Math 171 Group Worksheet over Trigonometry (1.4) Name KEY

Math 171 Group Worksheet over Trigonometry (1.4) September 5, 2017 Credit given for work shown.

Name

KEY

1. Let tan() = 2 and sec() = - 5.

(a) What quadrant is in? III (b) sin() = -2/ 5 (c) cos() = -1/ 5

(d) cot() = 1/2

(e) csc() = - 5/2

Useful identities: cos(x + y) = cos x cos y - sin x sin y sin(x + y) = sin x cos y + cos x sin y

2. Use the addition formula (above) and the triangle (below) to compute sin

2 3

+

exactly.

sin(2/3 + ) = sin(2/3) cos() + cos(2/3) sin()

3 12 -1 5

=

+

2 13 2 13

3. Find all solutions to 3 sin(x)-sin(2x) = 0 in the interval [0, 2]. (Hint: Use formula above.)

3 sin(x) - sin(x + x) = 0

3 sin(x) - (sin x cos x + cos x sin x) = 0

3 sin(x) - (2 sin x cos x) = 0

sin x( 3 - 2 cos x) = 0

sin x = 0 or 3 - 2 cos x = 0

sin x = 0 or cos x =

3 2

x = 0, , 2, /6, 11/6

4. How can we measure the height of a tall building or monument? The following technique used by surveyors is an interesting application of trig. The surveyor stands a fixed distance D from the monument and measures the angle to its top.

(a) Express the height H of the monument in terms of the angle of elevation and D. H tan = D H = D tan

(b) Express the distance D to the monument in

terms of the angle of elevation and height

H.

tan

=

H D

H

D=

= H cot

tan

(c) The height of the Statue of Liberty is 305 feet. If the measured angle of elevation is /3, then how far away is the surveyor? D = H = 305 ft tan 3

5. Determine the height of the triangle in terms of b and . Use this value to determine the area of the triangle in terms of a, b and .

h sin =

b h = b sin

A

=

1 2

?

base

?

height

A

=

1 2

a(b

sin

)

6. Steel is melted to fill a rectangular container measuring 2 ft x 2 ft x 1 ft and then poured into ingots by tilting the container at an angle as illustrated. In the process, it is important to control the rate of turning so that a constant pour rate into the ingots is achieved. Express the total volume V poured into the ingots as a function of the angle . You may have to express the volume using a piecewise function.

When 0 /4, the horizontal length of the top of the molten steel (l) satisfies the

equation

2

2

cos =

l =

l

cos

Using the area formula from the above question,

1

1

2

2 sin

Volume = Area ? 1 = (2)(l) sin = (2)

sin =

= 2 tan

2

2 cos

cos

When /4 /2, the horizontal length of the top of the molten steel (l) satisfies the

equation

2

2

cos(/2 - ) =

l =

l

cos(/2 - )

1

1

Volume = 4 - Area ? 1 = (2)(l) sin(/2 - ) = 4 - (2)

2

2

2 sin(/2 - )

=4-

= 4 - 2 tan(/2 - )

cos(/2 - )

2 cos(/2 - )

sin(/2 - )

Together Volume =

2 tan

if 0 /4

4 - tan(/2 - ) if /4 < /2

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