∣∣sin(z)

∣∣2 =∣∣sin(x+ iy)

∣∣2

= sin2(x)+ sinh2(y)

0 =∣∣sin(z)

∣∣2 =∣∣sin(x+ iy)

∣∣2 = sin2(x)+ sinh2(y)

0 =∣∣sin(z)

∣∣2 =∣∣sin(x+ iy)

∣∣2 = sin2(x)+ sinh2(y)

are y = 0

and x = nπ . So z = nπ where n is an integer are the zeros ofthe sine function.

0 =∣∣sin(z)

∣∣2 =∣∣sin(x+ iy)

∣∣2 = sin2(x)+ sinh2(y)

are y = 0 and x = nπ .

So z = nπ where n is an integer are the zeros ofthe sine function.

0 =∣∣sin(z)

∣∣2 =∣∣sin(x+ iy)

∣∣2 = sin2(x)+ sinh2(y)

0 =∣∣sin(z)

∣∣2 =∣∣sin(x+ iy)

∣∣2 = sin2(x)+ sinh2(y)

Identities for Hyperbolic Functions

sinh(z) = cosh(z)

cosh(z) = sinh(z)

3. sinh(z1 + z2) = sinh(z1)cosh(z2)+ cosh(z1)sinh(z2)4. cosh(z1 + z2) = cosh(z1)cosh(z2)+ sinh(z1)sinh(z2)5. cosh2(z)− sinh2(z) = 16. sinh(z+2πi) = sinh(z)7. cosh(z+2πi) = cosh(z)

... and more. And they can all be verified straight from the definition.

Identities for Hyperbolic Functions1.

sinh(z) = cosh(z)

cosh(z) = sinh(z)

sinh(z) = cosh(z)

cosh(z) = sinh(z)

sinh(z) = cosh(z)

cosh(z) = sinh(z)

3. sinh(z1 + z2) = sinh(z1)cosh(z2)+ cosh(z1)sinh(z2)

4. cosh(z1 + z2) = cosh(z1)cosh(z2)+ sinh(z1)sinh(z2)5. cosh2(z)− sinh2(z) = 16. sinh(z+2πi) = sinh(z)7. cosh(z+2πi) = cosh(z)

sinh(z) = cosh(z)

cosh(z) = sinh(z)

3. sinh(z1 + z2) = sinh(z1)cosh(z2)+ cosh(z1)sinh(z2)4. cosh(z1 + z2) = cosh(z1)cosh(z2)+ sinh(z1)sinh(z2)

5. cosh2(z)− sinh2(z) = 16. sinh(z+2πi) = sinh(z)7. cosh(z+2πi) = cosh(z)

sinh(z) = cosh(z)

cosh(z) = sinh(z)

3. sinh(z1 + z2) = sinh(z1)cosh(z2)+ cosh(z1)sinh(z2)4. cosh(z1 + z2) = cosh(z1)cosh(z2)+ sinh(z1)sinh(z2)5. cosh2(z)− sinh2(z) = 1

6. sinh(z+2πi) = sinh(z)7. cosh(z+2πi) = cosh(z)

sinh(z) = cosh(z)

cosh(z) = sinh(z)

3. sinh(z1 + z2) = sinh(z1)cosh(z2)+ cosh(z1)sinh(z2)4. cosh(z1 + z2) = cosh(z1)cosh(z2)+ sinh(z1)sinh(z2)5. cosh2(z)− sinh2(z) = 16. sinh(z+2πi) = sinh(z)

7. cosh(z+2πi) = cosh(z)... and more. And they can all be verified straight from the definition.

sinh(z) = cosh(z)

cosh(z) = sinh(z)

sinh(z) = cosh(z)

cosh(z) = sinh(z)

... and more.

And they can all be verified straight from the definition.

sinh(z) = cosh(z)

cosh(z) = sinh(z)

Theorem.

The only zeros of the complex hyperbolic sine function arethe numbers nπi, where n is an integer.

Proof. sinh(z) =−isin(iz) is equal to zero where sin(iz) is equal tozero. Hence the zeros of sinh(z) are at z = nπi, where n is an integer.

Theorem. The only zeros of the complex hyperbolic sine function arethe numbers nπi, where n is an integer.

Proof.

sinh(z) =−isin(iz) is equal to zero where sin(iz) is equal tozero. Hence the zeros of sinh(z) are at z = nπi, where n is an integer.

Proof. sinh(z) =−isin(iz)

is equal to zero where sin(iz) is equal tozero. Hence the zeros of sinh(z) are at z = nπi, where n is an integer.

