Find the Period of Trigonometric Functions - Grade 12 Math

Step 1: Identify full cycles on the graph.
Locate zeros that delimit a whole cycle or an integer number of cycles. In this example, observing the interval from $x = 0$ to $x = 1$, we can count exactly two full wave cycles.

Step 2: Calculate the experimental period.
Because two cycles fit into a horizontal distance of $1 - 0 = 1$, the period $P$ (the length of one cycle) is: $$P = \frac{1 - 0}{2} = \frac{1}{2}$$

Step 3: Solve for parameter $b$.
The formula for the period of a sine wave is $P = \frac{2\pi}{b}$. Equate this to our found period: $$\frac{2\pi}{b} = \frac{1}{2}$$ Cross-multiply and solve for $b$: $$b = 4\pi$$

Step 1: Find the distance of one full cycle.
Observe the zero crossings. The graph starts a cycle at $x = -\frac{\pi}{4}$, goes through a peak and a trough, and completes one full cycle at the zero crossing $x = \frac{\pi}{4}$.

Step 2: Calculate the period.
The period $P$ is the horizontal distance between these two points: $$P = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{2\pi}{4} = \frac{\pi}{2}$$

Step 3: Solve for parameter $b$.
Equate the graphical period to the algebraic formula $P = \frac{2\pi}{b}$: $$\frac{\pi}{2} = \frac{2\pi}{b}$$ Multiply both sides by $2b$: $$\pi \cdot b = 4\pi \implies \mathbf{b = 4}$$

Step 1: Identify key points to calculate distance.
The graph provides tick marks on the x-axis, but no direct coordinates for the peaks. However, we can find the exact locations of the zeros (x-intercepts) by taking the midpoint between the given tick marks.

Zero on the left (midpoint between $-\pi/4$ and $-\pi/8$): $$x_1 = \frac{-\pi/4 + (-\pi/8)}{2} = \frac{-3\pi/8}{2} = -\frac{3\pi}{16}$$

Zero on the right (midpoint between $0$ and $\pi/8$): $$x_2 = \frac{0 + \pi/8}{2} = \frac{\pi}{16}$$

Step 2: Calculate the period.
The distance between two consecutive zeros represents exactly *half* a cycle (half a period). $$\text{Half Period} = \frac{\pi}{16} - \left(-\frac{3\pi}{16}\right) = \frac{4\pi}{16} = \frac{\pi}{4}$$ Therefore, the full period $P$ is twice this distance: $$P = 2 \times \frac{\pi}{4} = \frac{\pi}{2}$$

Step 3: Solve for $b$.
$$\frac{2\pi}{b} = \frac{\pi}{2} \implies \mathbf{b = 4}$$

Step 1: Understand the relationship between extremes.
The horizontal distance along the x-axis between a consecutive maximum (Point A at $x = \frac{3\pi}{6}$) and minimum (Point B at $x = \frac{7\pi}{6}$) is exactly equal to half a period.

Step 2: Calculate the period.
First, find the distance between A and B: $$\text{Distance} = \frac{7\pi}{6} - \frac{3\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}$$ The full period $P$ is twice this distance: $$P = 2 \times \frac{2\pi}{3} = \frac{4\pi}{3}$$

Step 3: Solve for $b$.
Equate the algebraic formula to the numerical period: $$\frac{2\pi}{b} = \frac{4\pi}{3}$$ Cross multiply: $6\pi = 4\pi b \implies \mathbf{b = \frac{6\pi}{4\pi} = \frac{3}{2}}$

Step 1: Find the period directly.
The horizontal distance between two consecutive minimum points represents exactly one full cycle. Therefore, the period $P$ is simply the difference between their x-coordinates: $$P = 0.1 - (-0.3) = 0.4$$

Step 2: Solve for $b$.
Using the period formula $P = \frac{2\pi}{b}$: $$\frac{2\pi}{b} = 0.4$$ $$0.4b = 2\pi \implies b = \frac{2\pi}{0.4} = \frac{20\pi}{4} = \mathbf{5\pi}$$

