Math 110 Some answers to sample test 2
1. Do p. 760 #60. Solve sec( t ) = - 2 / sqrt( 3 ).
Invert to obtain cos( t ) = -sqrt( 3 ) / 2.
cosine is negative in quadrants 2 and 3. The reference angle is π/6.
So t = 5π/6 in quadrant 2 and t = 7π/6 in quadrant 4.
Thus all solutions are t = 5π/6 + 2kiπ or 7π/6 + 2kiπ where k is an integer.
2. Do p. 762 #80.
From the observation deck of the lighthouse at Sasquatch Point 50 feet above the surface of
Lake Ippizuti,a lifeguard spots a boat out on the lake sailing directly toward the lighthouse.
The first sighting had an angle of depression of 8.2 degrees and the second sighting had
an angle of 25.9 degrees. How far had the boat traveled between the sightings?
With reference to the figure, find PQ.
Here TB = 50 feet. Angle TPB is 8.2 degrees. Angle TQB is 25.9 degrees.
Find PQ as PB - QB.
tan( TQB ) = TB / QB. So QB = TB / tan( TQB ) = 50 / tan( 25.9◦ )
tan( TPB ) = TB / PB. So PB = TB / tan( TPB ) = 50 / tan( 8.2◦ ).
So PQ = PB - QB = 50 / tan( 25.9◦ ) - 50 / tan( 8.2◦ ) = 146.4 feet.
3. Do p. 762 #91. Show
cos(t)
Left = 1−sin(t)
2
cos(t)
1
= cos(t)2 = cos(t)
cos(t)
1−sin(t)2
= sec(t).
Use a Pythogoran identity on the bottom.
= sec(t)
1
1
4. Do p. 763 #116. Show csc(t)−cot(t)
− csc(t)+cot(t)
= 2cot(t).
1
B−A
1
Use the algebraic identity A − B = AB on the left side.
csc(t)+cot(t)−(csc(t)−cot(t))
1
1
Left = csc(t)−cot(t)
− csc(t)+cot(t)
= (csc(t)+cot(t))·(csc(t)−cot(t))
Simplify the top. FOIL the bottom.
=
=
=
2cot(t)
csc(t)2 −csc(t)cot(t)+cot(t)csc(t)−cot(t)2
2cot(t)
Use a Pythagorean
csc(t)2 −cot(t)2
2cot(t)
= 2cot(t) = right
1
Simplify the bottom. Use a Pythagorean identity.
identity.
5. Do p. 782 #20. Find the exact value of csc( 5 Pi / 12 ).
Notice 5/12 = 1/2 * 5/6. Use a half
qangle identity for sine.
1−cos( 5π )
5π
1 5π
6
csc( 5π
.
12 ) = sin( 12 ) = sin( 2 6 ) = ±
2
5π
Since 12 is in the first quadrant where the sine is positive, choose the plus sine.
q
q √
√
1− −2 3
1+ 23
5π
csc( 12 ) =
=
2
2
Multiply the
top
and
bottom
of the fraction inside the outer sqrt by 2.
√
q
csc( 5π
12 ) =
√
2+ 3
4
=
√
2+ 3
2
6. Do p. 782 #22.If A is a Quadrant IV angle with cos( A ) = √15 and sin( B ) = √110
where π/2 < B < π, then find a) cos( A + B ) b) sin( A + B ) c) tan( A + B )
d) cos( A - B ) e) sin( A - B ) f) tan( A - B ).
−2
By the Pythagoran Theorem sin( A ) = √
and cos( B ) = √−3
5
10
To find parts a b and c use sum formulas. To find parts d e and f use difference formulas.
−2 √1
√
cos( A + B ) = cos( A ) cos( B ) - sin( A ) sin( B ) = √15 √−3
−√
= 5−1
10
5 10
2
1
sin( A + B ) = sin( A ) cos( B ) + cos( A ) sin( B ) =
tan( A + B ) = sin( A + B )/cos( A + B ) =
tan( A + B ) =
7
√
5 2
−1
√
5 2
−2 √
−3
√
5 10
+
√1 √1
5 10
=
7
√
5 2
= −7 from the previous 2 lines.
tan(A)+tan(B)
1−tan(A)tan(B)
tan( A ) = sin( A ) / cos( A ) =
tan( B ) = sin( B ) / cos( B ) =
−2
√
5
1
√
5
√1
10
−3
√
10
= −2
=
−1
3
tan(A)+tan(B)
−2+(−1/3)
So tan( A + B ) = 1−tan(A)tan(B)
= 1−(−2)·(−1/3)
= −7/3
The two answers agree.
1/3 = −7
Now use the difference formulas.
