Trigonometry

COMPLEX FUNCTIONS AND TRIGONOMETRIC IDENTITIES Revision E By Tom Irvine Email: [email protected] September 14, 2006

Trigonometric Functions of Angle α

r y α x

sin (α ) =

y r

cos(α ) =

x r

tan (α ) =

y x

cot (α ) =

x y

sec(α ) =

r x

csc(α ) =

r y

(1) Trigonometric Expansion sin x = x −

cos x = 1 −

x3 x5 + −L 3! 5!

(2)

x2 x4 + −L 2! 4!

(3)

1

x3 x5 sinh x = x + + +L 3! 5!

(4)

x2 x4 cosh x = 1 + + +L 2! 4!

(5)

Exponential Expansion exp(x ) = 1 + x +

x 2 x3 + +L 2! 3!

(6)

Trigonometric Identities

cos 2 (α ) =

1 1 + cos(2α ) 2 2

(7)

sin 2 (α ) =

1 1 − cos(2α ) 2 2

(8)

cos(α )cos(β ) =

1 1 cos(α + β) + cos(α − β) 2 2

(9)

1 1 sin (α )sin (β) = − cos(α + β) + cos(α − β) 2 2

(10)

cos(α )sin (β) =

1 1 sin (α + β ) − sin (α − β ) 2 2

sin (α )cos(β) =

(11)

1 1 sin (α + β ) + sin (α − β) 2 2

(12)

sin 2 (α ) + cos 2 (α ) = 1

(13)

sec2 (α ) − tan 2 (α ) = 1

(14)

csc2 (α ) − cot 2 (α ) = 1

(15)

sin (2α ) = 2 sin (α )cos(α )

(16)

2

cos(2α ) = cos 2 (α ) − sin 2 (α )

(17)

sin (α ± β) = sin (α )cos(β) ± cos(α )sin (β)

(18)

cos(α ± β ) = cos(α )cos(β) m sin (α )sin (β )

(19)

tan (α ) ± tan (β) 1 m tan (α ) tan (β )

(20)

tan (α ± β ) =

A sin (αt ) + B cos(αt ) = A 2 + B2 [ sin (αt + θ) ]

where

⎛ B⎞ θ = arctan⎜ ⎟ ⎝A⎠ (21)

Euler's Equation exp(± jα ) = cos(α ) ± j sin (α )

(22)

sin (α ) =

exp( jα ) − exp(− jα ) 2j

(23)

cos(α ) =

exp( jα ) + exp(− jα ) 2

(24)

Hyperbolic Functions sinh (α ) =

exp(α ) − exp(− α ) 2

(25a)

cosh (α ) =

exp(α ) + exp(− α ) 2

(25b)

3

cosh 2 (α ) − sinh 2 (α ) = 1

(26)

Derivatives d (sin u ) = cos u du dx dx

(27)

d (cos u ) = − sin u du dx dx

(28)

d (tan u ) = sec2 u du dx dx

(29)

d (cot u ) = − csc u du dx dx

(30)

d (sec u ) = tan u sec u du dx dx

(31)

d (csc u ) = − csc u cot u du dx dx

(32)

d (sinh u ) = cosh u du dx dx

(33)

d (cosh u ) = sinh u du dx dx

(34)

d (tanh u ) = 1 2 du dx cosh u dx

(35)

Natural Logarithm of a Complex Number ⎛b⎞ ⎡ ⎤ ln (a + j b ) = ln ⎢ a 2 + b 2 exp( jθ)⎥ , θ = arctan⎜ ⎟ ⎣ ⎦ ⎝a⎠

4

(36)

⎡ ⎤ ln (a + j b ) = ln ⎢ a 2 + b 2 ⎥ + ln[ exp( jθ)] ⎣ ⎦

⎡ ⎤ ln(a + j b ) = ln ⎢ a 2 + b 2 ⎥ + jθ ⎣ ⎦

(37)

(38)

⎛b⎞ ⎡ ⎤ ln(a + j b ) = ln ⎢ a 2 + b 2 ⎥ + j arctan⎜ ⎟ ⎣ ⎦ ⎝a⎠

(39)

APPENDIX A The Square Root of a Complex Number. Consider x2 = a + jb

(A-1)

Thus x = ± a + jb

(A-2)

where a and b are real coefficients. Solve for x.

Let

x1 = (c + j d )

(A-3a)

x 2 = −(c + j d )

(A-3b)

where c and d are real coefficients. Substitute equation (A-3a) into (A-1).

