Sinusoidal Function Transformations

Worked example

1. 01

   Factor 3 out of the input: $3\theta - 3\pi /4 = 3(\theta - \pi /4)$. So the same function can be written as:

   $$2.5\sin\left(3\theta-\frac{3\pi}{4}\right)-1=2.5\sin\left(3\left(\theta-\frac{\pi}{4}\right)\right)-1$$
2. 02

   Read the constants: amplitude 2.5, midline $y = -1$, period $2\pi /3$, and a phase shift of $\pi /4$ to the RIGHT.

3. 03

   A sine graph with $a > 0$ begins a cycle on its midline, going up. After the shift, that happens at $\theta = \pi /4$ instead of $\theta = 0$.

4. 04

   Step by a quarter period, $(2\pi /3) \div 4 = \pi /6$, starting at $\pi /4$: $\theta = \pi /4$, $5\pi /12$, $7\pi /12$, $3\pi /4$, $11\pi /12$. Follow the pattern midline → max → midline → min → midline.

   θ	π/4	5π/12	7π/12	3π/4	11π/12
   y	−1	1.5	−1	−3.5	−1

   

   One period of $y = 2.5 \sin (3\theta - 3\pi /4) - 1$ begins at $\theta = \pi /4$.

Worked example

1. 01

   Choose a key point to be the start of your cycle. Each choice leads to a different — but correct — equation.

2. 02

   Start at the minimum $(2, -3)$. A cosine that starts at a minimum has $a < 0$ (3.6A); shift it right 2: $h(x) = -6 \cos ((\pi /5)(x - 2)) + 3$.

3. 03

   Start at the maximum $(7, 9)$. A cosine with $a > 0$, shifted right 7: $h(x) = 6 \cos ((\pi /5)(x - 7)) + 3$.

4. 04

   Start on the midline going up, which happens halfway between the minimum and the maximum, at $x = 4.5$. A sine with $a > 0$, shifted right 4.5: $h(x) = 6 \sin ((\pi /5)(x - 4.5)) + 3$.

   

   Three different starting points, three correct equations for the same graph.

   All three equations produce exactly the same graph. On a free-response question, any correct equation earns credit, so choose whichever key point is easiest to read.

Worked example

1. 01

   Midline: $(4.6 + 1.0) \div 2 = 2.8$. Amplitude: $(4.6 - 1.0) \div 2 = 1.8$.

2. 02

   High tide to the next low tide is half a period: $9.2 - 3 = 6.2$ hours. So the period is 12.4 hours, and $b = 2\pi \div 12.4 = 5\pi /31 \approx 0.507$.

3. 03

   Start the cycle at high tide, $(3, 4.6)$: a cosine with $a > 0$, shifted right 3. $D(t) = 1.8 \cos ((5\pi /31)(t - 3)) + 2.8$.

4. 04

   Noon is $t = 12$: $D(12) = 1.8 \cos ((5\pi /31) \cdot 9) + 2.8 \approx 2.527$ meters. If you round $b$ to 0.507 first, you get $1.8 \cos(0.507 \cdot 9) + 2.8 \approx 2.532$ instead. Both round to about 2.53 m, but they already differ in the third decimal place — so keep $b$ exact (or store it in your calculator) until the final step.

5. 05

   Low tide was at $t = 9.2$, and the next high tide comes half a period later, at $t = 9.2 + 6.2 = 15.4$. Noon lies between them, so the water is rising.

   

   $D(t) = 1.8 \cos ((5\pi /31)(t - 3)) + 2.8$ over one day.