Download Math 123 - College Trigonometry Euler`s Formula and Trigonometric

Math 123 - College Trigonometry
Euler’s Formula and Trigonometric Identities
Euler’s formula, named after Leonhard Euler, states that:
For any real number x,
eix = cos(x) + isin(x)
(1)
Where e is the base of the natural logarithm (e = 2.71828 . . .), and i is the imaginary unit.
Using this formula you can derive most of the trigonometric identities/formulas (sum and
difference formulas, double and half angle formulas, and even the Pythagorean formulas).
Pythagorean formula:
cos2 (x) + sin2 (x) = 1
By the rule of exponent we know that
eix · e−ix = 1
However, using the Euler’s formula, we know
eix = cos(x) + i sin(x)
and
e−ix = cos(x) − i sin(x)
Hence,
1 = eix · e−ix
= cos(x) + i sin(x) · cos(x) − i sin(x)
= cos2 (x) − i cos(x) sin(x) + i cos(x) sin(x) − i2 sin2 (x)
= cos2 (x) + sin2 (x)
Therefore, cos2 (x) + sin2 (x) = 1.
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Sum formulas:
Here we want to use the Euler’s formula to derive the formulas for:
cos(x + y), sin(x + y) and tan(x + y)
First, note that by the Euler’s formula we have:
ei(x+y) = cos(x + y) +i sin(x + y)
| {z }
| {z }
Real Part
(2)
Imaginary Part
Now, expanding ei(x+y) gives you:
ei(x+y) = eix · eiy
= cos(x) + i sin(x) · cos(y) + i sin(y)
= cos(x) cos(y) + i cos(x) sin(y) + i sin(x) cos(y) + i2 sin(x) sin(y)
= cos(x) cos(y) − sin(x) sin(y) +i sin(x) cos(y) + cos(x) sin(y)
{z
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{z
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Imaginary Part
Real Part
Now, by equation (2), ei(x+y) = cos(x + y) + i sin(x + y), you can easily see that:
cos(x + y) = cos(x) cos(y) − sin(x) sin(y)
and
sin(x + y) = sin(x) cos(y) + cos(x) sin(y)
To get the formula for tan(x + y), all we need to do is applying the definition of tangent.
tan(x + y) =
=
=
=
=
sin(x + y)
cos(x + y)
sin(x) cos(y) + cos(x) sin(y)
cos(x) cos(y) − sin(x) sin(y)
sin(x) cos(y) + cos(x) sin(y)
cos(x) cos(y)
÷
(divide by 1 in a clever way)
cos(x) cos(y) − sin(x) sin(y)
cos(x) cos(y)
sin(x) cos(y) + cos(x) sin(y) ÷ cos(x) cos(y)
cos(x) cos(y) − sin(x) sin(y) ÷ cos(x) cos(y)
tan(x) + tan(y)
(using algebra and definition of tangent)
1 − tan(x) tan(y)
Thus,
tan(x + y) =
tan(x) + tan(y)
1 − tan(x) tan(y)
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Difference formulas:
Here we want to use the Euler’s formula to derive formulas for: Here we want to use the
Euler’s formula to derive the formulas for:
cos(x − y), sin(x − y) and tan(x − y)
This is similar to the sum identities. Instead of expanding ei(x+y) , we will expand ei(x−y)
instead.
