Math / Trigonometry

Question $y=A(\sin B x+C)+D \quad y=A(\cos B x+C)+D$ 24. $\begin{array}{l} y=4 \cos (4 x+\pi) \\ A=4 \\ B=4 \\ c=\pi \\ D=0 \\ \text { Amplitude } \\ 4 \\ \begin{array}{l} \text { Phase } \frac{-\pi}{4} \\ \text { Shift } \end{array} \\ \text { Period } \frac{2 \pi}{4}=\frac{\pi}{2} \\ \frac{2 \pi}{4}=\frac{\pi}{2} \end{array}$ $\square$ $3$

Studdy Solution

STEP 1

1. We are given a trigonometric function of the form $y = A \cos(Bx + C) + D$ .
2. We need to identify the amplitude, phase shift, and period of the function.
3. The given values are $A = 4$ , $B = 4$ , $C = \pi$ , and $D = 0$ .

STEP 2

1. Identify the amplitude.
2. Calculate the phase shift.
3. Determine the period of the function.

STEP 3

The amplitude of the function $y = A \cos(Bx + C) + D$ is the absolute value of $A$ . Given:
$A = 4$
Therefore, the amplitude is:
$\text{Amplitude} = |A| = 4$

STEP 4

The phase shift of the function is given by the formula:
$\text{Phase Shift} = -\frac{C}{B}$
Substitute the given values $C = \pi$ and $B = 4$ :
$\text{Phase Shift} = -\frac{\pi}{4}$

STEP 5

The period of the function is calculated using the formula:
$\text{Period} = \frac{2\pi}{B}$
Substitute the given value $B = 4$ :
$\text{Period} = \frac{2\pi}{4} = \frac{\pi}{2}$
The amplitude is $4$ , the phase shift is $-\frac{\pi}{4}$ , and the period is $\frac{\pi}{2}$ .

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