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Find an equation for a sinusoidal function that has period $2\pi$ , amplitude $1$ , and contains the point $\left(-\frac{\pi}{2},-1\right)$ . Write your answer in the form $f(x) = A\sin(Bx + C) + D$ , where $A$ , $B$ , $C$ , and $D$ are real numbers. `f(x)` =______

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Write equations of sine functions using properties

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Q. Find an equation for a sinusoidal function that has period

2\pi

, amplitude

1

, and contains the point

\left(-\frac{\pi}{2},-1\right)

. Write your answer in the form

f(x) = A\sin(Bx + C) + D

, where

A

,

B

,

C

, and

D

are real numbers. `f(x)` =______

Period formula: The period of a sinusoidal function is given by the formula $2\pi/B$ , where $B$ is the frequency. Since we are given that the period is $2\pi$ , we can find $B$ by setting $2\pi/B$ equal to $2\pi$ and solving for $B$ .
Finding B: Solving $2\pi/B = 2\pi$ gives us $B = 1$ because any number divided by itself is $1$ .
Amplitude: The amplitude of the function is given as $1$ , which means $A = 1$ .
Phase shift: To find the phase shift $C$ , we need to consider the point $(-\frac{\pi}{2}, -1)$ . Since the amplitude is $1$ , the sinusoidal function at its minimum value $(-1)$ occurs at $-\frac{\pi}{2}$ for the sine function. Normally, the sine function has its minimum value at $-\frac{3\pi}{2}$ , $-\frac{\pi}{2}$ , $\frac{5\pi}{2}$ , etc. Since $-\frac{\pi}{2}$ is already a standard point for the minimum of the sine function, we do not need a phase shift, so $C = 0$ .
Vertical shift: The vertical shift $D$ can be determined by looking at the value of the function at the given point. Since the amplitude is $1$ and the function value at $x = -\frac{\pi}{2}$ is $-1$ , which is the minimum value of the sine function, there is no vertical shift. Therefore, $D = 0$ .
Final equation: Putting all the values together, we get the equation of the sinusoidal function: $f(x) = A\sin(Bx + C) + D$ , which becomes $f(x) = 1\sin(1x + 0) + 0$ , or simply $f(x) = \sin(x)$ .

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