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Math 110 Some answers to sample test 3
1. Do p. 782 #24.
sin( A ) = 3/5 and A is in quadrant 1. cos( B ) = 12/13 and B is in quadrant 4.
The triangle with angle A has an adjacent leg of 4 by the Pythagorean theorem.
The triangle with angle B has an opposite leg of -5 by the Pythagorean theorem.
a) Find sin( A + B ). Use the sine of a sum formula.
sin( A + B ) = sin( A ) cos( B ) + cos( A ) sin( B ) = 3/5 * 12/13 + 4/5 * (-5)/13 = 16 / 65
b) Find cos( A - B ). Use the cosine of a difference formula.
cos( A - B ) = cos( A ) cos( B ) + sin( A ) sin( B ) = 4/5 * 12/13 + 3/5 * ( -5)/13 = 33 / 65.
c) Find tan( A - B ). Use the tangent of a difference formula.
tan( A - B ) =
tan(A)−tan(B)
1+tan(A)tan(B)
=
−5
3
4 − 12
1+ 34 · −5
12
=
56
48
1− 15
48
=
56
33
2. Do p. 783 #34.
cos(A+B)
Show cos(A−B)
= 1−tan(A)tan(B)
1+tan(A)tan(B)
Start with the right side. Use tan( w ) =
right =
right =
right =
sin(A)
1− cos(A)
sin(A)
1+ cos(A)
sin(B)
cos(B)
sin(B)
cos(B)
sin(w)
cos(w) .
Multiply top and bottom by cos( A ) cos( B ).
cos(A)cos(B)−sin(A)sin(B)
cos(A)cos(B)+sin(A)sin(B)
cos(A+B)
cos(A−B) = lef t
Use cosine of sum and of difference identities.
3. Do p. 783 #46.
Find cos( π8 ). Use the half angle identity for cosine with π8 = 12 · π4 .
q
1+cos( π
4)
Choose plus since π8 is in quadrant 1.
cos( π8 ) = ±
2
q
√
1+ √12
cos( π8 ) =
Multiply top and bottom inside the outer sqrt by 2
q√ 2
√
2+1
√
cos( π8 ) =
Multiply top and bottom inside the outer sqrt by 2.
2
2
√ √
2
cos( π8 ) = 2+
2
4. Do p. 784 #54.
sin( t ) = -4/5 and t is in quadrant 3.
The side adjacent to angle t has length -3 by the Pythagorean theorem.
Use the double angle formulas. sin( 2t ) = 2 sin( t ) cos( t ) = 2 (-4/5) (-3/5) = 24/25.
cos(2t) = 1 − 2sin(t)2 = 1 − 2(−4/5)2 = −7
25
tan( 2t ) = sin( 2t ) / cos( 2t ) = -24/7.
π
3π
Use the half angle formulas. Since t is in quadrant 3, π < t < 3π
2 and 2 < t/2 < 4 and so t/2 is in quadrant 2.
q
sin( 2t ) = ± 1−cos(t)
Choose plus since t/2 is in quadrant 2 where sine is positive.
q −3 2 q
q
1− 5
8
4
√2
sin( 2t ) =
=
=
2
10
5 =
5
For cos( t/2q) choose minus since t/2 is in quadrant 2 where cosine is negative.
q
q
1+ −3
2
1
−1
5
√
cos( 2t ) = −
=
−
=
−
2
10
5 =
5
tan( 2t ) =
sin( 2t )
cos( 2t )
=
2
√
5
−1
√
5
= −2.
5. Do p. 784 #68.
Show 32sin(t)4 cos(t)2 = 2 − cos(2t) − 2cos(4t) + cos(6t)
Start by working with each term of the right side.
cos(6t) = cos(4t + 2t) = cos(4t)cos(2t) − sin(4t)sin(2t) by the cosine of a sum formula.
cos(6t) = (2cos(2t)2 − 1)cos(2t) − 2sin(2t)cos(2t)sin(2t) by double angle formulas for 4t.
cos(6t) = 2cos(2t)3 − cos(2t) − 2cos(2t)sin(2t)2
cos(6t) = 2cos(2t)3 − cos(2t) − 2cos(2t)(1 − cos(2t)2 ) since sin(w)2 = 1 − cos(w)2
cos(6t) = 4cos(2t)3 − 3cos(2t) by algebra.
Now cos(4t) = 2cos(2t)2 − 1 by double angle formula for cos(4t ).
