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E Math (IGCSE)

H1 Math

H2 Math

IB AA HL Math

IB AA SL Math

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IB AI SL Math

IP Math

N(A) A Math

N(A) E Math

PSLE

PSLE Foundation Mathematics

<ol class="abc-list"><li data-list-item-id="item-1773464876489-br9vyb0ln"><div class="list-item-paragraph">State the amplitude and period of \(2\sin3x+1\)<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1773464877125-1sfyhvu0e"><div class="list-item-paragraph">Sketch the graph of \(y=2\sin3x+1\) for&nbsp;\(0\leq x\leq \pi\).<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1773464946940-po939pu8x"><div class="list-item-paragraph">By drawing a suitable straight line on your sketch, determine the number of solutions of the equation \(\pi\sin 3x-x=0\).<span class="inline-marks">[3]</span></div></li></ol>

State the amplitude and period of \(2 \sin 3x + 1\).

Sketch the graph of \(y = 2 \sin 3x + 1\) for \(0 \le x \le \pi\).

By drawing a suitable straight line on your sketch, determine the number of solutions of the equation \(\pi \sin 3x - x = 0\).

Trigonometric Graphs and Parameters

Trigonometric Equations

2025 MSHS PRELIMS P1 Q10

Trigonometry

<ol class="abc-list"><li data-list-item-id="item-a"><div class="list-item-content"><span class="list-item-text">State the amplitude and period of the graph of \(y = 3\sin \frac{x}{2} + 1\).</span><span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b"><div class="list-item-content"><span class="list-item-text">Sketch the graph of \(y = 3\sin \frac{x}{2} + 1\) for&nbsp;\(0 \le x \le 4\pi\).</span><span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-c"><div class="list-item-content"><span class="list-item-text">By adding another suitable curve on your sketch, determine the number of solutions of the equation \(\sin \frac{x}{2} = -\frac{1}{3} - \frac{2}{3} \cos \frac{x}{2}\).</span><span class="inline-marks">[3]</span></div></li></ol>

State the amplitude and period of the graph of \(y = 3\sin \frac{x}{2}+1\).

Sketch the graph of \(y = 3\sin \frac{x}{2}+1\) for \(0 \le x \le 4\pi\).

By adding another suitable curve on your sketch, determine the number of solutions of the equation \(\sin \frac{x}{2} = -\frac{1}{3} - \frac{2}{3}\cos \frac{x}{2}\).

2025 SCGS PRELIMS P1 Q7

<p>The height of tide, \(y\) m, at a coast on a particular day can be approximated by<br>\(y = 2 + \cos \left( \frac{\pi}{6} t \right),\) where&nbsp;\(t\) is the time in hours after midnight.</p>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">Sketch the graph of \(y = 2 + \cos \left( \frac{\pi}{6} t \right)\) for&nbsp;\(0 \le t \le 24\).</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">Find the height of tide at \(7:30\) a.m. on that day.</span></div></li><li data-list-item-id="item-iii"><div class="list-item-content"><span class="list-item-text">Find the time(s) when the height of tide is \(2.6\) m on that day.</span></div></li><li data-list-item-id="item-iv"><div class="list-item-content"><span class="list-item-text">Find the interval(s) of time that the height of tide is below \(1.4\) m.</span></div></li></ol>

The height of tide, \(y\) m, at a coast on a particular day can be approximated by
\(y = 2 + \cos \left( \frac{\pi}{6} t \right),\) where \(t\) is the time in hours after midnight.

Sketch the graph of \(y = 2 + \cos \left( \frac{\pi}{6} t \right)\) for \(0 \le t \le 24\).

Find the height of tide at \(7:30\) a.m. on that day.

Find the time(s) when the height of tide is \(2.6\) m on that day.

Find the interval(s) of time that the height of tide is below \(1.4\) m.

