Sine

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ACT Math › Sine

Questions 1 - 10

A sine function has a period of , a -intercept of , an amplitude of and no phase shift. These describe which of these equations?

$f(x) = 2sin(\pi x)$

$f(x) = sin(\pi x)$

$f(x) = 2sin(2\pi x)$

$f(x) = 2sin(\pi x) + 1$

Explanation

Looking at this form of a sine function:

We can draw the following conclusions:

because the amplitude is specified as .
$b = \pi$ because of the specified period of since $\frac{2\pi}{\pi} = 2$ .
because the problem specifies there is no phase shift.
because the -intercept of a sine function with no phase shift is .

Bearing these in mind, $f(x) = 2sin(\pi x)$ is the only function that fits all four of those.

What is the period of the function $f(t) = -2\sin(4t)$ ?

$\frac{\pi}{2}$

$-\pi$

$2\pi$

Explanation

To find the period of Sine and Cosine functions you use the formula:
$\frac{2\pi}{\left | b\right |}$ where comes from $f(t) = a\sin(bt)$ . Looking at our formula you see b is 4 so
$\frac{2\pi}{\left | 4\right |}=\frac{\pi}{2}$

A sine function has a period of , a -intercept of , an amplitude of and no phase shift. These describe which of these equations?

$f(x) = 2sin(\pi x)$

$f(x) = sin(\pi x)$

$f(x) = 2sin(2\pi x)$

$f(x) = 2sin(\pi x) + 1$

Explanation

Looking at this form of a sine function:

We can draw the following conclusions:

because the amplitude is specified as .
$b = \pi$ because of the specified period of since $\frac{2\pi}{\pi} = 2$ .
because the problem specifies there is no phase shift.
because the -intercept of a sine function with no phase shift is .

Bearing these in mind, $f(x) = 2sin(\pi x)$ is the only function that fits all four of those.

What is the period of the function $f(t) = -2\sin(4t)$ ?

$\frac{\pi}{2}$

$-\pi$

$2\pi$

Explanation

Solve for :

if $\frac{-\pi}{2} \le 3x \le \frac{\pi}{2}$

$\frac{\pi}{18}$

$\frac{\pi}{6}$

$\frac{3\pi}{2}$

$\frac{\pi}{2}$

$3\pi$

Explanation

Recall that the standard triangle, in radians, looks like:

Rt1

Since $sin(x) = \frac{opposite}{hypotenuse}$ , you can tell that $sin(\frac{\pi}{6}) = \frac{1}{2}$ .

Therefore, you can say that must equal $\frac{\pi}{6}$ :

$3x = \frac{\pi}{6}$

Solving for , you get:

$x=\frac{\pi}{18}$

What is the domain of the given trigonometric function:

$z(t) = 2\sin(-2t)$

$(-\infty,\infty)$

$[-\pi,\pi]$

$[0,\infty]$

Explanation

For both Sine and Cosine, since there are no asymptotes like Tangent and Cotangent functions, the function can take in any value for . Thus the domain is:

$(-\infty,\infty)$

Solve for :

if $\frac{-\pi}{2} \le 3x \le \frac{\pi}{2}$

$\frac{\pi}{18}$

$\frac{\pi}{6}$

$\frac{3\pi}{2}$

$\frac{\pi}{2}$

$3\pi$

Explanation

Recall that the standard triangle, in radians, looks like:

Rt1

Since $sin(x) = \frac{opposite}{hypotenuse}$ , you can tell that $sin(\frac{\pi}{6}) = \frac{1}{2}$ .

Therefore, you can say that must equal $\frac{\pi}{6}$ :

$3x = \frac{\pi}{6}$

Solving for , you get:

$x=\frac{\pi}{18}$

What is the domain of the given trigonometric function:

$z(t) = 2\sin(-2t)$

$(-\infty,\infty)$

$[-\pi,\pi]$

$[0,\infty]$

Explanation

For both Sine and Cosine, since there are no asymptotes like Tangent and Cotangent functions, the function can take in any value for . Thus the domain is:

$(-\infty,\infty)$

Which of the following represents a sine wave with a range of $-5 \leq y \leq 5$ ?

Explanation

The range of a sine wave is altered by the coefficient placed in front of the base equation. So, if you have , this means that the highest point on the wave will be at and the lowest at . However, if you then begin to shift the equation vertically by adding values, as in, , then you need to account for said shift. This would make the minimum value to be and the maximum value to be . For our question, then, it is fine to use . The for the function parameter only alters the period of the equation, making its waves "thinner."

Which of the following represents a sine wave with a range of $-5 \leq y \leq 5$ ?

Explanation

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