Asked by Kyle
1. Graph the following. Show all your work for full marks. You must have a scale that uses the π symbol. π¦ = βsin(π₯ βπ3)+ 2
2. Solve for 0 β€ π₯ β€ 2π. Show all your work.
π ππ2π₯ + 2πππ 2π₯ = 0
3. Use the addition/subtraction formulas to find an exact value for the following.
πππ 5π/6
2. Solve for 0 β€ π₯ β€ 2π. Show all your work.
π ππ2π₯ + 2πππ 2π₯ = 0
3. Use the addition/subtraction formulas to find an exact value for the following.
πππ 5π/6
Answers
Answered by
Bosnian
1)
y = - sin ( x - Ο / 3 ) + 2
y = 2 - sin ( x - Ο / 3 )
For function:
y = A sin [ B ( x + C ) ] + D
amplitude is A
period is 2 Ο / B
phase shift is C
vertical shift is D
In this case:
A = - 1
B = 1
C = Ο / 3
D = 2
amplitude is -1
period is 2 Ο / 1 = 2 Ο
phase shift is ( Ο / 3 ) / 1 = Ο / 3
vertical shift is 2
2)
sin ( 2 x ) + 2 cos ( 2 x ) = 0
Subtract 2 cos ( 2 x ) to both sides
sin ( 2 x ) = - 2 cos ( 2 x )
Divide both sides by cos ( 2 x )
sin ( 2 x ) / cos ( 2 x ) = - 2
tan ( 2 x ) = - 2
Take the inverse tangent of both sides
2 x = Ο n + tan β» ΒΉ ( - 2 )
Since tan ( - x ) = - tan ( x )
2 x = Ο n - tan β» ΒΉ ( 2 )
Divide both sides by 2
x = 1 / 2 [ Ο n - tanβ»ΒΉ ( 2 ) ]
Since tanβ»ΒΉ ( 2 ) = 1.107148717794
x = 1 / 2 ( Ο n - 1.107148717794)
x = Ο n / 2 - 1.107148717794 / 2
x = Ο n / 2 - 0.553574359
For n = 0
x = Ο β 0 - 0.553574359 = 0 - 0.553574359 = - 0.553574359
x = - 0.553574359 radians is not in interval 0 β€ x β€ 2 Ο
For n = 1
x = Ο β 1 / 2 - 0.553574359 = Ο / 2 - 0.553574359 =
1.570796327 - 0.553574359
x = 1.017221968 rad
For n = 2
x = Ο β 2 / 2 - 0.553574359 = Ο - 0.553574359 =
3.141592654 - 0.553574359
x = 2.588018295 rad
For n = 3
x = Ο β 3 / 2 - 0.553574359 = 3 Ο / 2 - 0.553574359 =
4.712388980 - 0.553574359
x = 4.158814621rad
For n = 4
x = Ο β 4 / 2 - 0.553574359 = 2 Ο - 0.553574359 =
6.283185307 - 0.553574359 = 5.729610948 rad
For n = 5
x = Ο β 5 / 2 - 0.553574359 = 5 Ο / 2 - 0.553574359 =
7.853981634 - 0.553574359
x = 7.300407275 rad
7.300407275 radians is not in interval 0 β€ x β€ 2 Ο
So the solutions are:
x = 1.017221968 rad
x = 2.588018295 rad
x = 4.158814621rad
and
x = 5.729610948 rad
3)
5 Ο / 6 = 6 Ο / 6 - Ο / 6 = Ο - Ο / 6
Use identity:
cos ( Ο - ΞΈ ) = - cos ( ΞΈ )
cos ( 5 Ο / 6 ) = cos ( Ο - Ο / 6 ) = - cos ( Ο / 6 ) = - β3 / 2
y = - sin ( x - Ο / 3 ) + 2
y = 2 - sin ( x - Ο / 3 )
For function:
y = A sin [ B ( x + C ) ] + D
amplitude is A
period is 2 Ο / B
phase shift is C
vertical shift is D
In this case:
A = - 1
B = 1
C = Ο / 3
D = 2
amplitude is -1
period is 2 Ο / 1 = 2 Ο
phase shift is ( Ο / 3 ) / 1 = Ο / 3
vertical shift is 2
2)
sin ( 2 x ) + 2 cos ( 2 x ) = 0
Subtract 2 cos ( 2 x ) to both sides
sin ( 2 x ) = - 2 cos ( 2 x )
Divide both sides by cos ( 2 x )
sin ( 2 x ) / cos ( 2 x ) = - 2
tan ( 2 x ) = - 2
Take the inverse tangent of both sides
2 x = Ο n + tan β» ΒΉ ( - 2 )
Since tan ( - x ) = - tan ( x )
2 x = Ο n - tan β» ΒΉ ( 2 )
Divide both sides by 2
x = 1 / 2 [ Ο n - tanβ»ΒΉ ( 2 ) ]
Since tanβ»ΒΉ ( 2 ) = 1.107148717794
x = 1 / 2 ( Ο n - 1.107148717794)
x = Ο n / 2 - 1.107148717794 / 2
x = Ο n / 2 - 0.553574359
For n = 0
x = Ο β 0 - 0.553574359 = 0 - 0.553574359 = - 0.553574359
x = - 0.553574359 radians is not in interval 0 β€ x β€ 2 Ο
For n = 1
x = Ο β 1 / 2 - 0.553574359 = Ο / 2 - 0.553574359 =
1.570796327 - 0.553574359
x = 1.017221968 rad
For n = 2
x = Ο β 2 / 2 - 0.553574359 = Ο - 0.553574359 =
3.141592654 - 0.553574359
x = 2.588018295 rad
For n = 3
x = Ο β 3 / 2 - 0.553574359 = 3 Ο / 2 - 0.553574359 =
4.712388980 - 0.553574359
x = 4.158814621rad
For n = 4
x = Ο β 4 / 2 - 0.553574359 = 2 Ο - 0.553574359 =
6.283185307 - 0.553574359 = 5.729610948 rad
For n = 5
x = Ο β 5 / 2 - 0.553574359 = 5 Ο / 2 - 0.553574359 =
7.853981634 - 0.553574359
x = 7.300407275 rad
7.300407275 radians is not in interval 0 β€ x β€ 2 Ο
So the solutions are:
x = 1.017221968 rad
x = 2.588018295 rad
x = 4.158814621rad
and
x = 5.729610948 rad
3)
5 Ο / 6 = 6 Ο / 6 - Ο / 6 = Ο - Ο / 6
Use identity:
cos ( Ο - ΞΈ ) = - cos ( ΞΈ )
cos ( 5 Ο / 6 ) = cos ( Ο - Ο / 6 ) = - cos ( Ο / 6 ) = - β3 / 2
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