ei(x+y) = cos(x + y) +i sin(x + y)
(3)
| {z } | {z }
Real Part
Imaginary
Now, expanding ei(x+y) gives you:
ei(x−y) = eix · e−iy
= cos(x) + i sin(x) · cos(y) − i sin(y)
= cos(x) cos(y) − i cos(x) sin(y) + i sin(x) cos(y) − i2 sin(x) sin(y)
= cos(x) cos(y) + sin(x) sin(y) +i sin(x) cos(y) − cos(x) sin(y)
{z
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{z
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Imaginary Part
Real Part
Using equation (3), we can equate the real and imaginary part. Hence,
cos(x − y) = cos(x) cos(y) + sin(x) sin(y)
and
sin(x − y) = sin(x) cos(y) − cos(x) sin(y)
To get tan(x − y), you compute
sin(x−y)
,
cos(x−y)
tan(x − y) =
similar to what we did for tan(x + y). Hence,
tan(x) − tan(y)
1 + tan(x) tan(y)
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Double-Angle formulas:
The double angle formula falls out of the Euler’s formula very quickly. First note that,
ei2x = ei(x+x) = eix · eix . Then by f we expand this, we would have:
eix · eix = cos(x) + i sin(x) · cos(x) + i sin(x)
= cos2 (x) + i cos(x) sin(x) + i sin(x) cos(x) − sin2 (x)
(4)
= cos2 (x) − sin2 (x) +i 2 cos(x) sin(x)
{z
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{z
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Imaginary Part
Real Part
However, by the Euler’s formula we have that
(eix )2 = ei2x = cos(2x) +i sin(2x)
| {z } | {z }
(5)
cos(2x) = cos2 (x) − sin2 (x)
= cos2 (x) − (1 − cos2 (x))
= 2 cos2 (x) − 1
= 1 − 2 sin2 (x)
(6)
sin(2x) = 2 cos(x) sin(x)
(7)
Real
Imaginary
Therefore,
and
In fact, using the Euler’s formula, you can get n-angle formula. All you have to do is expand
(eix )n = einx . For example, to get a triple angle formula, I would expand
ei3x = eix · eix · eix
= eix · cos2 (x) − sin2 (x) + i2 cos(x) sin(x)
= cos(x) + i sin(x) · cos2 (x) − sin2 (x) + i2 cos(x) sin(x)
= cos3 (x) − cos(x) sin2 (x) + i2 cos2 (x) sin(x) + i sin(x) cos2 (x) − i sin3 (x) + i2 2 cos(x) sin2 (x)
= cos3 (x) − cos(x) sin2 (x) − 2 cos(x) sin2 (x) + i 2 cos2 (x) sin(x) + sin(x) cos2 (x) − sin3 (x)
= cos3 (x) − 3 cos(x) sin2 (x) + i 3 cos2 (x) sin(x) − sin3 (x)
However, from the Euler’s formula, we know that
ei3x = cos(3x) + i sin(3x)
Therefore,
cos(3x) = cos3 (x) − 3 cos(x) sin2 (x)
sin(3x) = 3 cos2 (x) sin(x) − sin3 (x)
Using this technique, you can easily derive formula cos(nx) and sin(nx). From the double
angle formulas, you can get half angle formulas quickly. From the triple angle formulas,
you can get third angle formulas but not so easy (Do you see why?).
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Product to Sum Formulas:
Product to sum formulas can be easily derived using the Euler’s formula.
First, note that
i(x+y)
i(x−y)
ix iy
−iy
ix
e
+e
= e (e + e ) = e
cos(y) + i sin(y) + cos(y) − i sin(y)
(8)
= 2eix cos(y) = 2 cos(x) cos(y) +i 2 sin(x) cos(y)
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Real Part
Imaginary Part
However, note that
ei(x+y) + ei(x−y) = cos(x + y) + i sin(x + y) + cos(x − y) + i sin(x − y)
= cos(x + y) + cos(x − y) +i sin(x + y) + sin(x − y)
|
{z
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{z
}
Real Part
(9)
Imaginary Part
Therefore, we have the followings
cos(x) cos(y) =
1
cos(x − y) + cos(x + y)
2
(10)
sin(x) cos(y) =
1
sin(x + y) + sin(x − y)
2
(11)
and
Now, consider following:
e
i(x+y)
−e
i(x−y)
ix
=e
iy
e −e
−iy
=e
ix
cos(y) + i sin(y) − cos(y) − i sin(y)
= 2ieix sin(y) = −2 sin(x) sin(y) +i 2 cos(x) sin(y)
{z
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{z
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Real Part
(12)
Imaginary Part
But again,
ei(x+y) − ei(x−y) = cos(x + y) + i sin(x + y) − cos(x − y) − i sin(x − y)
= cos(x + y) − cos(x − y) +i sin(x + y) − sin(x − y)
|
{z
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{z
}
Real Part
(13)
Imaginary Part
Thus, we have the followings:
sin(x) sin(y) =
1
cos(x − y) − cos(x + y)
2
(14)
cos(x) sin(y) =
1
sin(x + y) − sin(x − y)
2
(15)
and
Now to get the sum to product formulas all you have to do is make a change of variables.
I will let you think about this.
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