Trigonometric Modelling

Trigonometric Inequalities

<p>The London Eye is one of the world's most famous Ferris wheels, which has a radius of \(60\) m and its centre is \(75\) m above the ground. It makes a complete revolution in \(30\) minutes. Let&nbsp;\(h\) m be the height of a passenger above the ground \(t\) minutes after boarding at the bottom of the wheel.</p>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">Express \(h\) as a function of \(t\). Hence find the amplitude and period of \(h\).</span></div></li><li data-list-item-id="item-ii"><div class="list-item-paragraph">Sketch the graph of the function in <strong>(i)</strong> for&nbsp;\(0 \le t \le 30\).</div></li><li data-list-item-id="item-iii"><div class="list-item-content"><span class="list-item-text">Find the height of the passenger above the ground after</span></div><ol class="abc-list"><li data-list-item-id="item-a"><div class="list-item-content"><span class="list-item-text">\(6\) minutes, and</span></div></li><li data-list-item-id="item-b"><div class="list-item-content"><span class="list-item-text">\(25\) minutes.</span></div></li></ol></li><li data-list-item-id="item-iv"><div class="list-item-content"><span class="list-item-text">Find the interval of time that the passenger is above \(100\) m from the ground</span></div></li></ol>

The London Eye is one of the world's most famous Ferris wheels, which has a radius of \(60\) m and its centre is \(75\) m above the ground. It makes a complete revolution in \(30\) minutes. Let \(h\) m be the height of a passenger above the ground \(t\) minutes after boarding at the bottom of the wheel.

Express \(h\) as a function of \(t\). Hence find the amplitude and period of \(h\).

Find the height of the passenger above the ground after

Find the interval of time that the passenger is above \(100\) m from the ground

The daylength on the first day of each month from January \(2021\) to January \(2023\), \(D\) (in hours), for four cities of different latitude is recorded and sketched with respect to \(t\) (in months) using smooth curves in the same diagram below.

Explain with clear reasons why sine and cosine functions can be used to model the above data.

The curves representing the daylength of Singapore (respectively, Harbin and Reykjavik) can be approximated by \(D_S = a_1 \cos b_1 t + c_1\) (respectively, \(D_H = a_2 \cos b_2 t + c_2\) and \(D_R = a_3 \cos b_3 t + c_3\)). State, by giving clear reasons, the relationships between \(a_1\), \(a_2\), \(a_3\); \(b_1\), \(b_2\), \(b_3\) and \(c_1\), \(c_2\), \(c_3\), respectively.

Given that the daylength of Sydney on \(01\) January \(2021\) is \(14\) hours (nearest integer), suggest a suitable function to model the daylength of Sydney for \(0 \le t \le 24\).

The table below shows the latitudes at which the four cities locate. State the relationship between the amplitude of the daylength functions and the latitude of the city. Hence make a conjecture on the amplitude of the daylength function at the North Pole.

Data Source: [Sunrise Sunset Times Lookup - MAPLOGS](https://www.maplogs.com/)

<p>A radio device transmits signals by electromagnetic waves, where the electric field of the waves can be expressed as&nbsp;\(\text{E}\hspace{0.1em}(t) = E_0 \cos(2\pi f t)\) (in volt, \(\text{V}\)), with respect to time \(t\) (in seconds). \(f\) is called the frequency of the wave in hertz (\(\text{Hz}\)). Assume that the amplitude of the electric field is \(5\) volt, the frequency of the electric field is \(100\) MHz (\(1\text{ MHz} = 10^6\text{ Hz}\)).</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Find the specific expression of the electric field \(\text{E}\hspace{0.1em}(t)\) of this electromagnetic wave.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Find the period of \(\text{E}\hspace{0.1em}(t)\). Hence derive the relationship between frequency and period.</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">Sketch the graph of \(\text{E}\hspace{0.1em}(t)\), for&nbsp;\(0 \le t \le 4 \times 10^{-8}\).</div></li><li data-list-item-id="item-iv" data-list-item-path="item-iv"><div class="list-item-paragraph">In different circumstances, the graphs of \(\text{E}\hspace{0.1em}(t)\) are shown in the figures below, respectively. Briefly describe the possible operations on the radio device in terms of power and/or frequency tuning in each of the cases.</div></li></ol>
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A radio device transmits signals by electromagnetic waves, where the electric field of the waves can be expressed as \(\text{E}(t) = E_0 \cos(2\pi f t)\) (in volt, \(\text{V}\)), with respect to time \(t\) (in seconds). \(f\) is called the frequency of the wave in hertz (\(\text{Hz}\)). Assume that the amplitude of the electric field is \(5\) volt, the frequency of the electric field is \(100\) MHz (\(1\text{ MHz} = 10^6\text{ Hz}\)).

