Find dy/dx given x = cos t, y = t+ sin t
Find $\dfrac{dy}{dx}$ given $x = \cos t$, $y = t+\sin t$.
Explanation of Method
Half-angle identities collapse the ratio to a single cotangent.
Step 1: Differentiate with respect to $t$
\[ x'_{t} = -\sin t, \qquad y'_{t} = 1+\cos t. \]
Step 2: Form the ratio
\[ \frac{dy}{dx} = \frac{1+\cos t}{-\sin t}. \]
Step 3: Simplify with half-angle identities
Using $1+\cos t = 2\cos^{2}\dfrac{t}{2}$ and $\sin t = 2\sin\dfrac{t}{2}\cos\dfrac{t}{2}$, one factor of $\cos\dfrac{t}{2}$ cancels: \[\frac{dy}{dx} = \frac{2\cos^{2}\frac{t}{2}}{-2\sin\frac{t}{2}\cos\frac{t}{2}} = -\frac{\cos\frac{t}{2}}{\sin\frac{t}{2}} = -\cot\frac{t}{2}.\]
Step 4: Verification
\[t = \frac{\pi}{2}:\quad \frac{dy}{dx} = \frac{1+0}{-1} = -1; \qquad -\cot\frac{\pi}{4} = -1 \;\;\checkmark\]
\[t = \pi:\quad \frac{dy}{dx} = \frac{1-1}{0} \ \text{(indeterminate)}; \qquad -\cot\frac{\pi}{2} = 0 \;\;\checkmark\]
The simplified form settles the indeterminate case $t = \pi$, where numerator and denominator both vanish: the tangent there is horizontal.
Result
Final answer:
\[\boxed{\,\frac{dy}{dx} = -\dfrac{1+\cos t}{\sin t} = -\cot\dfrac{t}{2}, \qquad t \neq 2k\pi\,}\]
Remark
Remark. At $t = 2k\pi$ both $x'_{t}$ and $y'_{t}$ vanish simultaneously — a singular point of the parametrisation, where $-\cot\tfrac{t}{2}$ becomes infinite. This curve is a cycloid-like arch reflected; the singular points are its cusps.