I have a follow-up question from a problem I posted earlier:

Approximate sin(61π/360):
So, sin(60π/360 + π/360)
= sin(π/6 + (π/360))

Now the equation for linear approximation is f(x+a)=f(a)+f'(a)(x-a). To get the right answer, that means plugging in the numbers like this: sin(61π/360) = sin(π/6) + cos(π/6)*(π/360), where x=π/6 and a=π/360.

So my question is, why does the (x-a) part in the equation equal (π/360)? If it's (x-a), shouldn't it be ((π/6)-(π/360)) which would equal (59π/360)?

<<Now the equation for linear approximation is f(x+a)=f(a)+f'(a)(x-a). To get the right answer, that means plugging in the numbers like this: sin(61π/360) = sin(π/6) + cos(π/6)*(π/360), where x=π/6 and a=π/360. >>

Nope. the Linear approximation is
f(x)=f(a)+f'(a)(x-a)
here, a= 60PI/360 (pi/60) , x=61PI/360, so x-a=PI/360

The notation (x-a) in the equation f(x+a) = f(a) + f'(a)(x-a) represents the difference between the value you want to approximate (x) and the reference point (a) around which you are performing the linear approximation.

In this case, you are trying to approximate sin(61π/360), which can be rewritten as sin(π/6 + (π/360)). In this expression, x is π/6 and a is π/360.

So, the difference between the value you want to approximate (π/6) and the reference point (π/360) is given by (x-a) which is (π/6 - π/360).

When you substitute these values into the formula, you get sin(61π/360) = sin(π/6) + cos(π/6) * (π/360) or f(x+a) = f(a) + f'(a)(x-a).

Therefore, the correct expression for (x-a) in this case is (π/6 - π/360), not ((π/6) - (π/360)).