Find the specific expression of the electric field \(\text{E}(t)\) of this electromagnetic wave.

Find the period of \(\text{E}(t)\). Hence derive the relationship between frequency and period.

Sketch the graph of \(\text{E}(t)\), for \(0 \le t \le 4 \times 10^{-8}\).

In different circumstances, the graphs of \(\text{E}(t)\) are shown in the figures below, respectively. Briefly describe the possible operations on the radio device in terms of power and/or frequency tuning in each of the cases.

<p>The voltage, \(V\) (in volt, \(V\)), in an alternating current (A.C.) circuit, is given by \(V = 220 \sin(100\pi t)\), where&nbsp;\(t\) is the time in seconds.</p>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">Find the amplitude, period and frequency of the voltage. Note that frequency (in hertz, \(\text{Hz}\)) is defined as the reciprocal of the period.</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">Find the value of the voltage when&nbsp;\(t = \frac{1}{600}\) \(\text{s}\).</span></div></li><li data-list-item-id="item-iii"><div class="list-item-content"><span class="list-item-text">Sketch the graph of \(V = 220 \sin(100\pi t)\), for&nbsp;\(0 \le t \le 0.05\).</span></div></li><li data-list-item-id="item-iv"><div class="list-item-content"><span class="list-item-text">Given that the frequency is tuned to \(100\) \(\text{Hz}\), the amplitude is tuned to \(110\) \(\text{V}\), find the new expression of the voltage and sketch on the same diagram the graph of the voltage, for&nbsp;\(0 \le t \le 0.05\). Hence explain how the values of amplitude and frequency affect the curve.</span></div></li></ol>

The voltage, \(V\) (in volt, \(V\)), in an alternating current (A.C.) circuit, is given by \(V = 220 \sin(100\pi t)\), where \(t\) is the time in seconds.

Find the amplitude, period and frequency of the voltage. Note that frequency (in hertz, \(\text{Hz}\)) is defined as the reciprocal of the period.

Find the value of the voltage when \(t = \frac{1}{600}\) \(\text{s}\).

Sketch the graph of \(V = 220 \sin(100\pi t)\), for \(0 \le t \le 0.05\).

Given that the frequency is tuned to \(100\) \(\text{Hz}\), the amplitude is tuned to \(110\) \(\text{V}\), find the new expression of the voltage and sketch on the same diagram the graph of the voltage, for \(0 \le t \le 0.05\). Hence explain how the values of amplitude and frequency affect the curve.

<p>The Singapore Flyer is one of the world's most famous Ferris wheels, which has a radius of \(75\) m and its centre is \(90\) m above the ground. It makes a complete revolution in \(30\) minutes. Let&nbsp;\(h\) m be the height of a passenger above the ground \(t\) minutes after boarding at the bottom of the wheel.</p>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">Express \(h\) as a function of \(t\). Hence find the amplitude and period of \(h\).</span></div></li><li data-list-item-id="item-ii"><div class="list-item-paragraph">Sketch the graph of the function in <strong>(i)</strong> for&nbsp;\(0 \le t \le 30\).</div></li><li data-list-item-id="item-iii"><div class="list-item-content"><span class="list-item-text">Find the height of the passenger above the ground after</span></div><ol class="abc-list"><li data-list-item-id="item-a"><div class="list-item-content"><span class="list-item-text">\(7.5\) minutes,</span></div></li><li data-list-item-id="item-b"><div class="list-item-content"><span class="list-item-text">\(15\) minutes, and</span></div></li><li data-list-item-id="item-c"><div class="list-item-content"><span class="list-item-text">\(20\) minutes.</span></div></li></ol></li><li data-list-item-id="item-iv"><div class="list-item-content"><span class="list-item-text">Find the interval of time that the passenger is above \(90\) m from the ground.</span></div></li></ol>

The Singapore Flyer is one of the world's most famous Ferris wheels, which has a radius of \(75\) m and its centre is \(90\) m above the ground. It makes a complete revolution in \(30\) minutes. Let \(h\) m be the height of a passenger above the ground \(t\) minutes after boarding at the bottom of the wheel.

