Find the equation of the tangents to the curve \( y=\cos x \) at \( x=-\pi, \frac{\pi}{2} \), and \( -\frac{3 \pi}{2} \). Graph the curve over the interval \( \left[-\frac{3 \pi}{2}, 2 \pi\right] \) together with their tangents. Label each curve and tangent. What is the equation of the tangent (III) to the curve at \( x=-\frac{3 \pi}{2} \) ? \( y=\square \)

To find the equations of the tangents to the curve \( y = \cos x \) at the specified points, we first need to calculate the derivative of the function, which gives us the slope of the tangent line at any point \( x \). 1. **Find the derivative of \( y = \cos x \)**: \[ \frac{dy}{dx} = -\sin x \] 2. **Evaluate the slope and the function value at each specified point**: - **At \( x = -\pi \)**: \[ y(-\pi) = \cos(-\pi) = -1 \] \[ \text{slope} = -\sin(-\pi) = 0 \] The equation of the tangent line is: \[ y - (-1) = 0(x - (-\pi)) \implies y = -1 \] - **At \( x = \frac{\pi}{2} \)**: \[ y\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0 \] \[ \text{slope} = -\sin\left(\frac{\pi}{2}\right) = -1 \] The equation of the tangent line is: \[ y - 0 = -1\left(x - \frac{\pi}{2}\right) \implies y = -x + \frac{\pi}{2} \] - **At \( x = -\frac{3\pi}{2} \)**: \[ y\left(-\frac{3\pi}{2}\right) = \cos\left(-\frac{3\pi}{2}\right) = 0 \] \[ \text{slope} = -\sin\left(-\frac{3\pi}{2}\right) = 1 \] The equation of the tangent line is: \[ y - 0 = 1\left(x + \frac{3\pi}{2}\right) \implies y = x + \frac{3\pi}{2} \] 3. **Summary of the tangent equations**: - Tangent at \( x = -\pi \): \( y = -1 \) - Tangent at \( x = \frac{\pi}{2} \): \( y = -x + \frac{\pi}{2} \) - Tangent at \( x = -\frac{3\pi}{2} \): \( y = x + \frac{3\pi}{2} \) 4. **Final answer for the tangent (III) at \( x = -\frac{3\pi}{2} \)**: \[ y = x + \frac{3\pi}{2} \] Thus, the equation of the tangent (III) to the curve at \( x = -\frac{3\pi}{2} \) is: \[ y = x + \frac{3\pi}{2} \]