1. For $y = \sin(x)\cos(x) - 3$:
   Use the double-angle identity: $\sin(2x) = 2 \sin(x) \cos(x)$. Dividing by 2 yields $\sin(x)\cos(x) = \frac{1}{2}\sin(2x)$.
   Rewrite the function: $$y = \frac{1}{2} \sin(2x) - 3$$ The internal multiplier is $b=2$. Period: $$\mathbf{P = \frac{2\pi}{2} = \pi}$$
2. For $y = 2 + 5\cos^2(x)$:
   Use the power-reducing identity: $\cos^2(x) = \frac{1}{2}(\cos(2x) + 1)$.
   Rewrite the function: $$y = 2 + 5\left(\frac{1}{2}(\cos(2x)+1)\right) = 2 + \frac{5}{2}\cos(2x) + \frac{5}{2} = \frac{5}{2}\cos(2x) + \frac{9}{2}$$ The internal multiplier is $b=2$. Period: $$\mathbf{P = \frac{2\pi}{2} = \pi}$$
3. For $y = \cos(x) + \sin(x)$:
   Rewrite this as a single phase-shifted sine wave. Multiply and divide by $\sqrt{1^2 + 1^2} = \sqrt{2}$: $$y = \sqrt{2}\left(\frac{1}{\sqrt{2}}\cos(x) + \frac{1}{\sqrt{2}}\sin(x)\right)$$ $$y = \sqrt{2}\left(\sin(\frac{\pi}{4})\cos(x) + \cos(\frac{\pi}{4})\sin(x)\right)$$ Using the angle addition formula $\sin(A+B) = \sin A\cos B + \cos A\sin B$: $$y = \sqrt{2}\sin\left(x + \frac{\pi}{4}\right)$$ The internal multiplier is $b=1$. Period: $$\mathbf{P = \frac{2\pi}{1} = 2\pi}$$

By definition, if $p$ is the period of function $f$, then: $$f(x+p) = f(x) \quad \text{for all } x$$

Let's perform a substitution where $x = kX$, giving: $$f(kX+p) = f(kX)$$

Now, factor out the constant $k$ from the left side's argument: $$f\left(k\left(X+\frac{p}{k}\right)\right) = f(kX)$$

Because we defined $h(x) = f(kx)$, we can substitute $h$ into our equation: $$h\left(X+\frac{p}{k}\right) = h(X)$$

This statement defines a new periodic function $h$. It indicates that shifting the input by $\frac{p}{k}$ yields the original function value. Therefore, $h(x) = f(kx)$ is periodic with a period of: $$\mathbf{\frac{p}{k}}$$

Step 1: Find the period of each individual term.
For the first term, $y_1 = 3\sin(4x)$, the parameter $b = 4$. Its period is: $$P_1 = \frac{2\pi}{4} = \frac{\pi}{2}$$ For the second term, $y_2 = 2\cos(6x)$, the parameter $b = 6$. Its period is: $$P_2 = \frac{2\pi}{6} = \frac{\pi}{3}$$

Step 2: Find the Least Common Multiple (LCM).
The period of a sum of periodic functions is the Least Common Multiple (LCM) of their individual periods. We need to find the LCM of $\frac{\pi}{2}$ and $\frac{\pi}{3}$.
To find the LCM of fractions, express them with a common denominator. The least common denominator for $2$ and $3$ is $6$: $$P_1 = \frac{3\pi}{6}$$ $$P_2 = \frac{2\pi}{6}$$ The LCM of the numerators ($3\pi$ and $2\pi$) is $6\pi$. Divide this by the common denominator: $$\text{LCM} = \frac{6\pi}{6} = \mathbf{\pi}$$

Step 1: Find the period of the base function.
Ignore the absolute value first. The base function is $f(x) = \sin(2x)$. The period of this sine wave is: $$P_{\text{base}} = \frac{2\pi}{2} = \pi$$

Step 2: Analyze the effect of the absolute value.
A standard sine wave oscillates above and below the x-axis, completing a positive hump and a negative trough over its full period of $\pi$.
Taking the absolute value forces the negative trough to reflect upwards above the x-axis, becoming an identical positive hump. Because the first half of the cycle and the second half of the cycle now look perfectly identical, the pattern repeats twice as fast.
Therefore, applying an absolute value to a sine or cosine wave halves its fundamental period.

Step 3: Calculate the final period.
$$P = \frac{P_{\text{base}}}{2} = \mathbf{\frac{\pi}{2}}$$

Step 1: Recall the standard tangent period.
Unlike sine and cosine functions whose fundamental period is $2\pi$, the fundamental base period of the tangent function $y = \tan(x)$ is $\pi$. Therefore, the formula for the period of $y = a\tan(bx - c)$ is $P = \frac{\pi}{|b|}$.

Step 2: Identify the $b$ parameter.
Looking at the argument of the tangent function $\left(\frac{\pi x}{3} - \frac{\pi}{4}\right)$, the multiplier on the $x$ variable is $b = \frac{\pi}{3}$. (Note: the $-\frac{\pi}{4}$ is a phase shift and does not affect the period length).

Step 3: Calculate the period.
$$P = \frac{\pi}{\frac{\pi}{3}} = \pi \times \frac{3}{\pi} = \mathbf{3}$$

The period is exactly $3$ units long.