Similar Triangles, Right Triangles, And The Definition Of The Sine, Cosine And Tangent Functions Of Angles Of A Right Triangle Worksheets With Answer Key Page 35
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43π
5
θ
π
θ
=
−
=
2
.
2
1
3
π
π
5
θ
π
θ
θ
The only solutions in the interval 0 ≤
< 2
are
=
and
=
. We use these
1
2
3
3
angles to represent the general solution:
π
π
5
θ
π
θ
π
=
+
=
+
2
n
or
2
n
,
where n is any integer.
3
3
All of these solutions can be represented in a single expression as
π
θ
π
=
±
+
n an integer.
2
n
,
3
−
3
θ
=
Example: Find the general solution of the equation
tan
.
3
Solution: One solution is
−
π
−
3
θ
=
=
1 -
Tan
.
1
3
6
The general solution can be represented as
π
−
θ
π
=
+
n an integer.
n
,
6
However, it is customary to represent the general solution in terms of a solution
π
5
π
θ
θ
π
+
=
between 0 and
. The angle
is not in this interval, so we use
to represent
1
1
6
the general solution as
π
5
θ
π
=
+
n
,
n an integer.
6
Problems:
In problems 188-191, find the general solution to each trigonometric equation.
2
θ
θ
sin =
189. tan
= -1
188.
2
3
θ
θ
=
−
190. cos
= 0
191.
sin
2
Equations of the form
θ
θ
θ
sin B
= c, cos B
= c and tan B
= c,
where B and c are constants, are slightly more complicated than the equations just considered. Since
π
2
θ
θ
θ
θ
sin B
and cos B
have period
, to express the general solution to sin B
= c and cos B
= c, add
B
π
π
2
2
θ
to the solutions in the interval 0 ≤
(positive and negative) integer multiples of
<
. The
B
B
π
θ
θ
function tan B
has period
so the general solution to tan B
= c is expressed by adding integer
B
π
π
θ
to the solutions in the interval 0 ≤
multiples of
<
.
B
B
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