So −2cos(4t) = −4cos(2t)2 + 2
Then right = 2 − cos(2t) − 2cos(4t) + cos(6t)
right = 2 − cos(2t) − 4cos(2t)2 + 2 + 4cos(2t)3 − 3cos(2t)
So right = 4 − 4cos(2t) − 4cos(2t)2 + 4cos(2t)3 = 4(1 − cos(2t))(1 − cos(2t)2 )
right = 4(1 − cos(2t))(sin(2t)2 )2 by a Pythagorean identity.
right = 4(1 − (1 − 2sin(t)2 ))(2sin(t)cos(t))2
right = 4 · 2sin(t)2 · 4sin(t)2 cos(t)2
1
right = 32sin(t)4 cos(t)2 = lef t
6. Do p. 784 #71.
cos(t)
sin(t)
Show sec(2t) = cos(t)+sin(t)
+ cos(t)−sin(t)
Add the fractions on the right side. LCD = ( cos( t ) + sin( t ) )(cos( t ) - sin( t ) ).
right = cos(t)·(cos(t)−sin(t))+sin(t)·(cos(t)+sin(t))
Use the distributive law in the top.
(cos(t)+sin(t))·(cos(t)−sin(t))
right =
right =
right =
right =
cos(t)2 −cos(t)sin(t)+cos(t)sin(t)+sin(t)2
(cos(t)+sin(t))·(cos(t)−sin(t))
1
FOIL in the bottom.
(cos(t)+sin(t))·(cos(t)−sin(t))
1
Use
a
double
angle formula for cosine.
cos(t)2 −sin(t)2
1
cos(2t) = sec(2t) = lef t
7. Do p. 809 #3.
y = -2cos(x). Amplitude is 2. Period is 2π. Phase shift is zero. Vertical shift is zero.
Five points to plot are ( 0, -2 ), ( π2 , 0 ), ( π, 2 ), ( 3π
2 , 0), ( 2π, -2 ).
8. Do p. 809 #9.
y = sin(−x − π4 ) − 2
Use sin( -w ) = -sin( w ).
y = −sin(x + π4 ) − 2. Amplitude is 1, period is 2π. Phase shift = - −π
4 . Vertical shift is -2.
−π
2π
−π
2π
,
-2
),
(
+
,
-3
),
(
+
,
-2
),
Five points to plot are ( −π
4
4
4
4
2
3·2π
−π
( −π
4 + 4 , -1 ), ( 4 + 2π, -2 )
π
3π
5π
7π
Five points to plot are ( −π
4 , -2 ), ( 4 , -3 ), ( 4 , -2 ), ( 4 , -1 ), ( 4 , -2 )
9. Do p. 809 #13.
y = tan( x - π3 . Period is π. Phase shift is π3 .
Domain is all reals except where x - π3 = π2 + kπ for an integer k.
Domain is all reals except 5π
is all reals.
6 + kπ for an integer k. Range
√
√
Five points to plot are ( π3 , 0), ( π3 + π4 , 1 ), ( π3 + π3 , 3), ( π3 − π4 , -1 ), ( π3 − π3 , - 3)
√
√
2π
π
3), ( 12
, -1 ), ( 0, - 3)
After arithmetic the points are ( π3 , 0), ( 7π
12 , 1 ), ( 3 ,
10. Do p. 809 #18.
x
π
y = −1
3 sec( 2 + 3 )
Consider the reciprocal function. y = −3cos( x2 + π3 )
Amplitude is 3. Period is
2π
1
2
= 4π. Phase shift is
( −2π
3 ,-3),
Five points to plot are
After arithmetic the points are
For the secant curve. points to
−π
3
1
2
=
−2π
3 .
Vertical shift is zero
−2π
−2π
−2π
( −2π
3 + π, 0 ), ( 3 + 2π ,3), ( 3 + 3π, 0), ( 3 + 4π
−2π
π
4π
7π
10π
( 3 ,-3), ( 3 , 0 ), ( 3 ,3), ( 3 , 0), ( 3 ,-3)
√
√
2
−1
4π 1
10π −1
−π − 2
5π
plot are ( −2π
3 , 3 ), ( 3 , 3 ), ( 3 , 3 ), ( 6 , 3 , ( 6 , 3 )
11. Do p. 809 #30.
√
f (x) = 23 cos(2x) − 3 2 3 sin(2x) + 6. Write f as one sine and as one cosine function.
Use # 36 and 35 on p. 810.
√
√
Here b = 32 and a = − 3 2 3 So a2 + b2 = 49 + 27
a2 + b2 = 6.
4 = 9 Then
−1
5π
5π
tan(φ) = b/a = √3 . So φ = 6 . So S(x) = 6sin(2x + 6 ) + 6.
π
π
By #35 C(x) = 6cos(2x + 5π
6 − 2 ) + 6 = 6cos(2x + 3 ) + 6.
2
,-3)