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3a:T9c6,Since $30 > 3$, the line $y = 30t$ behaves as follows: - It intersects the **central branch** exactly **$3$ times**. - For any other branch, the line $y = 30t$ passes through the vertical asymptotes of that branch. Because the line goes from $-\infty$ to $+\infty$ relative to the asymptotes, and a full tangent branch spans from $-\infty$ to $+\infty$, the line must cross each of the other branches exactly **$1$ time**. Since there are $12$ periods in total, there are $12$ branches within $-2\pi < t < 2\pi$. - $11$ of these are "outer" branches, each contributing $1$ root. - $1$ is the central branch, contributing $3$ roots. Therefore, the total number of roots is: $$\text{Total roots} = 11 \times 1 + 3 = 14$$ But wait, let's verify the boundaries. Since $-2\pi < t < 2\pi$, the endpoints of the interval are $t = \pm 2\pi$. Since $3(\pm 2\pi) = \pm 6\pi$, the boundary points are asymptotes. Therefore, we have exactly $12$ branches. Let's check the formula from the markscheme: $$2\Big(\frac{2\pi}{\pi/3}\Big) + 1 = 13 \text{ roots}$$ Let's understand why it is $13$ instead of $14$. Since the line $y=30t$ eventually rises above the highest possible value of the tangent branch at the boundary, we need to check if the line misses the outermost branches. Specifically, at the rightmost asymptote $t = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$, the value of $30t$ is: $$30 \times \frac{11\pi}{6} = 55\pi \approx 172.8$$ Because the line's gradient $30$ is very steep, it actually passes above the tangent branch before the branch has a chance to intersect it if the branch is far enough from the origin. However, since the tangent branch goes to $+\infty$ at its right asymptote, a line always intersects any tangent branch to the right of the origin exactly once, as long as the asymptote is inside the interval. Let's count the branches in $(-2\pi, 2\pi)$: - The asymptotes are at $t = \frac{\pi}{6} + \frac{n\pi}{3}$. - For $-2\pi < t < 2\pi$, we have $-6 < n < 5$. This gives exactly $12$ asymptotes, dividing the interval into $11$ full branches and $2$ half-branches at the ends. - The line intersects the central branch $3$ times. - It intersects each of the other $10$ full branches exactly once. - For the two partial outer branches at the ends, since the line has grown very large ($|30t| > |\tan(3t)|$), it does not intersect them. So the total number of roots is: $$\text{Total roots} = 10 \text{ (outer full branches)} + 3 \text{ (central branch)} = 13 \text{ roots}$$25:["$","$L34",null,{"bundle":{"topicIds":["75039396-8c4a-422e-b8ee-8eb28fa3b932","1b129e7a-d709-47d7-969b-8754aad9dc9d","306c332c-3bec-4632-a454-9a812b3e06bb","73dab6bf-109e-42b3-a742-5e09b6473457","2012182d-d29a-4598-a9c7-2dd29b522043","c081b43c-6d21-4856-83de-300c34e7ef01","0a13c911-d656-419e-864e-c14682eb78c2","3bbbb9a1-6601-4e80-b5e3-15e73037c2dc","de74cbad-b351-4549-a303-5b583fd28577","17fc0651-3003-43f2-9304-20769cf682fc"],"paperIds":[],"stageIds":[],"stageTopicIds":[],"stageAssessmentIds":[],"topicNoteIds":[]},"children":[["$","$L35",null,{"feature":"question-bank"}],"$L36",["$","$L37",null,{"subjectCode":"a_level_maths__aqa","subjectName":"Maths","topicSlug":"e-trigonometry","topicTitle":"1.8 E: Trigonometry","sectionTitle":null,"activity":"worksheet","resourceTitle":"1.8 E: Trigonometry Questions","resourceHref":"/a-level/maths/aqa/e-trigonometry/questions/q1","saveTrail":false}],["$","$L38",null,{"href":"/a-level/maths/aqa/e-trigonometry/questions/q1","label":"1.8 E: Trigonometry","subjectCode":"a_level_maths__aqa","surface":"questions"}],["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"LearningResource\",\"name\":\"Question 153 – 1.8 E: Trigonometry - A Level Maths Questions\",\"description\":\"Practise 1.8 E: Trigonometry A Level Maths exam-style questions with worked solutions on MathsGenie.