130
Chapter 6
•
Additional Topics
§6.1
Example6.2
Solve the equation 2cos
2
θ
−
1
=
0.
Solution:
Isolating cos
2
θ
gives us
cos
2
θ
=
1
2
⇒
cos
θ
= ±
1
radicallow
2
⇒
θ
=
π
4
,
3
π
4
,
5
π
4
,
7
π
4
,
and since the period of cosine is 2
π
, we would add 2
π
k
to each of those angles to get the general
solution. But notice that the above angles differ by multiples of
π
2
. So since every multiple of 2
π
is
also a multiple of
π
2
, we can combine those four separate answers into one:
θ
=
π
4
+
π
2
k
for
k
=
0,
±
1,
±
2,
...
Example6.3
Solve the equation 2 sec
θ
=
1.
Solution:
Isolating sec
θ
gives us
sec
θ
=
1
2
⇒
cos
θ
=
1
sec
θ
=
2
,
which is impossible. Thus, there is no solution
.
Example6.4
Solve the equation cos
θ
=
tan
θ
.
Solution:
The idea here is to use identities to put everything in terms of a single trigonometric
function:
cos
θ
=
tan
θ
cos
θ
=
sin
θ
cos
θ
cos
2
θ
=
sin
θ
1
−
sin
2
θ
=
sin
θ
0
=
sin
2
θ
+
sin
θ
−
1
The last equation looks more complicated than the original equation, but notice that it is actually a
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 Fall '11
 Dr.Cheun
 Calculus, Addition, Angles, Sin, Cos, Quadratic equation, Elementary algebra

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