Problem 131

To find the maximum distance from point A $(\cos\theta, \sin\theta)$ to the line $3x+4y-4=0$, we start by calculating the distance from a point to a line. The formula for the distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

Applying this formula to our problem, we substitute $x_0 = \cos\theta$, $y_0 = \sin\theta$, $A = 3$, $B = 4$, and $C = -4$. This gives us:

$$\text{Distance} = \frac{|3\cos\theta + 4\sin\theta - 4|}{\sqrt{3^2 + 4^2}} = \frac{|3\cos\theta + 4\sin\theta - 4|}{\sqrt{9 + 16}} = \frac{|3\cos\theta + 4\sin\theta - 4|}{\sqrt{25}}$$

Simplifying the denominator, we get:

$$\text{Distance} = \frac{|3\cos\theta + 4\sin\theta - 4|}{5}$$

To maximize this distance, we look at the expression inside the absolute value. By using the trigonometric identity and considering the angle addition formula, we can rewrite $3\cos\theta + 4\sin\theta$ as $5\sin(\theta + \alpha)$, where $\tan\alpha = \frac{3}{4}$ and $\alpha$ is an acute angle. Thus, we have:

$$\text{Distance} = \frac{|5\sin(\theta + \alpha) - 4|}{5}$$

The maximum value of $|5\sin(\theta + \alpha) - 4|$ occurs when $\sin(\theta + \alpha) = -1$ because the absolute value of the expression is maximized when $\sin(\theta + \alpha)$ is at its minimum, which is $-1$. Substituting this into our equation, we get:

$$\text{Distance}_{\text{max}} = \frac{|-5 - 4|}{5} = \frac{9}{5}$$

Therefore, the maximum distance from point A $(\cos\theta, \sin\theta)$ to the line $3x+4y-4=0$ is $\boxed{\frac{9}{5}}$, which corresponds to option D.