Trigonometry
Maximum and Minimum
Values
Reminders
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The maximum value of sin (x)o is
The minimum value of cos (2x)o is
The maximum value of sin (10x)o is
The minimum value of 3cos (x)o is
The maximum value of 4sin (3x)o is
The period of sin (x)o is
The period of cos (2x)o is
The period of 5sin (3x)o is
1
-1
1
-3
4
360
180
120
Here are the graphs of y = sin( x)o, y = 2 sin (x)o
and y = 3 sin (2x)o
sin.ogr
4
y
3
y =2 sin xo
2
y =3 sin 2xo
1
y = sin xo
0
0
-1
-2
-3
-4
90
180
270
360
x
Example 1
Find the maximum value of f(x) = 5 + 2sin xo and the
corresponding value of x. 0 < x < 360.
5 +2 (1) = 7
f(x) = 5 + 2 sin xo
5 + 2 (-1) = 3
y= sin x
1
2sin xo = 2
x
0
-1
2sin xo = 7 – 5
Can use graph
instead of quadrants
for sin xo or cos xo =
1 or – 1 or 0.
0
90
180
270
360
Min
5 + 2sin xo = 7
Max = 7 when
y
Max
sin xo = 1
x = 90
Example 2
A Trig Graph Problem
d(t) = 2000 + 1000 cos 36to is the depth in metres of water in a tank t
minutes after the start of an experiment.
(a) What is the depth of water in the tank at the start?
(b) Does the depth ever reach 3500 metres? (Give a reason for your
answer).
(c) What is the minimum depth and when does it occur?
(d) Find the depth after 7 minutes.
d(t) = 2000 + 1000 cos 36to
(a) At the start t = 0,
d(0) = 2000 + 1000 cos 36(0)o
= 2000 + 1000 cos 0o
= 3000 metres
.
2000 + 1000( 1 ) = 3000
MAX
d(t) = 2000 + 1000 cos 36to
(b)
2000 + 1000 ( - 1 ) = 1000
MIN
Since maximum value of d = 3000 m, the depth never reaches 3500 m.
(c)
Min depth = 1000 metres
when
2000 + 1000 cos 36to = 1000
1000 cos 36to = 1000 – 2000
1000 cos 36to = –1000
cos 36to = –1
36t = 180
t = 180/36
t=5
y
y =cos x
1
0
x
0
90
180
270
360
-1
Therefore it occurs after 5 minutes
d(t) = 2000 + 1000 cos 36to
(d)
(e)
Period = 360/36 = 10 mins
t = 7, d(t) = 2000 + 1000 cos (36 (7) )o
= 1691 m
and now the graph ………….
(check your answer to (c))
Max = 3000
Period = 360/36 = 10
Min = 1000
Displacement = +2000 (
2000)
Amplitude a = 1000
d
d(t) = 2000 + 1000 cos (36t)o
or
d(t) = 1000 cos (36t)o +2000
(5, 1000)
t