**Unformatted text preview: **5/4/2019 Trigonometry Week 11 Test WEEK 11 ANALYTIC TRIGONOMETRY PART 1 - SECTIONS 6.1 - 6.2
Trigonometry Week 11 Test - Grade Report
Score: 91% (90.6 of 100 pts) Submitted: May 4 at 1:07pm Question: 1 Grade: 1.0 / 4.0 Question: 2 Grade: 0.0 / 4.0 Simplify.
(cos x − 2
3 sin x) 9 − 6
Comment: 2
cos x sin x + 9 sin x (0%) Solution Expand the power and then multiply the two binomials.
(cos x − 2
3 sin x) = (cos x − 3 sin x) (cos x − 2
= cos x 3 sin x) − 3 cos x sin x − 3 cos x sin x 2
+ 9 sin x Combine the like terms.
2
cos x − 3 cos x sin x − 2
= cos x − 3 cos x sin x 6 cos x sin x + 2
9 sin x Simplify using the Pythagorean Identity,
Substitute
2
cos x − 2
= cos x
= 2
cos x = 9 − 2
1 − cos x for 6 cos x sin x
− 2
cos x + 2
sin x = 1 . 2
sin x.
+ 2
9 sin x 6 cos x sin x + 9 (1 − − 6 cos x sin x + 9 6 cos x sin x 2
+ 9 sin x 2
cos x) 2
− 9 cos x 2
− 8 cos x Question: 3 Grade: 1.0 / 4.0 Question: 4 Grade: 1.0 / 4.0 Question: 5 Grade: 1.0 / 4.0 Question: 6 Grade: 1.0 / 4.0 1/5 5/4/2019 Trigonometry Week 11 Test
Question: 7 Grade: 0.0 / 4.0 Find all solutions of 8 sin2 2x +6 sin2x = 3 on [0, 2π). Round to the nearest thousandth. x ≈ 2.791, 0.35, 9.074, 6.633, 15.357 (0%) Solution Since 0 ≤ x < 2π , multiply by 2 to find the interval for 2x : 0
Let sin 2x = z , and solve the equation. ≤ 2x < 4π . = 3 on [0, 4π)
Set the equation equal to 0.
2
8 z +6 z 8 z 2 +6z −3 = 0 This expression cannot be factored so use the Quadratic Formula.
−
−−−−−−−−−−
2
−6±√6 −4(8) (−3)
z =
2(8) z = z = −
−
−
−
−−
−6±√ 36 +96
16
−
−
−
−6+√ 132
16 z ≈ 0.343 16 , therefore sin 2x ≈ 0.343 .
because of the range of the sine function. = sin 2x sin 2x ≠ −1.093
= −
−
−
−6−√ 132 , −1.093 Recall that z
2x , arcsin (0.343) and 2x ≈ π − 0.35 (Since sine is positive in quadrant 2)
2x ≈ 0.35 and 2x ≈ 2.791 .
Now determine whether these solutions are within the interval [0, 4π) and find any additional solutions within this interval by adding or
subtracting 2π (since 2π is the period of y = sin x ) to each solution.
Find all values of 0.35+ 2 πn that are within the interval.
0.35+ 2 π ≈ 6.633 , 6.633+ 2 π ≈ 12.916
Since 12.916 ≥ 4π , 12.916 is not a solution in the interval.
Find all values of 2.791 + 2 π n that are within the interval.
2.791+ 2 π ≈ 9.074 ,9.074+ 2 π ≈ 15.357
Since 15.357 ≥ 4π , 15.357 is not a solution in the interval.
Recall that θ = 2 x , so the solutions are 0.35/2, 6.633/2, 2.791/2, and 9.074/2.
Therefore, the solutions are x ≈ 0.175, 1.396, 3.317, 4.537.
2x ≈ 0.35 Question: 8 Grade: 1.0 / 4.0 2/5 5/4/2019 Trigonometry Week 11 Test
Question: 9 Grade: 0.75 / 4.0 Identify all equivalent forms of the expression.
1 2
sin t − cos2 t + tan 2 t − tan 2
(1 + tan
t) 2 2
t sin
t 2
2
cos
t (sec
t) , , 1 (75%) Solution 1 2
sin t − =1 + 2
sin t − tan + 2 t − tan 2 tan t (1 2 2
t sin
t − 2
sin
t) Factor out
tan = (1 − = cos2 t
= cos2 2
sin t) (1 (1 + tan 2 + tan 2 t) t) ( 1
2
cos
t t . Factor.
Pythagorean
identity
Pythagorean
identity 2
t (sec
t) = cos2 t 2 Reciprocal
identtity ) =1 Multiply. Question: 10 Grade: 1.0 / 4.0 Question: 11 Grade: 1.0 / 4.0 Question: 12 Grade: 1.0 / 4.0 Question: 13 Grade: 1.0 / 4.0 Question: 14 Grade: 1.0 / 4.0 3/5 5/4/2019 Trigonometry Week 11 Test
Question: 15 Grade: 0.9 / 4.0 Find all solutions to sin 3x
Select all that apply. x = 4π
3 , cos 2x 3π x = , 4 x = = π
4 0 , on [0, 2π). x = π , x = 2π
3 , x = π
3 , x = 7π
4 , x = 5π
3 , x = 5π
4 (90%) Solution Solve sin 3x cos 2x = 0 by setting each factor equal to zero. Solve cos 2x = 0 .
Since 0 ≤ x < 2π , multiply by 2 to find the interval for 2x .
for cos 2x = 0 .
and solve the equation cos 0 ≤ 2x < 4π Let θ = 2x Use the graph of y Therefore, θ
Because θ = π
2 = 2x, = cos θ 3π , 2 , 2 x = 5π
2
π
2 θ = 0 on [0, 4π) . . , , 7π
2
3π
2 , within [0, 4π) .
5π
2 7π , 2 , therefore x = π
4 , 3π
4 , 5π
4 , 7π
4 . Solve sin 3x = 0 .
Since 0 ≤ x < 2π , multiply by 3 to find the interval for 3x .
= 3x, for sin 3x = 0 .
and solve the equation sin Therefore, θ = 0, π, 2π, 3π, 4π, 5π 0 ≤ 3x < 6π Let θ Because θ x = 0 on [0, 6π) . Use the graph of y on [0, 6π) . = 3x, 3 x = 0, π, 2π, 3π, 4π, 5π Therefore x = 0, π
3 , 2π
3 , π, 4π
3 Finally, all of the solutions of sin , 5π
3 = sin θ. . . 3x cos 2x = 0 are 4/5 5/4/2019 Trigonometry Week 11 Test x = 0, π
4 , π
3 , 2π
3 , 3π
4 , π, 5π
4 , 4π
3 Question: 16 Grade: 1.0 / 4.0 Question: 17 Grade: 1.0 / 4.0 Question: 18 Grade: 1.0 / 4.0 Question: 19 Grade: 1.0 / 4.0 Question: 20 Grade: 1.0 / 4.0 Question: 21 Grade: 1.0 / 4.0 Question: 22 Grade: 1.0 / 4.0 Question: 23 Grade: 1.0 / 4.0 Question: 24 Grade: 1.0 / 4.0 Question: 25 Grade: 1.0 / 4.0 , 5π
3 , 7π
4 on [0, 2π). 5/5 ...

View
Full Document

- Winter '19
- Kathleen Harrington
- Trigonometry, Law Of Cosines, Euler's formula, 2π, Trigonometry Week