Q.Find a sine function whose period is $\dfrac{2}{3}$.

$\sin(bx)$ has period $\dfrac{2\pi}{b}$; setting this equal to $\dfrac{2}{3}$ gives $b=3\pi$.

A general sine function $f(x)=\sin(bx)$ has period $\dfrac{2\pi}{b}$ (for $b>0$).

We need $\dfrac{2\pi}{b} = \dfrac{2}{3} \Rightarrow b = 2\pi \times \dfrac{3}{2} = 3\pi$.

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The general sine function $\sin(kx)$ has period $2\pi/k$; setting that equal to $2/3$ pins down $k$.

Step 1. For $f(x) = \sin(kx)$, the period is $T = \dfrac{2\pi}{k}$.

Step 2. We want $T = \dfrac{2}{3}$:

$\dfrac{2\pi}{k} = \dfrac{2}{3} \;\Rightarrow\; k = 2\pi \cdot \dfrac{3}{2} = 3\pi$

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$\tan(kx)$ has period $\dfrac{\pi}{k}$ since the tangent function itself has period $\pi$.

The function $\tan x$ has fundamental period $\pi$. For $\tan(kx)$, the period scales by $\dfrac{1}{k}$, giving period $\dfrac{\pi}{k}$.

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$\cos(k\theta)$ has period $\dfrac{2\pi}{k}$; choose $k$ so this equals $7$.

Step 1. The general cosine function $f(\theta)=\cos(k\theta)$ has period $\dfrac{2\pi}{k}$.

Step 2. We require period $=7$:

$\dfrac{2\pi}{k}=7 \;\Rightarrow\; k=\dfrac{2\pi}{7}$

Step 3. Substitute back:

$f(\theta)=\cos\!\left(\dfrac{2\pi}{7}\theta\right)$

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