HSF.TF.B.7: Solve trigonometric equations — Practice

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A Ferris wheel model is $h(t) = 25\sin\!\left(\dfrac{\pi t}{4}\right) + 27$ .
You want to find when $h = 40$ .

Which equation correctly isolates the trigonometric expression as the
first step toward solving for $t$ ?

$\sin\!\left(\dfrac{\pi t}{4}\right) = 40$

$\sin\!\left(\dfrac{\pi t}{4}\right) = 0.52$

$\dfrac{\pi t}{4} = \arcsin(40)$

$25\sin\!\left(\dfrac{\pi t}{4}\right) = 40$

For the equation $\sin(\theta) = 0.6$ , how many solutions exist in
the interval $[0, 2\pi)$ ?

Enter a whole number.

A tidal model is $h(t) = 3.2\cos\!\left(\dfrac{\pi t}{6.25}\right) + 4.0$ ,
where $t$ is hours after midnight.

After isolating the cosine expression to solve $h = 6$ , the equation
becomes $\cos\!\left(\dfrac{\pi t}{6.25}\right) = {?}$ .

Enter the decimal value on the right side.

A temperature model gives $\sin\!\left(\dfrac{\pi t}{12}\right) = 0.5$
after isolation, where $t$ is hours after midnight.

A student applies arcsin and gets $\dfrac{\pi t}{12} \approx 0.5236$ .
They then report $t \approx 0.5236$ as the answer.

What error did the student make?

They used the wrong inverse function — they should have used arccos.

They forgot to solve for $t$ : the angle $\dfrac{\pi t}{12} \approx 0.5236$ ,
so $t = \dfrac{12(0.5236)}{\pi} \approx 2.0$ .

They found the principal value correctly; no error was made.

They should have added $2\pi$ to 0.5236 before solving for $t$ .

A student evaluates $\arcsin(0.52)$ on their calculator and gets $31.3$ .
They use this value as the angle in radians to continue solving.

What most likely went wrong?

The student should have used arccos instead of arcsin.

The calculator was in degree mode; the correct radian value is approximately 0.5464.

The value 0.52 is outside the domain of arcsin, so the calculation failed.

The student should have squared 0.52 before applying arcsin.

After isolating, you have $\sin\!\left(\dfrac{\pi t}{4}\right) = 0.52$ .

Using a calculator in radian mode, $\arcsin(0.52) \approx 0.5464$ .

Which expression gives the principal solution for $t$ ?

$t = 0.5464$

$t = \dfrac{4 \times 0.5464}{\pi}$

$t = \dfrac{\pi \times 0.5464}{4}$

$t = \arcsin(0.52) \times 4$

For $\sin\!\left(\dfrac{\pi t}{4}\right) = 0.52$ , the principal solution
is $t_1 \approx 0.696$ .

The period of the model is 8 minutes. Which expression gives the
second solution in the interval $[0, 8)$ ?

$t_2 = 8 - 0.696 \approx 7.304$

$t_2 = 2\pi - 0.696 \approx 5.587$

$t_2 = \dfrac{4(\pi - 0.5464)}{\pi} \approx 3.304$

$t_2 = 0.696 + 8 \approx 8.696$

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