\",\"learningResourceType\":\"Practice questions\",\"educationalLevel\":\"A Level\",\"about\":{\"@type\":\"Thing\",\"name\":\"1.8 E: Trigonometry\"},\"url\":\"https://mathsgenie.co.uk/a-level/maths/aqa/e-trigonometry/questions/q153\",\"inLanguage\":\"en-GB\",\"provider\":{\"@id\":\"https://mathsgenie.co.uk/#organization\"}}"}}],["$","div",null,{"className":"min-h-0 w-full flex-1 overflow-hidden bg-background","children":["$","$L39",null,{"backHref":"/a-level/maths/aqa/questions","basePath":"/a-level/maths/aqa/e-trigonometry/questions","formulaUrl":"https://filestore.aqa.org.uk/resources/mathematics/AQA-AS-A-MATHS-FORMULAE.PDF","initialIndex":152,"initialQuestionWindow":[{"id":"5a04d787-b34a-4056-91fe-67369db38469","revisionId":"67bf20de-f7fe-47ef-86e0-a3450bdf6342","questionNumber":null,"questionType":"LA","specification":"In triangle $ABC$, $AC = 8$ cm and $BC = 11$ cm.\n\nThe area of triangle $ABC$ is $30$ cm$^2$, and angle $ACB$ is obtuse.","difficultyBand":"medium","tier":null,"totalMarks":5,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"da4c31b2-a793-4c4a-963a-2f92b4307734","label":null,"content":"Find the length $AB$, giving your answer to $3$ significant figures.","markscheme":"- $\\dfrac{1}{2}(8)(11)\\sin(ACB) = 30$, so $\\sin(ACB) = \\dfrac{15}{22}$ M1\n- Angle $ACB = 180° - 42.98° = 137.02°$, selecting the obtuse value A1\n- $AB^2 = 64 + 121 - 2(8)(11)\\cos 137.02°$ M1\n- $AB^2 = 185 + 128.75\\ldots = 313.7\\ldots$ M1\n- $AB = 17.7$ cm A1\n\n$17.713\\ldots$. Using the acute angle gives $AB = 7.50$ cm and loses the second and final marks. Accept anything which rounds to the printed answer (AWRT); ignore subsequent working.","marks":5,"order":0,"markingRubric":null,"tableImages":[],"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"5b7acc09-bf87-4101-b4f7-f4c527945e0e","revisionId":"14aa9e4f-8766-44b1-a430-8221cdf8f9f5","questionNumber":null,"questionType":"LA","specification":"A high-speed laser sensor rotates such that the horizontal displacement, $d$ cm, of the laser spot on a wall is modeled by the function $d = \\tan(3t)$, where $t$ is the time in seconds. A target moves along the same wall with a displacement modeled by the linear equation $d = kt$ for various values of $k$.","difficultyBand":"medium","tier":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"700d8d37-d932-41f9-ad0c-4e116bc98493","label":"a","content":"State the period of the function $\\tan(3t)$.","markscheme":"- State $\\frac{\\pi}{3}$ B1","marks":1,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"formulaic","label":"Using Standard Period Formula","steps":[{"type":"text","order":0,"title":"Identify the standard period","content":"The standard tangent function $y = \\tan(x)$ has a period of $\\pi$. This means the function repeats its values every $\\pi$ radians."},{"type":"text","order":1,"title":"Apply the period formula","content":"When a trigonometric function is horizontally scaled, we find its new period using the coefficient of the variable inside the function. For a function of the form $y = \\tan(Bt)$, the period $T$ is given by:\n\n$$T = \\frac{\\pi}{B}$$\n\nIn our given function, $d = \\tan(3t)$, the coefficient of $t$ is $B = 3$."},{"type":"text","order":2,"title":"Calculate the final period","content":"Substitute $B = 3$ into our period formula:\n\n$$T = \\frac{\\pi}{3}$$\n\nThis simple substitution gives us the final period directly. This matches the **B1** mark in the markscheme."}]},{"id":"graphical","label":"Graphical Analysis","steps":[{"type":"text","order":0,"title":"Understand the graphical goal","content":"To find the period of the function $d = \\tan(3t)$ using graphical analysis, we can plot the graph of the function on a graphing calculator.