To find a second slope $m_2$, such that the angle of $m_2$ is half that of $m$, follow these steps:

1. Understanding the problem:

Given:

• The slope of a line is $m$, where $m = \tan(\theta)$.
• You are asked to find $m_2$, the slope of a new line, where the angle $\theta_2$ of $m_2$ is half of $\theta$, i.e., $\theta_2 = \frac{\theta}{2}$.

2. Use the relationship between slope and angle:

• The slope of a line is related to its angle by $m = \tan(\theta)$.
• For the new line, we know the angle is $\theta_2 = \frac{\theta}{2}$, so its slope is: $m_2 = \tan\left(\frac{\theta}{2}\right)$

3. Express $m_2$ in terms of $m$:

Now, we need to express $\tan\left(\frac{\theta}{2}\right)$ in terms of $m$, which is $\tan(\theta)$. To do this, we use the half-angle identity for the tangent function: $\tan\left(\frac{\theta}{2}\right) = \frac{1 - \cos(\theta)}{\sin(\theta)}$ Alternatively, using the expression involving $m$, the half-angle formula for tangent can be written as: $\tan\left(\frac{\theta}{2}\right) = \frac{m}{1 + \sqrt{1 + m^2}}$ Thus, the slope of the new line $m_2$ is: $m_2 = \frac{m}{1 + \sqrt{1 + m^2}}$

Conclusion:

The slope $m_2$ of the new line, where the angle is half of the original line's angle, is given by: $m_2 = \frac{m}{1 + \sqrt{1 + m^2}}$

Would you like more details on the derivation or any clarification?

Relative Questions:

1. How do we derive the half-angle identity for tangent from basic trigonometric principles?
2. What is the geometrical interpretation of halving the angle of a line's slope?
3. Can we generalize this concept to other trigonometric functions, like sine or cosine?
4. What happens to $m_2$ as $m$ approaches very large values or zero?
5. How does this formula behave in the case of negative slopes?

Tip:

For small angles, the slope of a line approximates the angle itself (in radians), so for very small values of $\theta$, halving the angle roughly halves the slope too!