# Trigonometry Week 11 Test Grade Report

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Subject
Trigonometry
Type
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Trigonometry Week 11 Test - Grade Report
Score:
90% (89.6 of 100 pts)
Submitted:
Jul 19 at 8:27pm
Question: 1
Identify ALL equivalent forms of the expression.
36+36costsint−27sin2t36+36costsint−27sin2t
Select all that apply.
(6cost+3sint)26cost+3sint2
, (6cost+3sint)(6cost+3sint)6cost+3sint6cost+3sint (100%)
Solution
Pythagorean
identity, cos2t+sin2t=1cos2t
+sin2t=1.
Expand and collect the like
terms.
Simplify.
Factor.
So, the correct expressions
are (6cost+3sint)(6cost+3sint)6cost+3sint6cost+3sint and (6cost+3sint)26cost+3sint2.
Question: 2
True or false? 1+2cot2x+cot4xcsc2x1+2cot2x+cot4xcsc2x= cot2xcot2xis an identity.
True (0%)
Comment:
Solution
1+2cot2x+cot4xcsc2x1+2cot2x+cot4xcsc2x
==
(1+cot2x)(1+cot2x)csc2x1+cot2x1+cot2xcsc2x
Factor.
==
csc2x⋅csc2xcsc2xcsc2x⋅csc2xcsc2x
Substitute csc2xcsc2xfor 1+cot2x1+cot2x(Pythagorean
Identity).
==
csc2xcsc2x
Simplify.

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Since csc2xcsc2x cot2xcot2xfor all values of x, the equation is not an identity.
Question: 3
Find all solutions of cot24x−2cot4x+1=0cot24x−2cot4x+1=0 on [0,π).0,π.
Select all that apply.
x=x=5π165π16, x=x=9π169π16, x=x=π16π16, x=x=13π1613π16 (100%)
Solution
Since 0≤x<π0≤x<π, multiply by 44 to find the interval for 4x4x, 0≤4x<4π0≤4x<4π.
Let t=4xt=4x and solve the equation
cot2t−2cott+1=0cot2t−2cott+1=0 on [0,4π)0,4π.
Factoring yields
(cott−1)2=0cott−12=0
Ifcott−1=0cott−1=0, then cott=1cott=1.
Since cot x = 1/(tan x), solvetan(x)=1tan⁡x=1.
Consider that the reference angle
for 11is π4π4 for tangent and tangent is positive
Therefore, t=t=π4π4, t=t=5π45π4, t=t=9π49π4 , t=t=13π413π4, ...
Because t=4x,t=4x,4x=4x=π4π4, 5π45π4, 9π49π4, 13π413π4, ...
Therefore, x=x=π16π16, 5π165π16, 9π169π16, 13π1613π16 within 0≤x<π0≤x<π.
Question: 4
Find all solutions of 6sin2x−2sinx−1=06sin2x−2sinx−1=0 on [0,2π).0,2π. Round to the
nearest thousandth.
Select all that apply.
x≈6.005x≈6.005, x≈3.420x≈3.420, x≈0.653x≈0.653, x≈2.489x≈2.489 (100%)
Solution
Let sinx=zsinx=z. Substitute.
6sin2x−2sinx−1=06sin2x−2sinx−1=0
6 z2−2z−1=06 z2−2z−1=0
This expression cannot be factored, so use the Quadratic Formula.

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Trigonometry Week 11 Test - Grade Report Score: 90% (89.6 of 100 pts) Submitted: Jul 19 at 8:27pm Question: 1 Grade: 1.0 / 4.0 Identify ALL equivalent forms of the expression. 36+36costsint−27sin2t36+36costsint−27sin2t Select all that apply. (6cost+3sint)26cost+3sint2 , (6cost+3sint)(6cost+3sint)6cost+3sint6cost+3sint (100%) Solution 36+36costsint−27sin2t36+36costsint−27sin2t Pythagorean = 36(cos2t+sin2t)+36costsint−27sin2t36cos2t+sin2t+36 identity, cos2t+sin2t=1cos2t costsint−27sin2t +sin2t=1. = 36cos2t+36costsint+(36−27)sin2t36cos2t+36costsint+ Expand and collect the like terms. 36−27sin2t = 36cos2t+36costsint+9sin2t36cos2t+36costsint+9sin2t Simplify. = (6cost+3sint)(6cost+3sint)6cost+3sint6cost+3sint Factor. = (6cost+3sint)26cost+3sint2 So, the correct expressions are (6cost+3sint)(6cost+3sint)6cost+3sint6cost+3sint and (6cost+3sint)26cost+3sint2. Question: 2 Grade: 0.0 / 4.0 True or false? 1+2cot2x+cot4xcsc2x1+2cot2x+cot4xcsc2x= cot2xcot2xis an identity. True (0%) Comment: Solution 1+2cot2x+cot4xcsc2x1+2cot2x+cot4xcsc2x == (1+cot2x)(1+cot2x)csc2x1+cot2x1+cot2xcsc2x Facto == csc2x⋅csc2xcsc2xcsc2x⋅csc2xcsc2x Subst Identi == csc2xcsc2x Simp Since csc2xcsc2x≠ cot2xcot2xfor all values of x, the equation is not an identity. Question: 3 Grade: 1.0 / 4.0 Find all solutions of cot24x−2cot4x+1=0cot24x−2cot4x+1=0 on [0,π).0,π. Select all that apply. x=x=5π165π16, x=x=9π169π16, x=x=π16π16, x=x=13π1613π16 (100%) Solution Since 0? ...
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