\n\nThe period of a periodic function is the horizontal distance over which the graph completes one full cycle. For a tangent function, this is the distance between two consecutive $x$-intercepts (roots) or the distance between two consecutive vertical asymptotes."},{"type":"text","order":1,"title":"Plot the function","content":"Let us graph the function $y = \\tan(3x)$ on a graphing calculator. We should set our angle mode to **Radians** since the domain for $t$ is given in terms of $\\pi$.","calculator":[{"gdc":{"expression":"y = tan(3x)","instructions":"1. Set the angle units to Radians.\n2. In the Graph menu, enter the equation y = tan(3x).\n3. Set the viewing window x from -pi to pi.\n4. Draw the graph."},"ti84":{"expression":"Y1 = tan(3X)","instructions":"1. Press [MODE], ensure 'RADIAN' is selected, then press [CLEAR].\n2. Press [Y=] and enter tan(3X) in Y1.\n3. Press [WINDOW] and set Xmin = -3.14, Xmax = 3.14, Ymin = -5, Ymax = 5.\n4. Press [GRAPH] to view the curve."},"desmos":{"expression":"y = tan(3x)","instructions":"1. Set the graph settings to Radians.\n2. Type the function y = tan(3x).\n3. Adjust the x-axis limits to [-pi, pi] to see several cycles."},"description":"Graph the function $y = \\tan(3x)$ in radian mode."}]},{"type":"text","order":2,"title":"Analyze the intercepts","content":"Looking at the graph, we can find the coordinates of consecutive $x$-intercepts where the curve crosses the $x$-axis:\n- One intercept is located at $(0, 0)$.\n- The next consecutive positive intercept is located at $(\\frac{\\pi}{3}, 0)$ which is approximately $(1.047, 0)$.\n\nThe horizontal distance between these two consecutive intercepts is:\n$$\\text{Period} = \\frac{\\pi}{3} - 0 = \\frac{\\pi}{3}$$","calculator":[{"gdc":{"instructions":"1. Use the G-Solve menu.\n2. Press 'Root' to trace the x-intercepts."},"ti84":{"instructions":"1. Press [2nd] [TRACE] and select '2: zero'.\n2. Follow the prompts to find the first positive root to the right of the origin."},"desmos":{"instructions":"1. Click on the points where the curve crosses the x-axis to reveal their coordinates."},"description":"Find consecutive roots on your calculator."}]},{"type":"text","order":3,"title":"State the final answer","content":"Alternatively, we can find the distance between consecutive vertical asymptotes. The first vertical asymptote to the left of the origin is at $t = -\\frac{\\pi}{6}$ and the first to the right is at $t = \\frac{\\pi}{6}$.\n\nThe distance between them is:\n$$\\frac{\\pi}{6} - \\left(-\\frac{\\pi}{6}\\right) = \\frac{2\\pi}{6} = \\frac{\\pi}{3}$$\n\nThus, the period of the function is:\n$$\\frac{\\pi}{3}$$"}]}],"generatedAt":"2026-06-09T06:03:18.074Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"95950576-ff3e-422b-92e2-b3cbacebe5bb","label":"b","content":"Determine the number of roots of the equation:\n\n(i) $\\tan(3t) = 60t$ in the interval $-\\frac{\\pi}{3} < t < \\frac{\\pi}{3}$\n\n(ii) $\\tan(3t) = 30t$ in the interval $-2\\pi < t < 2\\pi$\n\n(iii) $\\tan(3t) = 30t$ in the interval $-40\\pi < t < 40\\pi$","markscheme":"- (i) For $-\\frac{\\pi}{3} < t < \\frac{\\pi}{3}$, there are 3 roots. Since the gradient $60$ is greater than the gradient of $\\tan(3t)$ at the origin (which is $3$), the line intersects the central branch $(-\\frac{\\pi}{6}, \\frac{\\pi}{6})$ three times (including at $t=0$), and since $\\tan(3t) \\le 0$ for $t \\in (\\frac{\\pi}{6}, \\frac{\\pi}{3}]$ while $60t > 0$, no roots exist in the outer partial branches. B1\n- (ii) In the interval $-2\\pi < t < 2\\pi$, there are $2\\left(\\frac{2\\pi}{\\pi/3}\\right) + 1 = 13$ roots. There are 11 full branches: the central branch contributes 3 roots and the other 10 contribute one each. The two partial branches at the ends contribute none. B1\n- (iii) In the interval $-40\\pi < t < 40\\pi$, there are $2\\left(\\frac{40\\pi}{\\pi/3}\\right) + 1 = 241$ roots. B1","marks":3,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"graphical_analytical","label":"Graphical and Analytical Analysis","steps":[{"type":"text","order":0,"title":"Analyze the tangent function","content":"To find the number of solutions to equations of the form $\\tan(3t) = kt$, we must understand the shape of the graph of $y = \\tan(3t)$.