Find the interval of time that the passenger is above \(90\) m from the ground.

<p>The diagram shows the graphs of \(y = -2 \sin kx\) and \(y = -\cos mx - 3\) for&nbsp;\(0 \leq x \leq \pi\). The curves have a common turning point \(Q\) (maximum or minimum point). Find</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">the values of \(k\) and \(m\),</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">the coordinates of \(Q\), and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">the coordinates of the next common turning point of the two curves (in the positive \(x\)-direction).</div></li></ol>
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The diagram shows the graphs of \(y = -2 \sin kx\) and \(y = -\cos mx - 3\) for \(0 \leq x \leq \pi\). The curves have a common turning point \(Q\) (maximum or minimum point). Find

the coordinates of the next common turning point of the two curves (in the positive \(x\)-direction).

<p>The diagram shows the graphs of \(y = 3 \cos 2x\) and \(y = k - \cos x\) for&nbsp;\(0 \le x \le \frac{5\pi}{4}\). The curves have a common maximum point \(Q\). Find</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">the value of \(k\),</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">the coordinates of \(Q\), and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">the coordinates of the next common maximum point of the two curves (in the positive \(x\)-direction).</div></li></ol>
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The diagram shows the graphs of \(y = 3 \cos 2x\) and \(y = k - \cos x\) for \(0 \le x \le \frac{5\pi}{4}\). The curves have a common maximum point \(Q\). Find

the coordinates of the next common maximum point of the two curves (in the positive \(x\)-direction).

<p>The function \(f\) is defined, for&nbsp;\(0 \le x \le 2\pi\), by \(f(x) = -4\cos 2x + 1\).</p>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">Sketch the graph of \(y = f(x)\) for&nbsp;\(0 \le x \le 2\pi\).</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">By sketching a proper straight line on the same diagram, find the number of solutions to the equation \(2\pi \cos 2x - x + \pi = 0\) for&nbsp;\(0 \le x \le 2\pi\).</span></div></li><li data-list-item-id="item-iii"><div class="list-item-content"><span class="list-item-text">Find the range of the value of \(k\) for which the equation \(-4\cos 2x + 1 - k = 0\) has more than two unique solutions for&nbsp;\(0 \le x \le 2\pi\).</span></div></li></ol>

The function \(f\) is defined, for \(0 \le x \le 2\pi\), by \(f(x) = -4\cos 2x + 1\).

Sketch the graph of \(y = f(x)\) for \(0 \le x \le 2\pi\).

By sketching a proper straight line on the same diagram, find the number of solutions to the equation \(2\pi \cos 2x - x + \pi = 0\) for \(0 \le x \le 2\pi\).

Find the range of the value of \(k\) for which the equation \(-4\cos 2x + 1 - k = 0\) has more than two unique solutions for \(0 \le x \le 2\pi\).

<p>The diagram shows the graphs of \(y = a \tan bx\), where&nbsp;\(b &gt; 0\), for&nbsp;\(-3\pi &lt; x &lt; 3\pi\) and \(y = m\). The graph of \(y = a \tan bx\) passes through the point \(\left(\frac{\pi}{2}, 1\right)\) and intersects \(y = m\) at the points \(A\), \(B\) and \(C\). The coordinates of \(C\) is \((p, m)\).</p>
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<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Find the period of \(y\) and the values of \(a\) and \(b\).</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">State the coordinates of one possible point about which the graph of \(y = a \tan bx\) is symmetrical and the order of rotational symmetry.</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">Find, in terms of \(p\) and \(m\), the coordinates of \(A\) and \(B\).</div></li><li data-list-item-id="item-iv" data-list-item-path="item-iv"><div class="list-item-paragraph">By sketching a proper straight line on the same diagram, find the number of solutions to the equation \(k\left(x - \frac{\pi}{2}\right) + 1 - a \tan bx = 0\), where&nbsp;\(k \le 0\), for&nbsp;\(-3\pi &lt; x &lt; 3\pi\).</div></li></ol>

The diagram shows the graphs of \(y = a \tan bx\), where \(b > 0\), for \(-3\pi < x < 3\pi\) and \(y = m\). The graph of \(y = a \tan bx\) passes through the point \(\left(\frac{\pi}{2}, 1\right)\) and intersects \(y = m\) at the points \(A\), \(B\) and \(C\). The coordinates of \(C\) is \((p, m)\).

