Use a tangent line approximation at x=0 to estimate the value of sin(-0.1).

I got 0.1

you should have recalled that

sin (-x) = -sinx

so that sin(-.1) could not have been positive.

did you mean -.1 ?

To use a tangent line approximation, we can use the formula:

f(x) ≈ f(a) + f'(a)(x - a)

where f(x) is the function, f(a) is the value of the function at a, f'(a) is the derivative of the function at a, and (x - a) is the difference between the x-value where we want to approximate the function and the x-value a.

For this case, we want to estimate the value of sin(-0.1) using a tangent line approximation at x = 0. The sine function is given by f(x) = sin(x), and we want to find f(-0.1).

First, we need to find the derivative of the sine function. The derivative of sin(x) is cos(x), so f'(x) = cos(x).

Next, we set a = 0 and x = -0.1, as mentioned in the question.

Using the tangent line approximation formula, we have:

f(-0.1) ≈ f(0) + f'(0)(-0.1 - 0)

Plugging in the values, we have:

f(-0.1) ≈ sin(0) + cos(0)(-0.1)

Since sin(0) = 0 and cos(0) = 1, we can simplify:

f(-0.1) ≈ 0 + 1(-0.1)

f(-0.1) ≈ -0.1

Therefore, using the tangent line approximation at x = 0, the estimated value of sin(-0.1) is approximately -0.1.