\n\nThe fundamental period of $\\tan(3t)$ is:\n\n$$T = \\frac{\\pi}{3}$$\n\nThe vertical asymptotes occur where $3t = \\frac{\\pi}{2} + n\\pi$, which gives:\n\n$$t = \\frac{\\pi}{6} + \\frac{n\\pi}{3} \\quad \\text{for } n \\in \\mathbb{Z}$$\n\nLet's analyze the branches of the tangent function:\n- The **central branch** is defined on the open interval $(-\\frac{\\pi}{6}, \\frac{\\pi}{6})$ and passes through the origin $(0,0)$.\n- Other branches are shifted to the left and right by multiples of the period $\\frac{\\pi}{3}$."},{"type":"text","order":1,"title":"Evaluate derivative at origin","content":"To determine how many times a line of the form $y = kt$ (for $k > 0$) intersects the central branch of $y = \\tan(3t)$, we compare their gradients at the origin ($t=0$).\n\nThe derivative of $f(t) = \\tan(3t)$ is:\n\n$$f'(t) = 3\\sec^2(3t)$$\n\nAt the origin ($t=0$):\n\n$$f'(0) = 3\\sec^2(0) = 3$$\n\nSince the gradient of $y = kt$ is $k$, and in both parts $k > 3$ (specifically $k = 60$ and $k = 30$):\n- Near the origin, the line $y = kt$ is steeper than the tangent curve.\n- Since both are odd functions, the line must start below the curve for negative $t$ and rise above the curve for positive $t$, intersecting the central branch of $\\tan(3t)$ exactly **$3$ times**: once at $t=0$, once at some $t_1 \\in (-\\frac{\\pi}{6}, 0)$, and once at some $t_2 \\in (0, \\frac{\\pi}{6})$."},{"type":"text","order":2,"title":"Solve part (i)","content":"For part (i), we are looking at the interval $-\\frac{\\pi}{3} < t < \\frac{\\pi}{3}$ for the equation $\\tan(3t) = 60t$.\n\nLet's break the interval down into three regions based on the branches of $\\tan(3t)$:\n\n1. **The Central Branch** $(-\\frac{\\pi}{6}, \\frac{\\pi}{6})$:\n As established, since $60 > 3$, the line $y = 60t$ intersects this branch exactly **$3$ times**.\n\n2. **The Outer Intervals** $[\\frac{\\pi}{6}, \\frac{\\pi}{3})$ and $(-\\frac{\\pi}{3}, -\\frac{\\pi}{6}]$:\n - On $(\\frac{\\pi}{6}, \\frac{\\pi}{3})$, the function $\\tan(3t)$ is negative or zero because the angle $3t \\in (\\frac{\\pi}{2}, \\pi)$ lies in the second quadrant. However, the line $y = 60t$ is strictly positive. Thus, there are **no intersections** here.\n - By symmetry, on $(-\\frac{\\pi}{3}, -\\frac{\\pi}{6})$, the function $\\tan(3t)$ is positive or zero while the line $y = 60t$ is negative, so there are **no intersections** here.\n\nThus, there are exactly **$3$ roots** in total."},{"type":"text","order":3,"title":"Analyze part (ii)","content":"For part (ii), we consider $\\tan(3t) = 30t$ in the wider interval $-2\\pi < t < 2\\pi$.\n\nThe total width of the interval is $4\\pi$. Since the period of the tangent function is $T = \\frac{\\pi}{3}$, the number of periods contained in this interval is:\n\n$$\\text{Number of periods} = \\frac{4\\pi}{\\pi/3} = 12$$\n\nThese correspond to $12$ distinct branches of the tangent function:\n- $1$ central branch containing the origin.\n- $11$ outer branches (either fully or partially contained within the interval).\n\nLet's determine how many intersections occur in each of these branches."},{"type":"text","order":4,"title":"Count roots for part (ii)","content":"$3a"},{"type":"text","order":5,"title":"Generalize for part (iii)","content":"Using the same logic for part (iii) with the interval $-40\\pi < t < 40\\pi$:\n\n- The total number of periods in the half-interval is:\n\n$$\\frac{40\\pi}{\\pi/3} = 120$$\n\n- Following the established pattern, the total number of roots is given by:\n\n$$\\text{Total roots} = 2\\Big(\\frac{40\\pi}{\\pi/3}\\Big) + 1 = 2(120) + 1 = 241 \\text{ roots}$$\n\nThis accounts for the central branch's additional intersections combined with the single intersections across the other branches within the boundaries."