Proof. sinh(z) =−isin(iz) is equal to zero where sin(iz) is equal tozero.

Hence the zeros of sinh(z) are at z = nπi, where n is an integer.

More Hyperbolic Functions

1. tanh(z) :=sinh(z)cosh(z)

2. coth(z) :=cosh(z)sinh(z)

3. sech(z) :=1

cosh(z)

4. csch(z) :=1

sinh(z)Note that hyperbolic secant and cosecant are not very common(around the world) either.

More Hyperbolic Functions1. tanh(z) :=

sinh(z)cosh(z)

3. sech(z) :=1

cosh(z)

4. csch(z) :=1

sinh(z)cosh(z)

3. sech(z) :=1

cosh(z)

4. csch(z) :=1

sinh(z)cosh(z)

3. sech(z) :=1

cosh(z)

4. csch(z) :=1

sinh(z)cosh(z)

3. sech(z) :=1

cosh(z)

4. csch(z) :=1

sinh(z)

Note that hyperbolic secant and cosecant are not very common(around the world) either.

sinh(z)cosh(z)

3. sech(z) :=1

cosh(z)

4. csch(z) :=1

Remaining Derivatives

tanh(z) =1

cosh2(z)

coth(z) =− 1sinh2(z)

sech(z) =−sech(z) tanh(z)

csch(z) =−csch(z)coth(z)

Remaining Derivatives1.

tanh(z) =1

cosh2(z)

tanh(z) =1

cosh2(z)

tanh(z) =1

cosh2(z)

tanh(z) =1

cosh2(z)

tanh(z) =1

cosh2(z)

Theorem.

For all complex numbers z we have

arccos(z) =−i log(

z+ i(1− z2) 1

)where the right side is a multivalued function.

Theorem. For all complex numbers z we have

z+ i(1− z2) 1

where the right side is a multivalued function.

Theorem. For all complex numbers z we have

z+ i(1− z2) 1

)where the right side is a multivalued function.

Proof.

w = arccos(z)cos(w) = z

eiw + e−iw

eiw−2z+ e−iw = 0(eiw)2−2zeiw +1 = 0

eiw =2z±√

4z2−42

= z±√

z2−1

z±√

z2−1)

= −i log(

z± i√

1− z2)

Proof.

w = arccos(z)

cos(w) = zeiw + e−iw

eiw =2z±√

4z2−42

= z±√

z2−1

z±√

z2−1)

= −i log(

z± i√

1− z2)

Proof.

eiw + e−iw

eiw =2z±√

4z2−42

= z±√

z2−1

z±√

z2−1)

= −i log(

z± i√

1− z2)

Proof.

eiw + e−iw

eiw =2z±√

4z2−42

= z±√

z2−1

z±√

z2−1)

= −i log(

z± i√

1− z2)

Proof.

eiw + e−iw

eiw−2z+ e−iw = 0

(eiw)2−2zeiw +1 = 0

eiw =2z±√

4z2−42

= z±√

z2−1

z±√

z2−1)

= −i log(

z± i√

1− z2)

Proof.

eiw + e−iw

eiw =2z±√

4z2−42

= z±√

z2−1

z±√

z2−1)

= −i log(

z± i√

1− z2)

Proof.

eiw + e−iw

eiw =2z±√

4z2−42

= z±√

z2−1

z±√

z2−1)

= −i log(

z± i√

1− z2)

Proof.

eiw + e−iw

eiw =2z±√

4z2−42

= z±√

z2−1

z±√

z2−1)

= −i log(

z± i√

1− z2)

Proof.

eiw + e−iw

eiw =2z±√

4z2−42

= z±√

z2−1

z±√

z2−1)

= −i log(

z± i√

1− z2)

Proof.

eiw + e−iw

eiw =2z±√

4z2−42

= z±√

z2−1

z±√

z2−1)

= −i log(

z± i√

1− z2)

Proof.

eiw + e−iw

eiw =2z±√

4z2−42

= z±√

z2−1

z±√

z2−1)

= −i log(

z± i√

1− z2)

Similarly We Can Derive ...