(c + j d )2 = (a + j b )

(A-4)

5

(c + j d )(c + j d ) = (a + j b )

(A-5)

c 2 − d 2 + j (2cd ) = a + j b

(A-6)

Equation (A-6) implies two equations. The first is c2 − d 2 = a

(A-7)

The second implied equation is 2cd = b

(A-8)

Solve for d using equation (A-8). d=

b 2c

(A-9)

Substitute equation (A-9) into (A-8). 2 ⎛ b⎞ 2 c −⎜ ⎟ = a

(A-10)

2 ⎛ b⎞ 2 c −a −⎜ ⎟ = 0

(A-11)

⎝ 2c ⎠

⎝ 2c ⎠

Multiply through by 4c 2 .

4c 4 − 4ac2 − b 2 = 0

(A-12)

Apply the quadratic formula. c2 =

4a ± 16a 2 + 16b 2 8

(A-13)

c2 =

4a ± 4 a 2 + b 2 8

(A-14)

6

c2 =

c=

a ± a 2 + b2 2

(A-15)

a ± a 2 + b2 2

(A-16)

Require c to be real.

c=

a + a 2 + b2 2

(A-17)

Substitute equation (A-17) into (A-9). ⎧ ⎪ ⎪ ⎪ b d=⎨ ⎪ ⎡ a + a 2 + b2 ⎪2 ⎢ 2 ⎪ ⎢ ⎩ ⎣⎢

⎫ ⎪ ⎪ ⎪ ⎬ ⎤⎪ ⎥⎪ ⎥⎪ ⎦⎥ ⎭ (A-18)

⎧ ⎪ ⎪ ⎪ d=⎨ ⎪ ⎪ ⎪ ⎩

⎫ ⎪ ⎪ b ⎪ ⎬ ⎛ ⎞⎪ 2 2 4⎜ a + a + b ⎟ ⎝ ⎠⎪ ⎪ 2 ⎭ (A-19)

7

⎧ ⎪ ⎪ d=⎨ ⎪ ⎪⎩

⎫ ⎪ b ⎪ ⎬ ⎛ ⎞⎪ 2 2 2⎜ a + a + b ⎟ ⎝ ⎠ ⎪⎭

(A-20)

Substitute equations (A-20) and (A-17) into (A-3a). ⎧ 2 2 ⎪ a+ a +b x1 = ⎨ 2 ⎪ ⎩

⎫ ⎪ ⎬+ ⎪ ⎭

⎧ ⎪ ⎪ j⎨ ⎪ ⎪⎩

⎫ ⎪ b ⎪ ⎬ ⎞⎪ ⎛ 2 2 2⎜ a + a + b ⎟ ⎠ ⎪⎭ ⎝

(A-21a)

Substitute equations (A-20) and (A-17) into (A-3b). ⎧ 2 2 ⎪ a+ a +b x 2 = −⎨ 2 ⎪ ⎩

⎧ ⎫ ⎪ ⎪ ⎪ ⎬ − j⎨ ⎪ ⎪ ⎭ ⎪ ⎩

⎫ ⎪ b ⎪ ⎬ ⎛ ⎞⎪ 2 2 2⎜ a + a + b ⎟ ⎝ ⎠ ⎪⎭

(A-21b)

Note that equations (A-21a) and (A-21b) cannot be used for the special case: a < 0 and b = 0. For this special case, the roots are x =±j a

(A-21c)

Example x2 = 2 + j 7

(A-22)

x = ± 2+ j7

(A-23)

8

Solve for x. Use equation (A-21a). a=2

(A-24)

b=7

(A-25)

x 1 = 2.154 + j 1.625

(A-26)

x 2 = −2.154 − j 1.625

(A-27)

APPENDIX B Arbitrary Root of a Complex Number Let x n = [a + j b]

(B-1a)

x = [a + j b]1 / n

(B-1b)

The coefficients a and b are real numbers. The denominator of the exponent n is also real. Take the natural logarithm.

