Some formulas for trigonometry in Math 222
September 3, 2011
In Math 222, we come up with two formulas frequently:
x
x
x
x
cos(x) = cos2 ( ) − sin2 ( ) = 2 cos2 ( ) − 1 = 1 − 2 sin2 ( )
2
2
2
2
x
x
sin(x) = 2 sin( ) cos( )
2
2
(1)
(2)
And, we need to know their equivalent formulas also. For example: 1 +
cos(2θ) = 2 cos2 (θ). In the integrals below, for example, we need these
formulas:
Z p
1 + cos(4x)dx
Z p
1 − sin(x)dx
More general formulas are:
cos(α + β) = cos(α) cos(β) − sin(α) sin(β)
(3)
sin(α + β) = sin(α) cos(β) + cos(α) sin(β)
(4)
Let α = β = x2 , and you can get the first two formulas.
Replacing β with −β in (3) and (4), we have:
cos(α − β) = cos(α) cos(β) + cos(α) sin(β)
(5)
sin(α − β) = sin(α) cos(β) − cos(α) sin(β)
(6)
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Using these formulas, one can easily get:
(7)
sin(α) sin(β)
(8)
sin(α) cos(β)
cos(α) sin(β)
The following integral won’t
1
1
(cos(α − β) + cos(α + β))
2
1
=
(cos(α − β) − cos(α + β))
2
1
=
(sin(α + β) + sin(α − β))
2
1
=
(sin(α + β) − sin(α − β))
2
R
be hard: sin(2x) cos(3x)dx
cos(α) cos(β) =
(9)
(10)
One good way to derive
Following is a good way to derive the above formulas.
That is to Use Euler’s identity:
eix = cos(x) + i sin(x)
(11)
Taking the first two formulas for example, since e2ix = eix eix , using the
identity, we then have:
cos(2x) + i sin(2x) = (cos(x) + i sin(x))2
(12)
Comparing the real and imaginary parts on both sides, you’ll get the first
two formulas.
It’s also quite easy to obtain (3) and (4). Because eiα eiβ = ei(α+β) , we
have:
(cos(α) + i sin(α))(cos(β) + i sin(β)) = cos(α + β) + i sin(α + β)
(13)
Multiplying the left side out and compare the real and imaginary parts on
both sides, you can get the two formulas.
I don’t think this is the proof. We used the Euler’s identity there,
however, to get Euler’s identity, we must know the Taylor series of ex , sin(x)
and cos(x). To know the expansions of sin(x) and cos(x), we must know the
derivatives of sin(·) and cos(·). To get the conclusions sin0 (x) = cos(x) and
cos0 (x) = − sin(x), we must use . . . We’re back.
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2
The elementary proofs
According to what’s said above, we just need to prove (3) and (4). We
first do so for 0 ≤ α, β ≤ π2 .
If one of the angles is 0 or π/2, it’s obviously correct.
In the figure above, the circle is the unit circle. OB = cos(β). OC =
OB ∗cos(α) = cos(β) cos(α). EC = BD = AB ∗sin(∠BAD) = AB ∗sin(α) =
sin(β) sin(α). OE = OC − EC. OE = cos(α + β). The first is correct.
AD = AB ∗ cos(∠BAD) = AB ∗ cos(α) = sin(β) cos(α).
DE = BC = OB ∗ sin(α) = cos(β) sin(α)
Then, sin(α + β) = AE = DE + AD. We’re done.
For general case, using sin(x) = sin(x + 2kπ), cos(x) = cos(x + 2kπ),
sin(−x) = − sin(x), cos(−x) = cos(x), sin(π − x) = sin(x), cos(π − x) =
− cos(x), sin(π/2 − x) = cos(x), cos(π/2 − x) = sin(x) etc, one can reduce
them to the above case.
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