Download Discrete Geometry and Trigonometry

Trigonometry — §4.2.1, 4.2.2
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Trigonometry Identities
! sin2 x + cos2 x = 1
! sin(x + y ) = sin x cos y + cos x sin y
Remember key
values of sin, cos.
! cos(x − y ) = cos x cos y − sin x sin y
Example. Determine the value of cot(π/24).
(p. 233)
π
π
π
cos 24
2 cos2 24
1 + cos 12
1 + cos( π3 − π4 )
=
.
π =
π
π
π
π =
π
sin 24
2 sin 24 cos 24
sin 12
sin( 3 − 4 )
Applying the identities,
1
√
2
+ 4 + 46
√
√
6
2
−
4
4
√
√
√
√
= 2 + 2 + 3 + 6.
Trigonometry — §4.2.1, 4.2.2
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Trigonometry Identities
Example. Find all acute angles x satisfying the equation (p. 231)
2 sin x cos 40◦ = sin(x + 20◦ ).
Solution. Find a solution that works. Then apply identity, solve for x.
2 sin x cos 40◦ = sin(x + 20◦ ) = sin x cos 20◦ + sin 20◦ cos x.
tan x = sin 20◦ /(2 cos 40◦ − cos 20◦ ).
For 0◦ ≤ x < 90◦ , tan x = C can only have one solution.
Idea: You may have to make use of e i θ = cos θ + i sin θ.
Extension: e i (nx) = (e ix )n , so cos nx + i sin nx = (cos x + i sin x)n .
Example. Let f (x) = a + b cos 2x + c sin 5x + d cos 8x for a, b, c, d ∈ R.
Then we can write f (x) as the real part of a + be 2ix − ice 5ix + de 8ix .
Discrete Geometry — §4.1.2, 4.1.6
Discrete Geometry
Idea: Draw good pictures; choose good coordinates.
Example. Inscribe in a circle a trapezoid T w/one side as diameter
and a triangle % with sides parallel to the sides of the trapezoid.
Prove that T and % have the same area.
(p. 208)
Solution. Define T by coordinates (1, 0), (−1, 0), (b, a), (−b, a).
Find the coordinates of %: they are (0, 1), (b, −a), and (−b, −a).
Idea: Points in the plane can be represented as complex numbers.
Example. Let ABC and BCD be two equilateral triangles, and line
passing through D passes through AC at M and through AB at N.
(p. 210)
Prove that the angle between BM and CN is π3 .
√
√
Solution. Choose A = i 3, B = −1, C = 1, and D = −i 3.
Notice DBN is similar to DCM ⇒ can find the coordinates of M
c−n
= te i π/3
and N: if |MC | = 2t, then |BN| = 2t . Calculate b−m
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