# Use differentials (or equivalently, a linear approximation) to approximate sin(27 degrees) as follows: Let f(x) = sin(x) and find the equation of the tangent line to f(x) at a nice point near 27 degrees. Then use this to approximate sin(27 degrees).

## Well, I must say, using humor to teach calculus can be quite tricky! But fear not, I shall give it a shot.

Let's find a nice point near 27 degrees to approximate sin(27 degrees). How about we take the angle 30 degrees? It's friendly, even-tempered, and always pays its bills on time.

Now, we need to find the equation of the tangent line to the sine function at 30 degrees. It's like finding a line that shares the same hobbies as the sine function but is only interested in one point.

The slope of the tangent line is the derivative of the sine function, which is cos(x). Evaluating this at 30 degrees, we get cos(30 degrees) = √3/2. This line is all about square roots and fractions, always calculating things meticulously.

Since we have the slope (√3/2) and a point (30 degrees, sin(30 degrees) = 1/2), we can use the point-slope form of a linear equation to find the equation of the tangent line to the sine function near 27 degrees.

y - 1/2 = (√3/2)(x - 30)

Now, let's approximate sin(27 degrees) using this tangent line. We are looking for y when x is approximately 27 degrees, just like trying to predict someone's behavior based on their past actions.

Plugging x = 27 degrees into the equation, we get:

y - 1/2 = (√3/2)(27 - 30)
y - 1/2 = (√3/2)(-3)
y - 1/2 = -3√3/2
y ≈ -3√3/2 + 1/2

So, approximately, sin(27 degrees) is equal to -3√3/2 + 1/2. That's right, even in math, we can make approximate predictions, just like guessing what people might say next in a comedy show!

Remember, my friend, math can be a lot of fun when we add a pinch of humor to it. Keep calculating and keep laughing!

## To approximate sin(27 degrees) using differentials, we can start by noting that the derivative of f(x) = sin(x) is f'(x) = cos(x).

Next, we need to find the equation of the tangent line to f(x) at a nice point near 27 degrees. Let's choose x = 30 degrees (or π/6 radians) as our nice point since we know that sin(30 degrees) = 0.5.

To find the equation of the tangent line, we use the point-slope form: y - y1 = m(x - x1), where (x1, y1) is the point on the curve and m is the slope of the tangent line.

At x = 30 degrees, we have y1 = f(x1) = sin(30 degrees) = 0.5, and the slope is m = f'(x1) = cos(30 degrees) = √3/2.

Plugging these values into the point-slope form, we have:
y - 0.5 = (√3/2)(x - (π/6))

Simplifying, we get:
y - 0.5 = (√3/2)x - (√3/2)(π/6)

Rearranging, we obtain the equation of the tangent line:
y = (√3/2)x - (√3/2)(π/6) + 0.5

Now, we can use this equation to approximate sin(27 degrees).
Substitute x = 27 degrees (or π/6 radians) into the equation:
y ≈ (√3/2)(π/6) - (√3/2)(π/6) + 0.5
y ≈ 0.5

Therefore, sin(27 degrees) ≈ 0.5 is the approximation using differentials.

## To approximate sin(27 degrees) using differentials or linear approximation, we'll start by finding the equation of the tangent line to f(x) = sin(x) at a point near 27 degrees.

First, let's convert 27 degrees to radians since trigonometric functions usually work with radians. We know that π radians = 180 degrees. So, to convert 27 degrees to radians, we can use the fact that 1 degree = (π/180) radians.

27 degrees = (27 * π / 180) radians = (9π / 60) radians = (π / 20) radians.

Now, let's find the equation of the tangent line by using the derivative of f(x) = sin(x) at the point x = π / 20. The derivative of sin(x) is cos(x).

So, f'(x) = cos(x).

Now, let's evaluate f'(x) at x = π / 20 to find the slope of the tangent line:

f'(π / 20) = cos(π / 20).

Since cos(π / 20) is a decimal value, we'll use a calculator to find its approximate value:

cos(π / 20) ≈ 0.9848.

Now we have the slope of the tangent line at x = π / 20. We can use the point-slope form of a line to find the equation of the tangent line:

y - f(π / 20) = f'(π / 20) * (x - π / 20).

The point (x, y) is (π / 20, sin(π / 20)), and we already found that the slope is f'(π / 20) = 0.9848. Plugging these values into the equation, we have:

y - sin(π / 20) = 0.9848 * (x - π / 20).

Now, let's rearrange this equation to solve for y:

y = 0.9848 * (x - π / 20) + sin(π / 20).

This is the equation of the tangent line to f(x) = sin(x) at x = π / 20.

To approximate sin(27 degrees), we can substitute x = 27 degrees into the equation of the tangent line, since the tangent line is an approximation of the original function near the point x = π / 20.

Substituting x = 27 degrees = (27 * π / 180) radians, we get:

y = 0.9848 * ((27 * π / 180) - π / 20) + sin(π / 20).

Calculating this expression will give us an approximation for sin(27 degrees).

## f(x) = sin(x)

df = cos(x) dx
So, since sin(30°) = 1/2

sin 27° = 1/2 + (√3/2)(-π/60)