Find the period of \(y\) and the values of \(a\) and \(b\).

State the coordinates of one possible point about which the graph of \(y = a \tan bx\) is symmetrical and the order of rotational symmetry.

Find, in terms of \(p\) and \(m\), the coordinates of \(A\) and \(B\).

By sketching a proper straight line on the same diagram, find the number of solutions to the equation \(k\left(x - \frac{\pi}{2}\right) + 1 - a \tan bx = 0\), where \(k \le 0\), for \(-3\pi < x < 3\pi\).

Symmetry of Trigonometric Functions

<p>The diagram shows the graph of \(y = a \cos bx + c\), where&nbsp;\(b &gt; 0\), for&nbsp;\(-2\pi \le x \le 2\pi\). The graph has a maximum point at \(\left( \frac{\pi}{2}, 4 \right)\) and a minimum point at \((0, -8)\).</p>
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<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Find the amplitude, the period and the range of values of \(y\) corresponding to the interval. Hence find the values of \(a\), \(b\) and \(c\).</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Write down the coordinates of the other maximum point(s) and the other minimum point(s) of the graph in the given interval.</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">Find the range of the value of \(k\) for which the equation \(a \cos bx + c = k\) has more than four unique solutions for&nbsp;\(-2\pi \le x \le 2\pi\).</div></li></ol>

The diagram shows the graph of \(y = a \cos bx + c\), where \(b > 0\), for \(-2\pi \le x \le 2\pi\). The graph has a maximum point at \(\left( \frac{\pi}{2}, 4 \right)\) and a minimum point at \((0, -8)\).

Find the amplitude, the period and the range of values of \(y\) corresponding to the interval. Hence find the values of \(a\), \(b\) and \(c\).

Write down the coordinates of the other maximum point(s) and the other minimum point(s) of the graph in the given interval.

Find the range of the value of \(k\) for which the equation \(a \cos bx + c = k\) has more than four unique solutions for \(-2\pi \le x \le 2\pi\).

<p>The diagram shows the graph of \(y = a \sin bx + c\), where&nbsp;\(b &gt; 0\), for&nbsp;\(-180^\circ \le x \le 180^\circ\). The graph has a maximum point at \(\left( 45^\circ, \frac{9}{2} \right)\) and a minimum point at \(\left( 135^\circ, -\frac{1}{2} \right)\).</p>
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<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Find the amplitude, the period and the range of values of \(y\) corresponding to the interval. Hence find the values of \(a\), \(b\) and \(c\).</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Write down the coordinates of the other maximum point(s) and the other minimum point(s) of the graph in the given interval.</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">Find the range of the value of \(k\) for which the equation \(a \sin bx + c = k\) has four unique solutions for&nbsp;\(-180^\circ \le x \le 180^\circ\).</div></li></ol>

The diagram shows the graph of \(y = a \sin bx + c\), where \(b > 0\), for \(-180^\circ \le x \le 180^\circ\). The graph has a maximum point at \(\left( 45^\circ, \frac{9}{2} \right)\) and a minimum point at \(\left( 135^\circ, -\frac{1}{2} \right)\).

Find the range of the value of \(k\) for which the equation \(a \sin bx + c = k\) has four unique solutions for \(-180^\circ \le x \le 180^\circ\).

<p>On a separate diagram, sketch each of the following graphs for the given interval, stating the amplitude, the period and the range of values of \(y\) corresponding to the interval.</p>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">\(y = -4 \cos 8\pi x - 4\), \(-1 \leq x \leq -\frac{1}{2}\)</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">\(y = -\frac{1}{8} \cos \frac{\pi}{8} x - \frac{1}{8}\), \(-8 \leq x \leq 8\)</span></div></li></ol>

On a separate diagram, sketch each of the following graphs for the given interval, stating the amplitude, the period and the range of values of \(y\) corresponding to the interval.

