OMPT trigonometric functions practice: lesson and 10 problems

Answer: C — $\frac{\pi }{2}$

Only the coefficient of $x$ inside the sine affects the period. With $b=4$, the period is $\frac{2\pi }{b}=\frac{2\pi }{4}=\frac{\pi }{2}$. The amplitude $3$, phase shift, and vertical shift $2$ play no role here.

Answer: -0.87

The angle $\frac{5\pi }{6}$ sits in the second quadrant with reference angle $\frac{\pi }{6}$. Cosine is negative there, so $\cos \frac{5\pi }{6}=-\cos \frac{\pi }{6}=-\frac{\sqrt{3}}{2}\approx -0.87$.

Answer: B — Stretch vertically by 2, reflect in the x-axis, shift right $\frac{\pi }{3}$

The factor $-2$ does two things: the $2$ stretches the graph vertically and the minus sign flips it across the x-axis. Inside the argument, $x-\frac{\pi }{3}$ moves the graph right by $\frac{\pi }{3}$ — the inside shift always works opposite to its sign.

Answer: 2

I substitute directly: $\frac{\pi \cdot 9}{6}=\frac{3\pi }{2}$, and $\sin \frac{3\pi }{2}=-1$. So $d(9)=5+3(-1)=2$ metres — this moment happens to be dead low tide.

Answer: Step 3

Steps 1 and 2 are fine. In Step 3 the student added 4 to $-2$ and then slipped back to the original $1$ from Step 1 instead of the $2$ from Step 2. Adding 4 to the whole interval $[-2,2]$ gives $[2,6]$, so the range is $[2,6]$.

Answer: 0.6

A tangent of $\frac{3}{4}$ in size suggests a 3-4-5 triangle, so the hypotenuse is 5. In quadrant II sine is positive and cosine negative, which is exactly what makes the tangent negative. Therefore $\sin \theta =\frac{3}{5}=0.6$ (and $\cos \theta =-\frac{4}{5}$, consistent with $\tan \theta =-\frac{3}{4}$).

Answer: 6

Only the coefficient of $x$ inside the cosine matters, and here it is $b=\frac{\pi }{3}$. The period is $\frac{2\pi }{b}=2\pi \cdot \frac{3}{\pi }=6$. The $\pi$ cancels, which is why models with a $\pi$ inside the argument have whole-number periods. Check: $g(0)=5-1=4$, the maximum, and $g(6)=5\cos (2\pi )-1=4$ again, one full cycle later.

Answer: Step 2

Step 1 is right: a negative coefficient reflects the wave but the amplitude is its absolute value. Step 3 is right too: $1-3=-2$ and $1+3=4$. The slip is Step 2. The period is $\frac{2\pi }{b}$, and with $b=\frac{1}{2}$ that is $2\pi ÷\frac{1}{2}=4\pi$. A $b$ smaller than 1 stretches the wave, so the period must be longer than $2\pi$, not shorter. Check: $y(0)=1$ and $y(4\pi )=-3\sin (2\pi )+1=1$, one cycle later.

Answer: 8

I set $15-8\cos \left(\frac{\pi t}{12}\right)=19$, so $-8\cos \left(\frac{\pi t}{12}\right)=4$ and $\cos \left(\frac{\pi t}{12}\right)=-\frac{1}{2}$. The first non-negative angle with cosine $-\frac{1}{2}$ is $\frac{2\pi }{3}$ (second quadrant, reference angle $\frac{\pi }{3}$). So $\frac{\pi t}{12}=\frac{2\pi }{3}$ and $t=\frac{2\cdot 12}{3}=8$: 8 a.m. Check: $T(8)=15-8\cos \frac{2\pi }{3}=15-8\cdot \left(-\frac{1}{2}\right)=19$. It is the first time because the model starts at its minimum $T(0)=7$ and rises until $t=12$, and $8$ is before that.

Answer: C — 6

Divide by 2: $\cos (3x)=-\frac{1}{2}$. As $x$ runs through $[0,2\pi )$, the argument $3x$ runs through $[0,6\pi )$, which is three full cosine waves. In each wave, cosine equals $-\frac{1}{2}$ exactly twice, at $\frac{2\pi }{3}$ and $\frac{4\pi }{3}$ plus multiples of $2\pi$. Three waves times two crossings gives $6$ solutions: $x=\frac{2\pi }{9},\frac{4\pi }{9},\frac{8\pi }{9},\frac{10\pi }{9},\frac{14\pi }{9},\frac{16\pi }{9}$. Check one: $2\cos \left(3\cdot \frac{8\pi }{9}\right)=2\cos \frac{8\pi }{3}=2\cos \frac{2\pi }{3}=-1$.