1. arcsin(z) =−i log(

iz+(1− z2) 1

)2. arctan(z) =

i+ zi− z

)3. arsinh(z) = log

(z+(1+ z2) 1

)4. arcosh(z) = log

(z+(z2−1

)5. artanh(z) =

1+ z1− z

Similarly We Can Derive ...1. arcsin(z) =−i log

(1− z2) 1

2. arctan(z) =i2

i+ zi− z

)3. arsinh(z) = log

(z+(1+ z2) 1

)4. arcosh(z) = log

(z+(z2−1

)5. artanh(z) =

1+ z1− z

(1− z2) 1

)2. arctan(z) =

i+ zi− z

3. arsinh(z) = log(

z+(1+ z2) 1

)4. arcosh(z) = log

(z+(z2−1

)5. artanh(z) =

1+ z1− z

(1− z2) 1

)2. arctan(z) =

i+ zi− z

)3. arsinh(z) = log

(z+(1+ z2) 1

4. arcosh(z) = log(

z+(z2−1

)5. artanh(z) =

1+ z1− z

(1− z2) 1

)2. arctan(z) =

i+ zi− z

)3. arsinh(z) = log

(z+(1+ z2) 1

)4. arcosh(z) = log

(z+(z2−1

5. artanh(z) =12

1+ z1− z

(1− z2) 1

)2. arctan(z) =

i+ zi− z

)3. arsinh(z) = log

(z+(1+ z2) 1

)4. arcosh(z) = log

(z+(z2−1

)5. artanh(z) =

1+ z1− z

Theorem.

arcsin(z) =1

(1− z2)12

where the right side is

multivalued again.

Proof.ddz

arcsin(z) =ddz

(−i log

(1− z2) 1

))= −i

iz+(1− z2)12

(1− z2)12(−2z)

iz+(1− z2)12

i(1− z2

) 12 − z

(1− z2)12

Theorem.ddz

arcsin(z) =1

(1− z2)12

multivalued again.

Proof.ddz

arcsin(z) =ddz

(−i log

(1− z2) 1

))= −i

iz+(1− z2)12

(1− z2)12(−2z)

iz+(1− z2)12

i(1− z2

) 12 − z

(1− z2)12

Theorem.ddz

arcsin(z) =1

(1− z2)12

multivalued again.

Proof.ddz

arcsin(z) =ddz

(−i log

(1− z2) 1

))= −i

iz+(1− z2)12

(1− z2)12(−2z)

iz+(1− z2)12

i(1− z2

) 12 − z

(1− z2)12

Theorem.ddz

arcsin(z) =1

(1− z2)12

multivalued again.

Proof.

arcsin(z) =ddz

(−i log

(1− z2) 1

))= −i

iz+(1− z2)12

(1− z2)12(−2z)

iz+(1− z2)12

i(1− z2

) 12 − z

(1− z2)12

Theorem.ddz

arcsin(z) =1

(1− z2)12

multivalued again.

Proof.ddz

arcsin(z)

(−i log

(1− z2) 1

))= −i

iz+(1− z2)12

(1− z2)12(−2z)

iz+(1− z2)12

i(1− z2

) 12 − z

(1− z2)12

Theorem.ddz

arcsin(z) =1

(1− z2)12

multivalued again.

Proof.ddz

arcsin(z) =ddz

(−i log

(1− z2) 1

= −i1

iz+(1− z2)12

(1− z2)12(−2z)

iz+(1− z2)12

i(1− z2

) 12 − z

(1− z2)12

Theorem.ddz

arcsin(z) =1

(1− z2)12

multivalued again.

Proof.ddz

arcsin(z) =ddz

(−i log

(1− z2) 1

))= −i

iz+(1− z2)12

(1− z2)12(−2z)

iz+(1− z2)12

i(1− z2

) 12 − z

(1− z2)12

Theorem.ddz

arcsin(z) =1

(1− z2)12

multivalued again.

Proof.ddz

arcsin(z) =ddz

(−i log

(1− z2) 1

))= −i

iz+(1− z2)12

(1− z2)12(−2z)

iz+(1− z2)12

i(1− z2

) 12 − z

(1− z2)12

Theorem.ddz

arcsin(z) =1

(1− z2)12

multivalued again.

Proof.ddz

arcsin(z) =ddz

(−i log

(1− z2) 1

))= −i

iz+(1− z2)12

(1− z2)12(−2z)

iz+(1− z2)12

i(1− z2

) 12 − z

(1− z2)12

Theorem.ddz

arcsin(z) =1

(1− z2)12

multivalued again.

Proof.ddz

arcsin(z) =ddz

(−i log

(1− z2) 1

))= −i

iz+(1− z2)12

(1− z2)12(−2z)

iz+(1− z2)12

i(1− z2

) 12 − z

(1− z2)12

Similarly ...

arccos(z) =− 1

(1− z2)12

arctan(z) =1

Similarly ...1.

arccos(z) =− 1

(1− z2)12

arctan(z) =1

Similarly ...1.

arccos(z) =− 1

(1− z2)12

arctan(z) =1

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