{

ln x = ln [a + j b]1 / n

}

(B-2)

ln x =

1 ln [a + j b] n

(B-3)

ln x =

1 ⎡ 2 b ⎞⎤ ⎛ ln ⎢ a + b 2 exp⎜ j arctan ⎟⎥ n ⎣ a ⎠⎦ ⎝

(B-4)

ln x =

1 ⎡ 2 b ⎞⎤ ⎤ 1 ⎡ ⎛ ln ⎢ a + b 2 ⎥ + ln ⎢exp⎜ j arctan ⎟⎥ n ⎣ a ⎠⎦ ⎦ n ⎣ ⎝

9

(B-5)

1 ⎤ ⎡ 1 b 2 2 ⎢ 2 ln x = ln a + b n ⎥ + j arctan ⎢ ⎥ n a ⎢⎣ ⎥⎦

)

(

(B-6)

1 ⎤ ⎫ ⎧ ⎡ b 1 ⎪ ⎪ ⎢ 2 2 exp{ln x} = exp⎨ln a + b 2n ⎥ + j arctan ⎬ ⎢ ⎥ a⎪ n ⎪ ⎣⎢ ⎦⎥ ⎭ ⎩

)

(

(B-7)

1 ⎤⎫ ⎧ ⎡ b⎫ ⎪ ⎪ ⎢ 2 ⎧ 1 2 x = exp⎨ln a + b 2n ⎥ ⎬ exp⎨ j arctan ⎬ ⎥⎪ a⎭ ⎩ n ⎪ ⎢⎣⎢ ⎦⎥ ⎭ ⎩

)

(

)

(

(B-8)

1 b⎫ ⎧ 1 2 2 x = a + b 2n exp⎨ j arctan ⎬

(

)

(B-9)

a⎭

⎩ n

1 ⎧ ⎛1 b ⎞⎫ b⎞ ⎛1 2 2 x = a + b 2n ⎨cos⎜ arctan ⎟ + j sin ⎜ arctan ⎟⎬

⎩

⎝n

a⎠

⎝n

a ⎠⎭

(B-10)

Note that equation (B-10) could be used for the special case of a square root.

APPENDIX C Cube Root of a Complex Number Consider x3 = a + j b

(C-1)

x = [a + j b]1 / 3

(C-2)

Equation (C-1) has three roots. The method in Appendix B yields the following formula for one of the cube roots.

10

1 ⎧ ⎛1 b ⎞⎫ b⎞ ⎛1 x 1= a 2 + b 2 6 ⎨cos⎜ arctan ⎟ + j sin ⎜ arctan ⎟⎬

)

(

⎩

a⎠

⎝3

⎝3

a ⎠⎭

(C-3)

Rearrange equation (C-1). x3 − a − j b = 0

(C-4)

Devise an equation for finding the other two roots.

x 3 − a − j b = (x − x1 )(x − x 2 )(x − x 3 )

(C-5)

Expand the right-hand-side.

[

]

x 3 − a − j b = x 2 − (x1 + x 2 )x + x1x 2 (x − x 3 )

[ ] [ ] x 3 − a − j b = [x 3 − (x1 + x 2 )x 2 + x1x 2 x ]− [x 3x 2 − x 3 (x1 + x 2 )x + x1x 2 x 3 ] x 3 − a − j b = [x 3 − (x1 + x 2 )x 2 + x1x 2 x ]+ [− x 3x 2 + x 3 (x1 + x 2 )x − x1x 2 x 3 ] x 3 − a − j b = [x 3 − (x1 + x 2 + x 3 )x 2 + (x1x 2 + x1x 3 + x 2 x 3 )x − x1x 2 x 3 ] x 3 − a − j b = x 2 − (x1 + x 2 )x + x1x 2 (x ) + x 2 − (x1 + x 2 )x + x1x 2 (− x 3 )

(C-6) (C-7) (C-8) (C-9) (C-10)

Equation (C-10) implies three separate equations.

(x1 + x 2 + x3 ) = 0

(C-11)

(x1x 2 + x1x3 + x 2x3 ) = 0

(C-12)

− x1x 2 x 3 = −a − j b

(C-13)

Continue with equation (C-11). x 2 = − x1 − x 3

(C-14)

11

Substitute equation (C-14) into (C-12).

(x1(− x1 − x 3 ) + x1x3 + (− x1 − x3 )x3 ) = 0

(C-15)

− x12 − x1x 3 + x1x 3 − x1x 3 − x 32 = 0

(C-16)

− x12 − x1x 3 − x 32 = 0

(C-17)

x 32 + x1x 3 + x12 = 0

(C-18)

Use the quadratic formula. − x1 ± x12 − 4x12 x3 = 2 x3 =

(C-19)

− x1 ± − 3x12 2

(C-20)

− x1 ± x1 − 3 2

(C-21)

x3 =

⎡−1± j 3 ⎤ x 3 = x1 ⎢ ⎥ 2 ⎣ ⎦

(C-22)

