The curve y=a+bsin⁡cxy = a + b\sin cx is defined for 0≤x≤2π0 \le x \le 2\pi. The curve has a period of π\pi, and the minimum value of yy = −6-6. The curve also passes through the point (2π,−2)(2\pi, -2).

  1. ((i))

    Given that bb is a positive integer, find the values of aa, bb and cc.

  2. ((ii))

    Find the coordinates of the maximum points.

  3. ((iii))

    Sketch y=a+bsin⁡cxy = a + b\sin cx for 0≤x≤2π0 \le x \le 2\pi.

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Worked solution

9((i))

1. Use the given period to find $c$.

For y=a+bsin⁡cxy=a+b\sin cx, the period is
2πc.\frac{2\pi}{c}.
Given that the period is π\pi,
2πc=π\frac{2\pi}{c}=\pi
c=2.c=2.

2. Substitute the point $(2\pi,-2)$ to find $a$.

The curve passes through (2π,−2)(2\pi,-2), so
−2=a+bsin⁡(2⋅2π).-2=a+b\sin(2\cdot 2\pi).
Since sin⁡4π=0\sin 4\pi=0,
−2=a.-2=a.
Hence
a=−2.a=-2.

3. Use the minimum value to find $b$.

For y=a+bsin⁡cxy=a+b\sin cx with b>0b>0, the minimum value occurs when sin⁡cx=−1\sin cx=-1.

Thus the minimum value is
a−b=−6.a-b=-6.
Using a=−2a=-2,
−2−b=−6-2-b=-6
b=4.b=4.

Answer

a=−2, b=4, c=2a=-2,\ b=4,\ c=2

9((ii))

1. Use the maximum value of the sine function to form an equation for $x$.

From part (i),
y=−2+4sin⁡2x.y=-2+4\sin 2x.
The maximum occurs when
sin⁡2x=1.\sin 2x=1.
Then
ymax⁡=−2+4(1)=2.y_{\max}=-2+4(1)=2.

2. Solve for all $x$-coordinates of maximum points in the given interval.

Since 0≤x≤2π0\le x\le 2\pi,
0≤2x≤4π.0\le 2x\le 4\pi.
For sin⁡2x=1\sin 2x=1,
2x=π2, 5π2.2x=\frac{\pi}{2},\ \frac{5\pi}{2}.
Hence
x=π4, 5π4.x=\frac{\pi}{4},\ \frac{5\pi}{4}.

Answer

The maximum points are (π4,2)\left(\dfrac{\pi}{4},2\right) and (5π4,2)\left(\dfrac{5\pi}{4},2\right).

9((iii))

1. Write the equation to be sketched and identify the main levels of the graph.

From earlier parts,
a=−2,b=4,c=2.a=-2,\quad b=4,\quad c=2.
Hence
y=−2+4sin⁡2x.y=-2+4\sin 2x.
The midline is y=−2y=-2.

Since −1≤sin⁡2x≤1-1\le \sin 2x\le 1,
−2−4≤y≤−2+4,-2-4\le y\le -2+4,
so
−6≤y≤2.-6\le y\le 2.

2. Mark the key points of the sine curve in the given interval.

The maximum value is 22, occurring at
x=π4, 5π4.x=\frac{\pi}{4},\ \frac{5\pi}{4}.
So the maximum points are
(π4,2),(5π4,2).\left(\frac{\pi}{4},2\right),\quad \left(\frac{5\pi}{4},2\right).

The minimum value is −6-6, occurring when sin⁡2x=−1\sin 2x=-1:
2x=3π2, 7π22x=\frac{3\pi}{2},\ \frac{7\pi}{2}
x=3π4, 7π4.x=\frac{3\pi}{4},\ \frac{7\pi}{4}.
So the minimum points are
(3π4,−6),(7π4,−6).\left(\frac{3\pi}{4},-6\right),\quad \left(\frac{7\pi}{4},-6\right).

The curve passes through the midline y=−2y=-2 at
x=0, π2, π, 3π2, 2π.x=0,\ \frac{\pi}{2},\ \pi,\ \frac{3\pi}{2},\ 2\pi.

3. Draw a smooth sine curve through the key points with the correct shape and interval.

Draw axes for 0≤x≤2π0\le x\le 2\pi.

Show horizontal guide levels y=2y=2, y=−2y=-2 and y=−6y=-6.

Plot and join smoothly:
(0,−2)→(π4,2)→(π2,−2)→(3π4,−6)→(π,−2)\left(0,-2\right)\to \left(\frac{\pi}{4},2\right)\to \left(\frac{\pi}{2},-2\right)\to \left(\frac{3\pi}{4},-6\right)\to \left(\pi,-2\right)
then repeat the same pattern:
(π,−2)→(5π4,2)→(3π2,−2)→(7π4,−6)→(2π,−2).\left(\pi,-2\right)\to \left(\frac{5\pi}{4},2\right)\to \left(\frac{3\pi}{2},-2\right)\to \left(\frac{7\pi}{4},-6\right)\to \left(2\pi,-2\right).

Answer

Sketch of y=−2+4sin⁡2xy=-2+4\sin 2x for 0≤x≤2π0\le x\le 2\pi, with midline y=−2y=-2, maximum points (π4,2)\left(\dfrac{\pi}{4},2\right) and (5π4,2)\left(\dfrac{5\pi}{4},2\right), and minimum points (3π4,−6)\left(\dfrac{3\pi}{4},-6\right) and (7π4,−6)\left(\dfrac{7\pi}{4},-6\right).

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