1.Β Β Β Verify the identity 1 - sin2 x cot2 x = sin2 x. Is there more than one way to verify the identity? If so, tell which way you think is easier and why.Β
2.John said πππ π + πππ π = π has no solution. Do you agree with John? Explain why or why not?Β
3.
SKILL CHECK: Verify each identity.Β Β
1.πππΒ² π½+1 / cotΒ²π½=secΒ²π½
2.(πππππ½ β π)πππππ½ β‘ πππππ½Β
3.8. π β πππ πΆ πππ πΆΒ β‘ πππ πΆ πππ πΆ β πΒ
4.tan A + cot A / sec A csc A =1
5.π + π πππ ππ½ β‘ πππππ½ β πππ ππ½Β
Trigonometric equation
6.cos x + βπ = - cos xΒ
7.sin2 x β tan x cos x = 0
8.sin x + βπ = β sin xΒ
9.Β 2 cos2 x β 5 cos x = 3
10.βπ csc x + 2 = 0
solve the worded problem
1.Β On which days of the year are there 10 hours of sunlight in Prescott, Arizona?
2.The tide, or depth of the ocean near the shore, changes throughout the day. The depth of the Bay of Fundy can be modeled by....... where d is the water depth in feet and t is the time in hours. Consider a day in which t = 0 represents 12:00 A.M. At what time(s) is the water depth 3 1 2 feet?
"\\displaystyle \n1).\\\\ 1 - \\sin^2{x}\\cot^2{x} \\\\\n\n\\textsf{There is more than one way} \\\\\n\n\\begin{aligned}\n1 - \\sin^2{x}\\cot^2{x} &= 1 - \\sin^2{x}\\cdot\\frac{cos^2{x}}{\\sin^2{x}}\\\\\n&= 1 - \\cos^2{x} = \\sin^2{x}\n\\end{aligned} \\\\\n\n\\begin{aligned}\n\\cos^2{x} + \\sin^2{x} &= 1\\\\\n\\cot^2{x} + 1 &= \\cosec^2{x} \\\\\n\\cot^2{x} &= \\cosec^2{x} - 1\n\\end{aligned} \\\\\n\n\\begin{aligned}\n1 - \\sin^2{x}\\cot^2{x} &= 1 - \\sin^2{x}\\left(\\cosec^2{x} - 1\\right)\\\\\n&= 1 - 1 + \\sin^2{x} = \\sin^2{x}\n\\end{aligned}\\\\\n\n\n\\textsf{The first way is easier because it}\\\\\n\\textsf{involves a short process and}\\\\\n\\textsf{is straightfoward.}\\\\\n\n\n2).\\\\ \\sin^2{x} + \\cos^2{x} = 2 \\\\\n\n\\textsf{It has no solution.}\n\n\\textsf{The maximum value of}\\\\\n\\cos^2{x}\\,\\, \\textsf{is}\\,\\, 1 \\,\\, \\textsf{at}\\,\\, x = 0.\\\\\n\n\\textsf{The maximum value of}\\,\\, \\sin^2{x}\\\\\n\\textsf{is}\\,\\, 1\\,\\, \\textsf{at}\\,\\, x = \\frac{\\pi}{2}.\\\\\n\n\\textsf{The values of}\\,\\, x\\,\\, \\textsf{do not coincide}\\\\\n\\textsf{at the maximum values and as such,}\\\\\n\\textsf{there is no solution to the given equation.}\\\\\n\n3).\\\\ \\begin{aligned}\n\\frac{\\cot^2{\\theta} + 1}{\\cot^2{\\theta}} &= 1 + \\tan^2{\\theta}\n\\\\&= \\sec^2{\\theta}\n\\end{aligned} \\\\\n\n4).\\\\ \\begin{aligned}\n(\\cosec^2{\\theta} - 1)\\sin^2{\\theta} &= 1 - \\sin^2{\\theta}\n\\\\&= \\cos^2{\\theta}\n\\end{aligned} \\\\\n\n5).