\(y = -4 \cos 8\pi x - 4\), \(-1 \leq x \leq -\frac{1}{2}\)

\(y = -\frac{1}{8} \cos \frac{\pi}{8} x - \frac{1}{8}\), \(-8 \leq x \leq 8\)

<p>On a separate diagram, sketch each of the following graphs for the given interval, stating the amplitude, the period and the range of values of \(y\) corresponding to the interval.</p>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">\(y = 8 \cos 8x + 8\), \(\frac{\pi}{2} \leq x \leq \pi\)</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">\(y = -\frac{1}{3} \cos 3x + 2\), \(\pi \leq x \leq 2\pi\)</span></div></li></ol>

\(y = 8 \cos 8x + 8\), \(\frac{\pi}{2} \leq x \leq \pi\)

\(y = -\frac{1}{3} \cos 3x + 2\), \(\pi \leq x \leq 2\pi\)

<p>On a separate diagram, sketch each of the following graphs for the given interval, stating the amplitude, the period and the range of values of \(y\) corresponding to the interval.</p>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">\(y = \frac{1}{5} \sin \pi x + 1, -3 \leq x \leq 3\)</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">\(y = -6 \sin \frac{4\pi}{3} x - 3, -3 \leq x \leq 0\)</span></div></li></ol>

\(y = \frac{1}{5} \sin \pi x + 1, -3 \leq x \leq 3\)

\(y = -6 \sin \frac{4\pi}{3} x - 3, -3 \leq x \leq 0\)

<p>On a separate diagram, sketch each of the following graphs for the given interval, stating the amplitude, the period and the range of values of \(y\) corresponding to the interval.</p>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">\(y = -\frac{1}{2} \sin 6x + \frac{7}{2}, -\frac{2\pi}{3} \le x \le \frac{2\pi}{3}\)</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">\(y = \frac{5}{6} \sin \frac{1}{3} x + \frac{5}{3}, -3\pi \le x \le 3\pi\)</span></div></li></ol>

\(y = -\frac{1}{2} \sin 6x + \frac{7}{2}, -\frac{2\pi}{3} \le x \le \frac{2\pi}{3}\)

\(y = \frac{5}{6} \sin \frac{1}{3} x + \frac{5}{3}, -3\pi \le x \le 3\pi\)

<p>The diagram shows the graph of \(y = \tan bx\), where&nbsp;\(b &gt; 0\), for&nbsp;\(-\frac{2\pi}{3} &lt; x &lt; \frac{2\pi}{3}\).</p>
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<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Find the period of \(y\). Hence find the value of \(b\).</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Explain why the amplitude of \(y = \tan bx\) is undefined.</div></li></ol>

The diagram shows the graph of \(y = \tan bx\), where \(b > 0\), for \(-\frac{2\pi}{3} < x < \frac{2\pi}{3}\).

Find the period of \(y\). Hence find the value of \(b\).

Explain why the amplitude of \(y = \tan bx\) is undefined.

<p>The diagram shows part of the graph of \(y = a \cos bx\), where&nbsp;\(b &gt; 0\), for&nbsp;\(0 \le x \le \frac{4\pi}{3}\).</p>
<div class="graph-block" data-graph-config="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" data-width="277" data-height="204" data-proportional style="width:277px;height:204px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:277px;--graph-aspect:277 / 204;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Find the amplitude, the period and the range of values of \(y\) corresponding to the interval. Hence find the values of \(a\) and \(b\).</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Write down the coordinates of the maximum point(s) and the minimum point(s) of the graph in the given interval.</div></li></ol>

The diagram shows part of the graph of \(y = a \cos bx\), where \(b > 0\), for \(0 \le x \le \frac{4\pi}{3}\).

Find the amplitude, the period and the range of values of \(y\) corresponding to the interval.\nHence find the values of \(a\) and \(b\).

Write down the coordinates of the maximum point(s) and the minimum point(s) of the graph in the given interval.

Math Question Bank