Sorry ok I was talking about on the unit circle how come the slope of the reference line (say if you were to angle pi/2, the reference line would be the positive y axis from y equals zero to y equals one, I think this line is called this, like if you were to draw pi/4 on the unit circle you would indicate this by drawing a line that is not on the circle arleady... this line is the line I'm referencing) how come the slope of this line or the first deriviative of it is equal to tangent
tan theta = (dsine(theta))/(dcos(theta))
???
tangent theta is equal to
the derivative of sine theta
over
the derivative of cosine theta?
I think you are greatly confused.
The slope of any line at an angle theta from the reference line is equal to tangent theta.
tangenttheta is NOT equal to derivative of sin over the derivative of cosine.
Why is slope equal to tangent?
slope is defined as deltaY/deltax where x is the reference point for the angle. but deltaY/deltaX is by definition the tangent of the central angle, theta.
The reference line you are referring to, starting from the origin and ending at a specific point on the unit circle, is called the terminal side of an angle.
To understand why the slope of this line or the first derivative (rate of change) of it is equal to the tangent of the angle, we need to consider the relationship between the coordinates of the point on the unit circle and the trigonometric functions sine and cosine.
Let's take an angle θ on the unit circle and consider the coordinates of the corresponding point (cos θ, sin θ). The x-coordinate (cos θ) represents the horizontal distance from the origin to the point on the circle, and the y-coordinate (sin θ) represents the vertical distance.
Now, if we want to find the slope of the reference line, we can use the concept of calculus and take the derivative of the y-coordinate with respect to the x-coordinate. In other words, we want to find dy/dx.
Using the trigonometric identity for tangent (tan θ = sin θ / cos θ), we can rewrite the y-coordinate (sin θ) in terms of the x-coordinate (cos θ) by multiplying both numerator and denominator of the tangent identity by cos θ:
sin θ = (tan θ) * (cos θ)
Now, taking the derivative of both sides with respect to x and using the chain rule, we get:
dy/dx = d/dx [(tan θ) * (cos θ)]
Applying the product rule on the right side, the derivative becomes:
dy/dx = (d/dx)(tan θ) * cos θ + tan θ * (d/dx)(cos θ)
The derivative of tangent with respect to x is zero because tangent is a constant function with respect to x. Thus, (d/dx)(tan θ) = 0. The derivative of cosine with respect to x is given by:
(d/dx)(cos θ) = -sin θ
Substituting these values back into the equation, we have:
dy/dx = 0 * cos θ + tan θ * (-sin θ) = -tan θ * sin θ
Simplifying further, we can write the result as:
dy/dx = -tan θ * sin θ = -(sin θ / cos θ) * sin θ = -sin^2 θ / cos θ
Recalling that the ratio sin θ / cos θ is equal to tan θ, we have:
dy/dx = -sin^2 θ / cos θ = -tan^2 θ
Therefore, the slope of the reference line or the first derivative of it is equal to -tan^2 θ.