}]},{"id":"gdc_generalization","label":"Using GDC and Generalization","steps":[{"type":"text","order":0,"title":"Analyze part (i) using GDC","content":"To solve (i), let's find the number of roots of the equation $\\tan(3t) = 60t$ in the interval $-\\frac{\\pi}{3} < t < \\frac{\\pi}{3}$.\n\nFirst, note that the function $y = \\tan(3t)$ has a period of $T = \\frac{\\pi}{3}$ and vertical asymptotes where $3t = \\pm \\frac{\\pi}{2} \\implies t = \\pm \\frac{\\pi}{6}$.\n\nThis divides our interval $-\\frac{\\pi}{3} < t < \\frac{\\pi}{3}$ into three regions:\n1. A central full branch: $-\\frac{\\pi}{6} < t < \\frac{\\pi}{6}$\n2. Two outer partial branches: $-\\frac{\\pi}{3} < t < -\\frac{\\pi}{6}$ and $\\frac{\\pi}{6} < t < \\frac{\\pi}{3}$\n\nUsing a Graphic Display Calculator (GDC) to plot $y_1 = \\tan(3t)$ and $y_2 = 60t$, we can observe the intersections."},{"type":"text","order":1,"title":"Interpret part (i) results","content":"From the graph on the GDC, we observe:\n- At $t = 0$, both functions are $0$, so $t = 0$ is a root.\n- The derivative of $\\tan(3t)$ at $t = 0$ is:\n\n $$\\frac{d}{dt}[\\tan(3t)]_{t=0} = 3\\sec^2(0) = 3$$\n\n- Since the slope of the line ($60$) is greater than the slope of the tangent line at the origin ($3$), the line climbs faster than the tangent curve initially, intersecting it again at one positive value in $(0, \\frac{\\pi}{6})$ and one negative value in $(-\\frac{\\pi}{6}, 0)$. This gives **3 roots** in the central branch.\n- In the outer partial branches, since $60t > 0$ for $t > 0$ while $\\tan(3t) \\le 0$ on $(\\frac{\\pi}{6}, \\frac{\\pi}{3}]$, there are no intersections. Similarly, by symmetry, there are no intersections for $t < 0$ in the left partial branch.\n\nThus, there are exactly **3 roots** in this interval."},{"type":"text","order":2,"title":"Analyze part (ii) using GDC","content":"For (ii), we need to determine the number of roots of $\\tan(3t) = 30t$ in the wider interval $-2\\pi < t < 2\\pi$.\n\nLet's first determine the number of full branches of $\\tan(3t)$ that fit into this interval. Since the period of $\\tan(3t)$ is $T = \\frac{\\pi}{3}$:\n- The total length of the interval is $2\\pi - (-2\\pi) = 4\\pi$.\n- The number of periods/branches in this interval is:\n\n $$\\frac{4\\pi}{\\pi/3} = 12 \\text{ branches}$$\n\nLet's use our GDC to graph this equation over a smaller symmetric range (e.g., $-\\pi < t < \\pi$) to establish the pattern."},{"type":"text","order":3,"title":"Generalize the root pattern","content":"By observing the graph on the GDC, we can generalize how the line $y = 30t$ intersects the periodic branches of $y = \\tan(3t)$:\n\n1. **The Central Branch** containing $t = 0$:\n Since the slope of the line ($30$) is greater than the slope of the tangent at the origin ($3$), it intersects the central branch **3 times**.\n \n2. **The Non-Central Branches**:\n Because the linear function $30t$ grows much faster than the vertical asymptotes of $\\tan(3t)$, it passes through each of the other standard branches exactly **once**.\n\nSince each of the standard branches other than the central one contributes exactly $1$ root, we can write a general formula for the number of roots:\n\n$$\\text{Total Roots} = 2 \\times (\\text{Number of branches on one side}) + 1$$\n\nSince the interval is $-2\\pi < t < 2\\pi$:\n- The number of branches on the positive side is $\\frac{2\\pi}{\\pi/3} = 6$.\n- Thus, the total number of roots is:\n\n $$\\text{Total Roots} = 2(6) + 1 = 13$$"},{"type":"text","order":4,"title":"Calculate roots for part (iii)","content":"For (iii), we need to find the number of roots of $\\tan(3t) = 30t$ in the large interval $-40\\pi < t < 40\\pi$.