⎡−1− j 3 ⎤ x 3 = x1 ⎢ ⎥ 2 ⎣ ⎦

(C-23)

Choose

Recall equation (C-14). x 2 = − x1 − x 3

(C-24)

⎡−1− j 3 ⎤ x 2 = − x1 − x1 ⎢ ⎥ 2 ⎣ ⎦

(C-25)

12

⎧⎪ ⎡ − 1 − j 3 ⎤ ⎫⎪ x 2 = x1⎨−1 − ⎢ ⎥⎬ 2 ⎪⎩ ⎣ ⎦ ⎪⎭

(C-26)

⎧⎪ ⎡1 + j 3 ⎤ ⎫⎪ x 2 = x1⎨−1 + ⎢ ⎥⎬ ⎪⎩ ⎣ 2 ⎦ ⎪⎭

(C-27)

⎧ 2 1+ j 3 ⎫ x 2 = x1 ⎨− + ⎬ 2 ⎭ ⎩ 2

(C-28)

⎡−1+ j 3 ⎤ x 2 = x1 ⎢ ⎥ 2 ⎣ ⎦

(C-29)

The roots x2 and x3 thus form a complex conjugate pair. Summarize the roots. 1 ⎧ ⎛1 b ⎞⎫ b⎞ ⎛1 2 2 x 1= a + b 6 ⎨cos⎜ arctan ⎟ + j sin ⎜ arctan ⎟⎬

(

)

⎩

⎝3

⎝3

a⎠

a ⎠⎭

(C-30)

⎡−1+ j 3 ⎤ x 2 = x1 ⎢ ⎥ 2 ⎣ ⎦

(C-31)

⎡−1− j 3 ⎤ x 3 = x1 ⎢ ⎥ 2 ⎣ ⎦

(C-32)

Example Solve for x. x3 = 2 + j 7

(C-33)

x = [2 + j 7]1 / 3

(C-34)

a=2

(C-35)

13

b=7

(C-36)

n=3

(C-37)

There are three roots. The first root is 1 ⎧ ⎛1 b ⎞⎫ b⎞ ⎛1 2 2 x 1= a + b 6 ⎨cos⎜ arctan ⎟ + j sin ⎜ arctan ⎟⎬

(

)

⎩

⎝3

⎝3

a⎠

a ⎠⎭

(C-38)

x1 = 1.938 [ 0.909 + j 418]

(C-39)

x1 = 1.761 + j 0.809

(C-40)

The second root is a coordinate transformation of the first root. ⎡−1+ j 3 ⎤ x 2 = x1 ⎢ ⎥ 2 ⎣ ⎦

(C-41)

⎡−1+ j 3 ⎤ x 2 = [1.762 + j 0.809] ⎢ ⎥ 2 ⎣ ⎦

(C-42)

x 2 = − 1.581 + j1.120

(C-43)

⎡−1− j 3 ⎤ x 3 = x1 ⎢ ⎥ 2 ⎣ ⎦

(C-44)

⎡−1− j 3 ⎤ x 3 = [1.761 + j 0.809]⎢ ⎥ 2 ⎣ ⎦

(C-45)

x 3 = −0.180 − j1.930

(C-46)

14

In summary, the cube roots of (2 + j 7) are x1 = 1.761 + j 0.809

(C-47)

x 2 = − 1.581 + j1.120

(C-48)

x 3 = −0.180 − j1.930

(C-49)

APPENDIX D Derivation of the Quadratic Formula

ax 2 + bx + c = 0

(D-1)

x 2 + (b / a )x + (c / a ) = 0

(D-2)

b ⎞2 b2 c ⎛ x + + =0 ⎜ ⎟ − 2 2a ⎠ a ⎝ 4a

(D-3)

b ⎞2 b2 c ⎛ x + = − ⎜ ⎟ 2 2a ⎠ a ⎝ 4a

(D-4)

b ⎞ 2 b 2 − 4ac ⎛ ⎜x + ⎟ = 2a ⎠ ⎝ 4a 2

(D-5)

b ⎞ b 2 − 4ac ⎛ ⎜x + ⎟ = ± 2a ⎠ ⎝ 4a 2

(D-6)

15

b ⎞ ± b 2 − 4ac ⎛ ⎜x + ⎟ = 2a ⎠ 2a ⎝

(D-7)

b 2 − 4ac 2a

(D-8)

− b ± b 2 − 4ac 2a

(D-9)

b x= − ± 2a

x=

16