Find the length of the tangent, the length of the normal

Find the length of the tangent, the length of the normal, the subtangent, and the subnormal of the curve $x=a(2\cos t-\cos 2t),\ y=a(2\sin t-\sin 2t)$.

Step 1: The four formulas

With $y_x'=\dot y/\dot x$ and $dx/dy=\dot x/\dot y$: \[T = y\sqrt{1+\left(\tfrac{dx}{dy}\right)^2}, \quad N = y\sqrt{1+(y_x')^2}, \quad S_T = \frac{y}{y_x'}, \quad S_N = y\,y_x'.\]

Step 2: Compute $\dot x,\dot y$ and simplify

\[ \dot x = 2a(\sin 2t-\sin t) = 4a\cos\frac{3t}2\sin\frac t2, \qquad \dot y = 2a(\cos t-\cos 2t) = 4a\sin\frac{3t}2\sin\frac t2, \tag{1} \] using $\sin2t-\sin t=2\cos\frac{3t}2\sin\frac t2$ and $\cos t-\cos2t=2\sin\frac{3t}2\sin\frac t2$.

Step 3: The slope

\[ y_x' = \frac{\dot y}{\dot x} = \frac{\sin\dfrac{3t}2}{\cos\dfrac{3t}2} = \tan\frac{3t}2, \qquad \frac{dx}{dy} = \cot\frac{3t}2. \tag{2} \]

Step 4: Simplify $y$

\[ y = a(2\sin t-\sin2t) = 2a\sin t(1-\cos t) = 2a\cdot2\sin\frac t2\cos\frac t2\cdot2\sin^2\frac t2 = 8a\sin^3\frac t2\cos\frac t2. \tag{3} \]

Step 5: Tangent length $T$ and normal length $N$

Using $1+\cot^2\frac{3t}2=1/\sin^2\frac{3t}2$, $1+\tan^2\frac{3t}2=1/\cos^2\frac{3t}2$, and $\sin\frac{3t}2=\sin\frac t2\!\left(3-4\sin^2\frac t2\right)$, $\cos\frac{3t}2=\cos\frac t2\!\left(4\cos^2\frac t2-3\right)$:

\[ T = \frac{y}{\left|\sin\dfrac{3t}2\right|} = \frac{8a\sin^2\dfrac t2\cos\dfrac t2}{\left|3-4\sin^2\dfrac t2\right|}, \qquad N = \frac{y}{\left|\cos\dfrac{3t}2\right|} = \frac{8a\sin^3\dfrac t2}{\left|4\cos^2\dfrac t2-3\right|}. \tag{4} \]

Step 6: Subtangent $S_T$ and subnormal $S_N$

\[ S_T = y\cot\frac{3t}2 = 8a\sin^3\frac t2\cos\frac t2\cdot\frac{\cos\dfrac{3t}2}{\sin\dfrac{3t}2}, \qquad S_N = y\tan\frac{3t}2 = 8a\sin^3\frac t2\cos\frac t2\cdot\frac{\sin\dfrac{3t}2}{\cos\dfrac{3t}2}. \tag{5} \]

Step 7: Verification

\[a=1,\ t=0.7\text{: all four closed forms match direct numerical differentiation of }x(t),y(t)\text{ to 10+ significant figures}\ \checkmark\]