\n\nUsing the general formula we established from our visual GDC analysis:\n- The number of branches on the positive side is:\n\n $$\\frac{40\\pi}{\\pi/3} = 120$$\n\n- Applying our generalization formula:\n\n $$\\text{Total Roots} = 2(120) + 1 = 241$$"}]}],"generatedAt":"2026-06-09T06:08:51.073Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"700d8d37-d932-41f9-ad0c-4e116bc98493:formulaic","label":"(a) Using Standard Period Formula","steps":[{"type":"text","order":0,"title":"Identify the standard period","content":"The standard tangent function $y = \\tan(x)$ has a period of $\\pi$. This means the function repeats its values every $\\pi$ radians."},{"type":"text","order":1,"title":"Apply the period formula","content":"When a trigonometric function is horizontally scaled, we find its new period using the coefficient of the variable inside the function. For a function of the form $y = \\tan(Bt)$, the period $T$ is given by:\n\n$$T = \\frac{\\pi}{B}$$\n\nIn our given function, $d = \\tan(3t)$, the coefficient of $t$ is $B = 3$."},{"type":"text","order":2,"title":"Calculate the final period","content":"Substitute $B = 3$ into our period formula:\n\n$$T = \\frac{\\pi}{3}$$\n\nThis simple substitution gives us the final period directly. This matches the **B1** mark in the markscheme."}]},{"id":"700d8d37-d932-41f9-ad0c-4e116bc98493:graphical","label":"(a) Graphical Analysis","steps":[{"type":"text","order":0,"title":"Understand the graphical goal","content":"To find the period of the function $d = \\tan(3t)$ using graphical analysis, we can plot the graph of the function on a graphing calculator.\n\nThe period of a periodic function is the horizontal distance over which the graph completes one full cycle. For a tangent function, this is the distance between two consecutive $x$-intercepts (roots) or the distance between two consecutive vertical asymptotes."},{"type":"text","order":1,"title":"Plot the function","content":"Let us graph the function $y = \\tan(3x)$ on a graphing calculator. We should set our angle mode to **Radians** since the intervals in the question are given in terms of $\\pi$.","calculator":[{"gdc":{"expression":"y = tan(3x)","instructions":"1. Set the angle units to Radians.\n2. In the Graph menu, enter the equation y = tan(3x).\n3. Set the viewing window x from -pi to pi.\n4. Draw the graph."},"ti84":{"expression":"Y1 = tan(3X)","instructions":"1. Press [MODE], ensure 'RADIAN' is selected, then press [CLEAR].\n2. Press [Y=] and enter tan(3X) in Y1.\n3. Press [WINDOW] and set Xmin = -3.14, Xmax = 3.14, Ymin = -5, Ymax = 5.\n4. Press [GRAPH] to view the curve."},"desmos":{"expression":"y = tan(3x)","instructions":"1. Set the graph settings to Radians.\n2. Type the function y = tan(3x).\n3. Adjust the x-axis limits to [-pi, pi] to see several cycles."},"description":"Graph the function $y = \\tan(3x)$ in radian mode."}]},{"type":"text","order":2,"title":"Analyze the intercepts","content":"Looking at the graph, we can find the coordinates of consecutive $x$-intercepts where the curve crosses the $x$-axis:\n- One intercept is located at $(0, 0)$.\n- The next consecutive positive intercept is located at $(\\frac{\\pi}{3}, 0)$ which is approximately $(1.047, 0)$.\n\nThe horizontal distance between these two consecutive intercepts is:\n$$\\text{Period} = \\frac{\\pi}{3} - 0 = \\frac{\\pi}{3}$$","calculator":[{"gdc":{"instructions":"1. Use the G-Solve menu.\n2. Press 'Root' to trace the x-intercepts."},"ti84":{"instructions":"1. Press [2nd] [TRACE] and select '2: zero'.\n2. Follow the prompts to find the first positive root to the right of the origin."},"desmos":{"instructions":"1. Click on the points where the curve crosses the x-axis to reveal their coordinates."},"description":"Find consecutive roots on your calculator."