How to Find the Tangent of a Point on the Unit Circle

Given a point on the unit circle, say $(\cos(\theta), \sin(\theta))$, we need to find the tangent line at this point.

Step 1: The equation of the unit circle is $x^2 + y^2 = 1$.

Step 2: To find the slope of the tangent, we differentiate implicitly with respect to $x$:

$2x + 2y \frac{dy}{dx} = 0$

Step 3: Rearrange to solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = -\frac{x}{y}$

Step 4: Substitute $x = \cos(\theta)$ and $y = \sin(\theta)$:

$\frac{dy}{dx} = -\frac{\cos(\theta)}{\sin(\theta)} = -\cot(\theta)$

So the slope of the tangent line at $(\cos(\theta), \sin(\theta))$ is $-\cot(\theta)$.

Consider the point on the unit circle given by $(cos( heta), sin( heta))$.

Step 1: The unit circle equation is $x^2 + y^2 = 1$.

Step 2: Differentiate implicitly with respect to $x$:

$frac{d}{dx}(x^2 + y^2) = frac{d}{dx}(1)$

Step 3: Apply the chain rule:

$2x + 2y frac{dy}{dx} = 0$

Step 4: Solve for $frac{dy}{dx}$:

$frac{dy}{dx} = -frac{x}{y}$

Step 5: Substitute $x = cos( heta)$ and $y = sin( heta)$:

$frac{dy}{dx} = -frac{cos( heta)}{sin( heta)} = -cot( heta)$

The tangent line at $(cos( heta), sin( heta))$ has a slope of $-cot( heta)$.

For point $(cos( heta), sin( heta))$ on the unit circle:

1. The unit circle: $x^2 + y^2 = 1$.

2. Differentiate implicitly:

$2x + 2y frac{dy}{dx} = 0$

3. Solve for $frac{dy}{dx}$:

$frac{dy}{dx} = -frac{x}{y}$

4. Substitute $x = cos( heta)$, $y = sin( heta)$:

$frac{dy}{dx} = -cot( heta)$

The slope of the tangent is $-cot( heta)$.

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