−2 √1
−1
√ = √
+√
= 5−5
cos( A - B ) = cos( A ) cos( B ) + sin( A ) sin( B ) = √15 √−3
10
5 10
2
2
−2 √
−3
5
sin( A - B ) = sin( A ) cos( B ) - cos( A ) sin( B ) = √
− √15 √110 = 5√
= √12
5 10
2
tan( A - B ) = sin( A - B )/cos( A - B ) = -1 from the previous 2 lines.
tan(A)−tan(B)
tan( A - B ) = 1+tan(A)tan(B)
=
−2−(−1/3)
1+(−2)·(−1/3)
=
−5/3
5/3
= −1
The two answers agree.
sin(A)cos(A)+sin(B)cos(B)
7. Do p. 783 #35. Show tan(A+B)
tan(A−B = sin(A)cos(A)−sin(B)cos(B)
Write left sidesin(A+B)
in terms of sines and cosines.
tan(A+B)
tan(A−B
cos(A+B)
sin(A−B)
cos(A−B)
=
Invert the bottom and multiply.
tan(A+B)
sin(A+B) cos(A−B)
Use sum and difference formulas for sine and cosine.
tan(A−B = cos(A+B) · sin(A−B)
tan(A+B)
(sin(A)cos(B)+cos(A)sin(B)·(cos(A)cos((B)+sin(A)sin(B))
FOIL the top and the
tan(A−B = (cos(A)cos(B)−sin(A)sin(B))·(sin(A)cos(B)−cos(A)sin(B))
cos(A)sin(A)cos(B)2 +sin(A)2 cos(B)sin(B)+cos(A)2 cos(B)sin(B)+cos(A)sin(A)sin(B)2
= cos(A)sin(A)cos(B)2 −cos(A)2 cos(B)sin(B)−sin(A)2 cos(B)sin(B)+cos(A)sin(A)sin(B)2
bottom.
In the top and in the bottom combine the first and last terms and the middle 2 terms.
Use xcos(t)2 + xsin(t)2 = x by the Pythagorean identity.
cos(A)sin(A)+cos(B)sin(B)
lef t = cos(A)sin(A)−cos(B)sin(B)
= right
8. Do p. 783 #47. Find the exact value of sin(5π/8)
Note 5/8 = (1/2) ( 5/4 ). Use the half angle formula for sine.
5π/8 is in quadrant 2 where sine
choose plus
q is−1positive.
q So
q
1
5π
√
√
1−
1+
1−cos( 4 )
2
2
sin(5π/8) = +
=
=
2
2
2
√
Multiply top q
and bottom of the fraction by 2
√
sin(5π/8) =
sin(5π/8) =
2+1
√
√ 2 √2
2+ 2
2
Multiply top and bottom of the fraction by
√
2
9. Do p. 784 #56. If sin( t ) = 5/13 and π/2 < t < π, then find the exact values of
a) sin( 2t ) b) cos( 2t ) c) tan( 2t )
d) sin( t/2 ) e) cos( t/2 ) f) tan( t / 2 )
For parts a b and c use the double angle formulas.
For parts d e and f use the half angle formulas.
By the Pythagorean formula, cos( t ) = -12/13 since the cosine is negative in the second quadrant.
a) sin( 2t ) = 2 sin( t ) cos( t ) = 2 ( 5/13 )( -12/13 ) = -120/169.
b) cos( 2t ) = 1 − 2sin(t)2 = 1 − 2(5/13)2 = 1 − 50/169 = 119/169
c) tan( 2t ) = sin( 2t ) / cos( 2t ) = -120/119 from the previous 2 lines. Check with the tangent formula.
tan( 2t ) =
2tan(t)
1−tan(t)2
=
2· −5
12
1− −5
12
2
=
−5
6
25
1− 144
=
−5
6
119
144
=
−5
6
·
144
119
= −5 ·
24
119
=
−120
119
which agrees with the previous answer.
Since t is in the second quadrant, t/2 is in the first quadrant.
In the first quadrant each of sine, cosine and tangent is positive.
So in the halfqangle formulas
plus.
q choose
q 12
q
1− −12
1+ 13
1−cos(t)
25
13
√5
sin( t/2 ) =
=
=
=
2
2
26 =
26
q 2−12
q
1+ 13
1+cos(t)
1
cos( t/2 ) =
=
= √26
2
2
tan( t/2) = sin( t/2 ) / cos( t/2 ) = 5 from the previous two lines.
10. Do p. 784 #69. Show cos(4t) = 8cos(t)4 − 8cos(t)2 + 1.
Write 4t = 2 · 2t. Use the double angle formula for cosine. cos( 2A ) = 2cos(A)2 − 1
Left = cos( 4t ) = 2cos(2t)2 − 1 = 2(2cos(t)2 − 1)2 − 1 Square the first term.
= 2(4cos(t)4 − 4cos(t)2 + 1) − 1 = 8cos(t)4 − 8cos(t)2 + 1 = right
11. Do p. 784 #73. Show
See today’s class notes.
1
cos(t)−sin(t)
−
1
cos(t)+sin(t)
=
2sin(t)
cos(2t)
2