}]},{"type":"text","order":3,"title":"State the final answer","content":"Alternatively, we can find the distance between consecutive vertical asymptotes. The first vertical asymptote to the left of the origin is at $t = -\\frac{\\pi}{6}$ and the first to the right is at $t = \\frac{\\pi}{6}$.\n\nThe distance between them is:\n$$\\frac{\\pi}{6} - \\left(-\\frac{\\pi}{6}\\right) = \\frac{2\\pi}{6} = \\frac{\\pi}{3}$$\n\nThus, the period of the function is:\n$$\\frac{\\pi}{3}$$"}]},{"id":"95950576-ff3e-422b-92e2-b3cbacebe5bb:graphical_analytical","label":"(b) Graphical and Analytical Analysis","steps":[{"type":"text","order":0,"title":"Analyze the tangent function","content":"To find the number of solutions to equations of the form $\\tan(3t) = kt$, we must understand the shape of the graph of $y = \\tan(3t)$.\n\nThe fundamental period of $\\tan(3t)$ is:\n\n$$T = \\frac{\\pi}{3}$$\n\nThe vertical asymptotes occur where $3t = \\frac{\\pi}{2} + n\\pi$, which gives:\n\n$$t = \\frac{\\pi}{6} + \\frac{n\\pi}{3} \\quad \\text{for } n \\in \\mathbb{Z}$$\n\nLet's analyze the branches of the tangent function:\n- The **central branch** is defined on the open interval $(-\\frac{\\pi}{6}, \\frac{\\pi}{6})$ and passes through the origin $(0,0)$.\n- Other branches are shifted to the left and right by multiples of the period $\\frac{\\pi}{3}$."},{"type":"text","order":1,"title":"Evaluate derivative at origin","content":"To determine how many times a line of the form $y = kt$ (for $k > 0$) intersects the central branch of $y = \\tan(3t)$, we compare their gradients at the origin ($t=0$).\n\nThe derivative of $f(t) = \\tan(3t)$ is:\n\n$$f'(t) = 3\\sec^2(3t)$$\n\nAt the origin ($t=0$):\n\n$$f'(0) = 3\\sec^2(0) = 3$$\n\nSince the gradient of $y = kt$ is $k$, and in both parts $k > 3$ (specifically $k = 60$ and $k = 30$):\n- Near the origin, the line $y = kt$ is steeper than the tangent curve.\n- Since both are odd functions, the line must start below the curve for negative $t$ and rise above the curve for positive $t$, intersecting the central branch of $\\tan(3t)$ exactly **$3$ times**: once at $t=0$, once at some $t_1 \\in (-\\frac{\\pi}{6}, 0)$, and once at some $t_2 \\in (0, \\frac{\\pi}{6})$."},{"type":"text","order":2,"title":"Solve part (i)","content":"For part (i), we are looking at the interval $-\\frac{\\pi}{3} < t < \\frac{\\pi}{3}$ for the equation $\\tan(3t) = 60t$.\n\nLet's break the interval down into three regions based on the branches of $\\tan(3t)$:\n\n1. **The Central Branch** $(-\\frac{\\pi}{6}, \\frac{\\pi}{6})$:\n As established, since $60 > 3$, the line $y = 60t$ intersects this branch exactly **$3$ times**.\n\n2. **The Outer Intervals** $[\\frac{\\pi}{6}, \\frac{\\pi}{3})$ and $(-\\frac{\\pi}{3}, -\\frac{\\pi}{6}]$:\n - On $(\\frac{\\pi}{6}, \\frac{\\pi}{3})$, the function $\\tan(3t)$ is negative or zero because the angle $3t \\in (\\frac{\\pi}{2}, \\pi)$ lies in the second quadrant. However, the line $y = 60t$ is strictly positive. Thus, there are **no intersections** here.\n - By symmetry, on $(-\\frac{\\pi}{3}, -\\frac{\\pi}{6})$, the function $\\tan(3t)$ is positive or zero while the line $y = 60t$ is negative, so there are **no intersections** here.\n\nThus, there are exactly **$3$ roots** in total."},{"type":"text","order":3,"title":"Analyze part (ii)","content":"For part (ii), we consider $\\tan(3t) = 30t$ in the wider interval $-2\\pi < t < 2\\pi$.\n\nThe interval has width $4\\pi$, equal to $12$ periods of $\\tan(3t)$. Its endpoints are zeros of the tangent function, not asymptotes. There are $11$ complete branches between the asymptotes in the interval, plus two partial branches at the ends. The central complete branch contains the origin."},{"type":"text","order":4,"title":"Count roots for part (ii)","content":"Since $30 > 3$, the line $y = 30t$ intersects the central branch exactly **3 times**.