Trigonometry Continued solutions
ο»ΏTrigonometry Continued solutions
1)Prove the identities: a)tan()sin() + cos() = sec()
Solution: tan()sin() + cos()
=
() ()
()
+
()
=
2() ()
+
2() ()
=
2()+2() ()
=
1 ()
= ().
b)1+22() = sin(2)
Solution:
2 () 1+2()
=
2 () sec2()
= 2 () 2()
= 2(()) 2()
= 2sin(x)cos(x) = sin(2x)
c)1+(())
+
cos() sin()
=
csc()
Solution:
() 1+()
+
() ()
=
2()+()[1+()] ()[1+()]
=
2()+2()+() ()[1+()]
=
1+() ()[1+()]
=
1 ()
= csc()
d)sec()
+
tan()
=
() 1-()
Solution: sec() + tan()
=
1 ()
+
() ()
=
1+() ()
=
[1+()][1-()] ()[1-()]
=
1-2() ()[1-()]
=
2() ()[1-()]
= 1-(()).
2)In oblique triangle ABC, A = 34?, B = 68? and a = 4.8. Find b to the nearest tenth.
Solution:
b
=
() ()
=
4.8sin(68) sin(34)
8.0
Make sure your calculator is in degree mode!
3)Find all values of x such that sin(2x) = sin(x) and 0 2.
Solution: sin(2x) = sin(x)
2sin(x)cos(x) = sin(x)
2sin(x)cos(x) ? sin(x) = 0
sin(x)[2cos(x) ? 1] = 0
sin(x)
=
0
or
2cos(x)
?
1
=
0
cos(x)
=
1 2
x
=
n
or
x
=
3
+
2,
x
=
5 3
+
2
for
n
.
The
subset
of
these
solutions
that
lie
in
the
interval
[0,
2]
are
x
=
0,
3
,
,
5 3
,
2.
4)If the rectangular coordinates of a point are (1, -1), find a polar representation of the point.
Solution: There are infinitely many polar representations of the given point. We will find the polar coordinates of (1, -1) that has r > 0 and 0 < 2. The polar point is (2, 74).
5)Sketch the graph of the function y = 1 + sin(2x) without using a calculator.
Solution: Amplitude = 1
Period =
Phase shift = 0
Midline: y = 1
Increment
=
4
Special points: (0, 1), (4, 2), (2, 1), (34, 0), (, 1)
6)Perform the indicated operation: 5[cos(15?) + isin(15?)] 10[cos(5?) + isin(5?)]
Solution: 5(10)[cos(15)cos(5) + icos(15)sin(5) + isin(15)cos(5) + i2sin(15)sin(5)] = 50[(cos(15)cos(5) ? sin(15)sin(5)) + i(sin(15)cos(5) + cos(15)sin(5))] = 50[cos(15 + 5) + isin(15 + 5)] = 50cos(20) + i50sin(20).
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