AMS 151 (Fall, 2009)
Joe Mitchell
Applied Calculus I
Practice Problems for Quiz # 3 – Solution Notes
1. Evaluate the following expressions:
(a). cos(tan
−
1
(
5
9
)) =
Draw a right triangle and let
θ
be one of the two angles that is not
π/
2. Then,
θ
= tan
−
1
(
5
9
) if the side opposite
θ
is of length 5 and the side adjacent is of length 9. Then, the hypotenuse is of length
√
5
2
+ 9
2
=
√
106. Thus,
cos(tan
−
1
(
5
9
)) = cos
θ
=
9
√
106
.
(b). tan(sin
−
1
(
√
3
2
)) =
Draw a right triangle and let
θ
be one of the two angles that is not
π/
2. Then,
θ
= sin
−
1
(
√
3
2
) if the side opposite
θ
is of length
√
3 and the hypotenuse is of length 2. Then, the side adjacent is of length
r
2
2
−
(
√
3)
2
= 1. Thus,
tan(sin
−
1
(
√
3
2
)) = tan
θ
=
√
3. (Note that
θ
=
π/
3, 60 degrees.)
(c). sin(
20
π
3
) =
Note that
20
π
3
= 2
π
+ 2
π
+ 2
π
+
2
π
3
(i.e., it is 3 revolutions, plus an additional angle 2
π/
3 (120 degrees). Thus,
sin(
20
π
3
) = sin(2
π/
3) =
√
3
2
.
(d). sec(cos
−
1
(
2
3
)) =
Draw a right triangle and let
θ
be one of the two angles that is not
π/
2. Then,
θ
= cos
−
1
(
2
3
) if the side adjacent
to
θ
is of length 2 and the hypotenuse is of length 3. Then, the side opposite is of length
√
3
2
−
2
2
=
√
5. Thus,
sec(cos
−
1
(
2
3
)) = sec
θ
= 1
/
(cos
θ
) = 1
/
(2
/
3) = 3
/
2.
2. The intensity of radiation,
I
, oscillates sinusoidally between a low of 100 and a high of 1200. The time between
one peak and the next peak is 5 minutes. At time 30 seconds after noon today, the radiation level is measured to be
650, and it is rising. Write the function
I
(
t
) for intensity as a function of
t
, the number of seconds after noon today.
Since
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 Fall '08
 Zhang
 Limit of a function, +∞

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