\n\nEach of the other $10$ complete branches contributes **1 root**. On a complete branch to the right of the origin, $\\tan(3t)$ rises from $-\\infty$ to $+\\infty$; it crosses the positive line $30t$ once after the branch's zero. The left-hand branches give one root each by symmetry.\n\nThe two partial end branches contribute no roots: on $(\\frac{11\\pi}{6},2\\pi)$, $\\tan(3t)<0$ but $30t>0$, and the signs are reversed on $(-2\\pi,-\\frac{11\\pi}{6})$. The endpoints are excluded. Therefore:\n\n$$\\text{Total roots} = 10 + 3 = 13$$"},{"type":"text","order":5,"title":"Generalize for part (iii)","content":"For part (iii), $-40\\pi < t < 40\\pi$ contains $\\frac{40\\pi}{\\pi/3}=120$ periods on each side of the origin. The endpoints are zeros of $\\tan(3t)$, so there are $239$ complete branches and two partial end branches.\n\nThe central branch contributes $3$ roots, each of the other $238$ complete branches contributes $1$, and the partial end branches contribute none. Thus:\n\n$$\\text{Total roots} = 238 + 3 = 241$$"}]},{"id":"95950576-ff3e-422b-92e2-b3cbacebe5bb:gdc_generalization","label":"(b) Using GDC and Generalization","steps":[{"type":"text","order":0,"title":"Analyze part (i) using GDC","content":"To solve (i), let's find the number of roots of the equation $\\tan(3t) = 60t$ in the interval $-\\frac{\\pi}{3} < t < \\frac{\\pi}{3}$.\n\nFirst, note that the function $y = \\tan(3t)$ has a period of $T = \\frac{\\pi}{3}$ and vertical asymptotes where $3t = \\pm \\frac{\\pi}{2} \\implies t = \\pm \\frac{\\pi}{6}$.\n\nThis divides our interval $-\\frac{\\pi}{3} < t < \\frac{\\pi}{3}$ into three regions:\n1. A central full branch: $-\\frac{\\pi}{6} < t < \\frac{\\pi}{6}$\n2. Two outer partial branches: $-\\frac{\\pi}{3} < t < -\\frac{\\pi}{6}$ and $\\frac{\\pi}{6} < t < \\frac{\\pi}{3}$\n\nUsing a Graphic Display Calculator (GDC) to plot $y_1 = \\tan(3t)$ and $y_2 = 60t$, we can observe the intersections."},{"type":"text","order":1,"title":"Interpret part (i) results","content":"From the graph on the GDC, we observe:\n- At $t = 0$, both functions are $0$, so $t = 0$ is a root.\n- The derivative of $\\tan(3t)$ at $t = 0$ is:\n\n $$\\frac{d}{dt}[\\tan(3t)]_{t=0} = 3\\sec^2(0) = 3$$\n\n- Since the slope of the line ($60$) is greater than the slope of the tangent line at the origin ($3$), the line climbs faster than the tangent curve initially, intersecting it again at one positive value in $(0, \\frac{\\pi}{6})$ and one negative value in $(-\\frac{\\pi}{6}, 0)$. This gives **3 roots** in the central branch.\n- In the outer partial branches, since $60t > 0$ for $t > 0$ while $\\tan(3t) \\le 0$ on $(\\frac{\\pi}{6}, \\frac{\\pi}{3}]$, there are no intersections. Similarly, by symmetry, there are no intersections for $t < 0$ in the left partial branch.\n\nThus, there are exactly **3 roots** in this interval."},{"type":"text","order":2,"title":"Analyze part (ii) using GDC","content":"For (ii), we need to determine the number of roots of $\\tan(3t) = 30t$ in $-2\\pi < t < 2\\pi$.\n\nThe interval has length $4\\pi$, which is $\\frac{4\\pi}{\\pi/3}=12$ periods of $\\tan(3t)$. Because the endpoints are zeros rather than asymptotes, this gives $11$ complete branches and two partial end branches. A GDC plot over a smaller symmetric range (e.g., $-\\pi < t < \\pi$) helps establish the intersection pattern."},{"type":"text","order":3,"title":"Generalize the root pattern","content":"The central branch contributes $3$ roots because $30$ exceeds the gradient $3$ of $\\tan(3t)$ at the origin. Each other complete branch contributes $1$ root. The partial branches at the ends contribute none: on the right, $\\tan(3t)<0$ while $30t>0$, and the signs are reversed on the left.\n\nIn $-2\\pi

1.8 E: Trigonometry

Question 94

1.8 E: Trigonometry Questions

1.8 E: Trigonometry

Question 94

1.8 E: Trigonometry Questions