\\\\ \\begin{aligned}\n1 - \\sec\\alpha\\cos\\alpha &= 1- \\frac{\\cos\\alpha}{\\cos\\alpha} \n\\\\&= 1 - 1 = 0\n\\end{aligned} \\\\\n\n\\begin{aligned}\n\\tan\\alpha\\cot\\alpha - 1 &= \\frac{\\tan\\alpha}{\\tan\\alpha} - 1\n\\\\&= 1 - 1 = 0\n\\end{aligned} \\\\\n\\therefore 1 - \\sec\\alpha\\cos\\alpha =\\tan\\alpha\\cot\\alpha - 1\\\\ \n6).\\\\ \\begin{aligned}\n\\frac{\\tan A + \\cot A}{\\sec A \\cosec A} &= (\\tan A + \\cot A)\\sin A \\cos A \n\\\\&= \\left(\\frac{\\sin A}{\\cos A}+ \\frac{\\cos A}{\\sin A}\\right)\\sin A \\cos A = \n\\\\&= \\sin^2 A + \\cos^2 A = 1\n\\end{aligned} \\\\\n\n\n7).\\\\ 1 + 2\\tan^2{\\theta} = \\sec^2{\\theta} + \\tan^2{\\theta}\\\\\n\n\\textsf{It is known that}\\,\\, \\cos^2{\\theta} + \\sin^2{\\theta} = 1\\\\\n\n\\tan^2{\\theta} + 1 = \\sec^2{\\theta}\\\\\n\n\\sec^2{\\theta} - \\tan^2{\\theta}\\\\\n\n\\begin{aligned}\n1 + 2\\tan^2{\\theta} &= \\sec^2{\\theta} + \\tan^2{\\theta}\n\\\\&= (\\sec^2{\\theta} + \\tan^2{\\theta})(1) = (\\sec^2{\\theta} + \\tan^2{\\theta})(\\sec^2{\\theta} - \\tan^2{\\theta}) \n\\\\&= \\sec^4{\\theta} - \\tan^4{\\theta}\n\\end{aligned} \\\\\n\n\n8).\\\\ \\cos{x} + \\sqrt{3} = -\\cos{x}\\\\\n2\\cos{x} = \\sqrt{3}\\\\\n\\cos{x} = \\frac{\\sqrt{3}}{2},\\,\\, x = \\frac{\\pi}{6}\\\\\n\n\n9).\\\\ \\sin(2x) - \\tan{x}\\cos{x} = 0\\\\\n\n2\\sin{x}\\cos{x} - \\sin{x} = 0\\\\\n\n\\sin{x}(2\\cos{x} - 1) = 0\\\\\n\n\\sin{x} = 0, x = n\\pi\\\\\n\n2\\cos{x} - 1 = 0\\\\\n\\cos{x} = \\frac{1}{2}\\\\\nx = \\frac{\\pi}{6} + 2n\\pi\\\\\n\n\\therefore x = n\\pi,\\,\\, \\frac{\\pi}{6} + 2n\\pi \\,\\, \\forall n \\in \\mathbb{Z}\\\\\n\n10).\\\\ \\sin{x} + \\sqrt{2} = - \\sin{x}\\\\\n\n2\\sin{x} = \\sqrt{2}\\\\\n\\sin{x} = \\frac{\\sqrt{2}}{2},\\,\\, x = \\frac{\\pi}{4}\\\\\n\n\n11).\\\\ 2\\cos^2{x} - 5\\cos{x} = 3\\\\\n\n2\\cos^2{x} - 5\\cos{x} - 3 = 0\\\\\n\n\\cos{x} = \\frac{5 \\pm 7}{4} \\\\\n\n\\cos{x} = 3,\\,\\, \\cos{x} = -\\frac{1}{2}\\\\\n\n\n\\cos(\\pi + x) = \\cos\\left(\\frac{\\pi}{3}\\right)\\\\\nx = -\\frac{2\\pi}{3} + 2n\\pi \\,\\, \\forall n \\in \\mathbb{Z}\\\\\n\n\n\\cos{x} = \\cos\\left(\\frac{2\\pi}{3}\\right) \\\\\n\nx = \\frac{2\\pi}{3} + 2n\\pi \\,\\, \\forall n \\in \\mathbb{Z}\\\\\n\n\n12).\\\\ \\sqrt{3}\\cosec{x} + 2 = 0 \\\\\n\n\\cosec{x} = \\frac{-2}{\\sqrt{3}} \\\\\n\n\\sin{x} = -\\frac{\\sqrt{3}}{2} \\\\\n\n\n\\sin(\\pi + x) = \\frac{\\sqrt{3}}{2} = \\sin\\left(\\frac{\\pi}{3}\\right) \\\\\n\n\nx = -\\frac{2\\pi}{3} + 2n\\pi \\,\\, \\forall n \\in \\mathbb{Z} \\\\\n\n\n\\sin(2\\pi - x) = \\frac{\\sqrt{3}}{2} = \\sin\\left(\\frac{\\pi}{3}\\right) \\\\\n\n\nx = \\frac{5\\pi}{3} + 2n\\pi \\,\\, \\forall n \\in \\mathbb{Z}"
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