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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(cos θ + i sin θ) 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 = 21 , b = 23 . Then a2 + b2 = 1, so r = 1. Because cos 60 = 21 , 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.
Part 1: A formula for sin(x + y), sin(x − y).
Let a = cos x, b = sin x, c = cos y, d = sin y. Then cisx · cisy = cis(x + y). This also means
(a + bi)(c + di) = cos(x + y) + i sin(x + y). Then if we compare the imaginary parts on each side,
bc + ad = sin(x + y). This means sin(x + y) = cos x sin y + sin x cos y.
If we plug in −y for y and use the facts that cos y = cos(−y), − sin y = sin(−y) then it can be seen that
sin(x − y) = bc − ad = sin x cos y − sin y cos x.
Part 2: A formula for cos(x + y), cos(x − y).
Remember that (a + bi)(c + di) = cos(x + y) + i sin(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) = cos x cos y − sin x sin y. If we plug in −y for y,
we get cos(x − y) = cos x cos y + sin x sin y.
Try some simple values of x and y to convince you that these identities are valid.
Example 1: Find sin 2x and cos 2x in terms of sin x, cos x. Solution:
sin 2x = sin(x + x) = sin x cos x + sin x cos x = 2 sin x cos x. Meanwhile,
cos(x + x) = cos x cos x − sin x sin x = cos2 x − sin2 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) + i sin(x − y).
Part 3: More sines and cosines Do you notice anything if you add sin(x + y) + sin(x − y)? You should end up with
2 sin x cos y. If you add cos(x + y) + cos(x − y) you get 2 cos x cos y.
These are known as the SUM-TO-PRODUCT identities. They aren’t used much, but it’s good to know them.
Example: Find 2 cos 37.5 cos 7.5.
Solution: We recognize that this equals 2 cos 45 cos 30. 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 = rn 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 cos 2x = cos2 x − sin2 x, sin 2x = 2 sin x cos x.
We can use DeMoivre’s on this: (cos x + i sin x)2 = cos2 x + 2 cos x sin xi − sin2 x = cos 2x + i sin 2x.
Comparing real and imaginary parts, we get that cos2 x − sin2 x = cos 2x, sin 2x = 2 sin x cos x.
The reason this is helpful is that it goes beyond 2x. We can find sin 3x, cos 3x.
(cos x + i sin x)3 = cos3 x − 3 cos x sin2 x + 3 cos2 x sin xi − i sin3 x = cos 3x + i sin 3x. Then we compare real and
imaginary parts:
cos 3x = cos3 x − 3 cos x sin2 x, sin 3x = 3 cos2 x sin x − sin3 x
Example: Find cos 60 in terms of a, b if a = cos 15, b = sin 15.
Solution: We can find that cos 4x = cos2 2x − sin2 2x = 1 − 2 sin2 2x. Now cos 4x = 1 − 8 cos2 x sin2 x. Then
cos 60 = 1 − 8a2 b2 .
Part 5: Tangent
How do we find a formula for tan(x + y)? We use the definition of tangent: tan x =
sin x
cos x .
We can use the formulas we already have for sin(x+y), cos(x+y). Then tan(x + y) =
If we divide the numerator and denominator by cos x cos y then we get tan(x + y) =
tan x−tan y
we also have tan(x − y) = 1+tan
x tan y .
Now tan(x + y) =
sin(x+y)
cos(x+y)
sin x cos y+sin y cos x
cos x cos y−sin x sin y .
tan x+tan y
1−tan x tan y .
If you are curious,
Example: Find tan(x + y) if tan x = 2, tan y = 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 tan x = ab . (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
tan x = 2, tan y = 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 tan N can be evaluated to be −1. (remember, tan x = ab . )
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
x+y
x+y
tangent formula: tan(A(x) + A(y)) = 1−xy
. This is because once again, A(tan x) = x. Now A(x) + A(y) = A( 1−xy
).
Now we can do the problem: A(2) + A(1) = A(−3).
Part 6: Mega Mega Problems List (not that big)
1. Find sin 82.5 degrees.
2. Find formulas for sin( x2 ), cos( x2 ).
3. Evaluate 16 cos x sin x cos 2x cos 4x cos 8x.
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 tan2 θ can be written in the form pq 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!