,θ = 60 degrees. So w = 1cis60.

Size: px

Start display at page:

Download ",θ = 60 degrees. So w = 1cis60."

Amice Byrd
5 years ago
Views:

1 Proving Trig Identities with Complex Numbers Introduction Complex numbers are numbers in the form a+bi, where i = 1. They can be expressed in the form r(cosθ +isinθ) for appropriate r,θ. THis is abbreviated as rcisθ, and it is helpful to know that cisa cisb = cis(a+b). It is assumed that the reader knows the definitions of sin,cos,tan and their inverses. Example 1: Convert w = 1 2 +i 3 2 into polar form. Solution: We see that in this case, a = 1 2,b = 3 2. Then a2 +b 2 = 1, so r = 1. Because cos60 = 1 2,θ = 60 degrees. So w = 1cis60. Example 2: Convert x = 2+2i 3 into polar form. Note that x = 4w from our previous example. Then all we need is to multiply r by 4. So x = 4cis60. Hopefully any confusion regarding complex numbers and polar form is cleared.

2 Part 1: A formula for sin(x+y),sin(x y). Let a = cosx,b = sinx,c = cosy,d = siny. Then cisx cisy = cis(x+y). This also means (a+bi)(c+di) = cos(x+y)+isin(x+y). Then if we compare the imaginary parts on each side, bc+ad = sin(x+y). This means sin(x+y) = cosxsiny +sinxcosy. If we plug in y for y and use the facts that cosy = cos( y), siny = sin( y) then it can be seen that sin(x y) = bc ad = sinxcosy sinycosx. Part 2: A formula for cos(x+y),cos(x y). Remember that (a+bi)(c+di) = cos(x+y)+isin(x+y) from last part. Then if we compare the real coefficients of each side, we get ac bd = cos(x+y). Then we have cos(x+y) = cosxcosy sinxsiny. If we plug in y for y, we get cos(x y) = cosxcosy +sinxsiny. Try some simple values of x and y to convince you that these identities are valid. Example 1: Find sin2x and cos2x in terms of sinx,cosx. Solution: sin2x = sin(x+x) = sinxcosx+sinxcosx = 2sinxcosx. Meanwhile, cos(x+x) = cosxcosx sinxsinx = cos 2 x sin 2 x. These are known as the DOUBLE ANGLE FORMULAS. They are handy to know, but you can easily derive them whenever you want. Extra info: The identity cisx cisy = cis(x y) holds. If we didn t want to derive formulas for cos(x y),sin(x y) by substituting y for y, we could use complex numbers again. The only difference would be that we would have (a+bi) (c+di) = cos(x y)+isin(x y).

3 Part 3: More sines and cosines Do you notice anything if you add sin(x+y)+sin(x y)? You should end up with 2sinxcosy. If you add cos(x+y)+cos(x y) you get 2cosxcosy. These are known as the SUM-TO-PRODUCT identities. They aren t used much, but it s good to know them. Example: Find 2cos37.5cos7.5. Solution: We recognize that this equals 2cos45cos30. This is easy to evaluate: It is 6 4. It s much easier than finding cosine of 37.5 and 7.5 Part 4: DeMoivre s and applications with trig DeMoivre s Theorem tells us that (rcisθ) n = r n cis(nθ). It can be proved by induction on n for integers. (We ll only focus on integer n here). Anyway, this is very helpful with trig. You may recall that cos2x = cos 2 x sin 2 x,sin2x = 2sinxcosx. We can use DeMoivre s on this: (cosx+isinx) 2 = cos 2 x+2cosxsinxi sin 2 x = cos2x+isin2x. Comparing real and imaginary parts, we get that cos 2 x sin 2 x = cos2x,sin2x = 2sinxcosx. The reason this is helpful is that it goes beyond 2x. We can find sin3x,cos3x. (cosx+isinx) 3 = cos 3 x 3cosxsin 2 x+3cos 2 xsinxi isin 3 x = cos3x+isin3x. Then we compare real and imaginary parts: cos3x = cos 3 x 3cosxsin 2 x,sin3x = 3cos 2 xsinx sin 3 x Example: Find cos60 in terms of a,b if a = cos15,b = sin15. Solution: We can find that cos4x = cos 2 2x sin 2 2x = 1 2sin 2 2x. Now cos4x = 1 8cos 2 xsin 2 x. Then cos60 = 1 8a 2 b 2.

4 Part 5: Tangent How do we find a formula for tan(x+y)? We use the definition of tangent: tanx = sinx sin(x+y) cosx. Now tan(x+y) = cos(x+y) We can use the formulas we already have for sin(x+y), cos(x+y). Then tan(x+y) = sinxcosy+sinycosx cosxcosy sinxsiny. If we divide the numerator and denominator by cosxcosy then we get tan(x+y) = tanx+tany 1 tanxtany. If you are curious, we also have tan(x y) = tanx tany 1+tanxtany. Example: Find tan(x+y) if tanx = 2,tany = 3. Solution: This is straightforward and we should get tan(x+y) = 1. How does this relate to complex numbers? Remember that if cisx = a+bi then tanx = b a. (You can envision this, it s the definition of tangent). Now if there are two complex numbers w = 1+2i,z = 1+3i and they have arguments x,y then tanx = 2,tany = 3. Now we multiply w,z. We get wz = 1+2i+3i 6 = 5+5i. If this new complex number has argument N, then N = x+y. This results from the fact that when you multiply two complex numbers you add their angles. Now tann can be evaluated to be 1. (remember, tanx = b a. ) Part 5.5 : Arctangent The arctangent function is the inverse of tangent. That means that arctan(tan x) = x. We will denote arctangent as A(x) in this article. (example: A(1) = 45 degrees). The arctangent function has the nicest form among the arcsine, arccosine, and arctangent functions and it comes up in problems a lot. For example: Find A(2)+A(1). First we need a formula for A(x)+A(y). How do we do that? Note that A(tan(A(x)+A(y))) = A(x)+A(y) because the A(x)s and tangents cancel. Then we simplfiy with the tangent formula: tan(a(x)+a(y)) = x+y x+y 1 xy. This is because once again, A(tanx) = x. Now A(x)+A(y) = A( 1 xy ). Now we can do the problem: A(2)+A(1) = A( 3).

5 Part 6: Mega Mega Problems List (not that big) 1. Find sin 82.5 degrees. 2. Find formulas for sin( x 2 ),cos(x 2 ). 3. Evaluate 16cosxsinxcos2xcos4xcos8x. 4. Find a formula for A(x)+A(y)+A(z). 5. Evaluate cos(105) cos( 15) sin 105 sin( 15). 6. You guys will love this problem :). Let w and z be complex numbers with θ equal to the argument of w z z. Then the maximum value of tan 2 θ can be written in the form p q where p,q are relatively prime positive integers. Find p+q. Conclusion: This article covers the basic trig identities. Complex numbers provide such an easy way to derive and re-derive these identities. We hope you enjoyed it!

MATH 15a: Linear Algebra Exam 1, Solutions

MATH 15a: Linear Algebra Exam 1, Solutions MATH 5a: Linear Algebra Exam, Solutions. Let T : R 3 R 4 be the linear transformation with T( e ) = 2 e + e 2 + 3 e 3 4 e 4, T( e 2 ) = e e 2 +2 e 3 +6 e 4, and T( e 3 ) = 4 e e 2 +7 e 3 +8 e 4. (a) (6