TRIGONOMETRY. Units: π radians rad = 180 degrees = 180 full (complete) circle = 2π = 360

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1 TRIGONOMETRY Units: π radians rad 180 degrees 180 full (complete) circle 2π 360 Special Values: 0 30 (π/6) 45 (π/4) 60 (π/3) 90 (π/2) sin(θ) 0 ½ 1/ 2 3/2 1 cos(θ) 1 3/2 1/ 2 ½ 0 tan(θ) 0 1/ ± erivatives: [sin(x)]/ x cos(x) [cos(x)]/ x sin(x) [tan(x)]/ x sec 2 (x) Chain rule: [sin(u)]/ x cos(u) u/ x example: [sin(2x)]/ x 2 cos(2x) Sum Rules: sin(a ± B) sin(a) cos(b) ± sin(b) cos(a) cos(a ± B) cos(a) cos(b) # sin(a) sin(b) tan(a ± B) [tan(a) ± tan(b)]/[1 # tan(a) tan(b)] sin 2 (A) + cos 2 (B) 1 sin(2a) 2 sin(a) cos(a) cos(2a) 2 cos 2 (A) sin 2 (A) cos 2 (A) sin 2 (A) Taylor Series: f(x) f(0) + f (0)x + f (0)x 2 /(2!) + where prime denotes a derivative Examples: sin(x) x x 3 /(3!) + x 5 /(5!) + cos(x) 1 x 2 /(2!) + x 4 /(4!) + tan(x) x + x 3 /(3!) + 2x 5 /15 + (1 + x) ½ 1 + x/2 x 2 /8 + ; (1 + x) b 1 + bx + b(b-1)x 2 /(2!) + exp(x) 1 + x + x 2 /(2!) + x 3 /(3!) + These are especially important when x is small (typically x << 1). In this case, sin(x) x, cos(x) 1, etc.

2 COMPLEX NUMBERS efinition: Z a + jb is a Complex Number if: 1) a and b are real numbers a Re(Z) is the real part of Z b Im(Z) is the imaginary part of Z 2) j is a solution to j or j 2 1 Geometry: Any complex number can be described by a point in a two-dimensional coordinate system. Consider the complex number Z j5.0;,pjlqu\ $[LV < E ME M 5HO $[LV ;

3 We could also use polar coordinates to locate Z:,PJLQU\ $[LV ME M θ 5HO $[LV with Z R cos(θ) + j R sin(θ) R [cos(θ) + j sin(θ)], where R (a 2 + b 2 ) ½ and θ tan 1 (b/a). erivatives: Consider the derivative of the function f(θ) cos(θ) + j sin(θ): f(θ)/ θ sin(θ) + j cos(θ) j [cos(θ) + j sin(θ)] Thus, we find the derivative of f(θ) with respect to θ gives jf(θ). Can you think of another function that satisfies f(θ)/ θ jf(θ)? Consider the exponential function exp(cθ) e cθ, (e cθ )/ θ c e cθ. So, if we let c j, then this has the same property upon differentiation as f(q) above.

4 It can be rigorously proven (by expanding both sides below in a Taylor series) that: e jθ cos(θ) + j sin(θ) (which is Euler s identity) Thus, we have three ways of writing a complex number Z: Z a + jb Z R[cos(θ) + j sin(θ)] Z R e jθ Some Operations with Complex Numbers: Addition and Subtraction Z 1 ± Z 2 (a 1 + jb 1 ) ± (a 2 + jb 2 ) (a 1 ± a 2 ) + j (b 1 ± b 2 ) Multiplication Z 1 Z 2 (a 1 + jb 1 ) (a 2 + jb 2 ) (a 1 a 2 b 1 b 2 ) + j (a 1 b 2 + b 1 a 2 ) or Z 1 Z 2 R 1 e jθ1 R 2 e jθ2 R 1 R 2 e j(θ1 + θ2) Note how much easier multiplication is using the exponential form. Now consider Euler s identity above and replace θ by θ to give e jθ cos(θ) j sin(θ) Adding these two equations together yields cos(θ) ½( e jθ + e jθ ) Subtracting these two equations yields sin(θ) ½ j ( e jθ e jθ ).

5 Examples: For what θ is e jθ 1? Euler s identity tells us that for θ 0, 2π, 2π, 4p, 4π, ; e jθ 1. For what values of θ will e jθ equal (a) 1, (b) j, and (c) j? (Try and figure it out on your own). The answers are: (a) π, π, 3π, 3π, (b) π/2, 3π/2, 5π/2, (c) 3π/2, π/2, 7π/2, 5π/2, You know how to graph complex numbers such as Z 1 2 e jπ/3 now (I hope!). But how about Z 2 2 e jπ/3? If we write Z 1 a + jb, then Z 2 a jb or we could write Z 2 as For instance: Z 2 ( 1) 2 e jπ/3 (e π ) 2 e jπ/3 2 e j(π + π/3) (This tells us to add π to the old angle π/3 to find the new direction) Z 2 2 e j4π/3,pjlqu\ $[LV E M 5HO $[LV M E

6 OR,P M θ π 5H θ θ M Now, how about Z 2 jz 1? Here, jz 1 j (a + jb) -b + ja OR jz 1 j 2 e jπ/3 jz 1 (e jπ/2 ) 2 e jπ/3 j(π/2 + π/3) 2 e (This tells us to add π/2 to the old angle of π/3 to find the new direction) jz 1 2 e j5π/6

7 For instance:,p M M M E E E 5H So, 1) Multiplication by 1 rotates the vector by 180º (or π) 2) Multiplication by j rotates the vector by 90º (or π/2)

8 Complex Conjugation: The complex conjugate of Z a + jb is defined to be Z* a jb OR Z R e jθ and Z* R e jθ. Thus, to form the complex conjugate of Z, just change the sign of j ( i.e., j j ). Note that Z Z* is always real. When we write Z in exponential form, Z R e jθ, powers of Z are easy to find: Z n R n e jnθ For instance, to find the 1/3 power of 1, we recognize that 1 1/3 e j(2π)1/3 e j2π/3. But, 1 is also equal to e 0 and e j4π, so 1 1/3 also equals 1 and e j4π/3. All together there are three possible answers, which is another way of saying that 1 has three cube-roots! (similarly four 4 th -roots, five 5 th -roots, and so on),p π π 5H