Trigonometry Continued solutions

Trigonometry 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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