Trigonometry Handout

Trigonometry
Now, back to our problem, If
1
sin θ = , then what is θ ?
2
One way to come up with the solution is to look at the unit circle and find out
which points have a y-coordinate of 1 π
2
. You will see that this occurs when θ = and
6
5π
when θ = . Since sine is periodic, we know that we have infinite solutions to our
6
problem, namely θ = π + 2 nπ
and θ 5 π
= + 2n
π .
6 6
1
Now, say instead that we wanted to find (
1
sin − )
. This will only give us one answer, but
2
1
which one will it be? Well, by our restrictions, we need to find θ such that sin θ = and
2
π π
−1
− ≤ θ ≤ . Looking at all of our possible solutions, only one works, (
1
)
π
θ = sin =
2 2
2 6
In general, there are infinitely many solutions to problems where you know the value of
the trigonometric function but not the angle. Using inverse trigonometric functions gives
us what is called the principal value of the relation. The principal value of the relation is
the angle that is a solution to the problem and that has the smallest absolute value. If
positive and negative values both satisfy, then we use the positive angle.
Example: Find tan −1
1
π π
Solution: We want to find x such that tan x = 1 and − ≤ x ≤ . If we use the
2 2
5
unit circle, tan x = 1 when cos x = sin x , which occurs when x =
π , π . Only one of these
4 4
−
values is in our interval, so 1 π
tan 1 = .
4
.
Trigonometric Identities:
1) Consequences of the Definitions
1
1
1) csc θ = 2) sec θ = 3)
sinθ
cosθ
cot θ =
1
tanθ
4)
sinθ
tan θ = 5)
cosθ
cosθ
cot θ =
sinθ
2) Some fundamental identities
2 2
6) cos θ + sin θ = 1
Proof: Refer back to Figure 9. The distance formula gives that
2 2 2
2 2 ⎛ x ⎞ ⎛ y ⎞ x y x + y r
x + y = r . Thus, cos θ + sin θ = ⎜ ⎟ + ⎜ ⎟ = + = = = 1
2 2 2 2
⎝ r ⎠ ⎝ r ⎠ r r r r
2
2
2
2
2
2
2
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