The ridiculous approximation for
using a degree 3 Maclaurin polynomial is
.
The Maclaurin series expansion for
up to the degree 3 is given by:
![\[ \sin(x) \approx x - (x^3)/(6) \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/hyl5g549xohsota97jiqbdsvgcz4918gkp.png)
To approximate
using this polynomial, we substitute
into the expression:
![\[ \sin(100) \approx 100 - (100^3)/(6) \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/kcl52ayfn6lovp75io91zw2tfla9avyep5.png)
Calculating this, we get:
![\[ \sin(100) \approx 100 - (1000000)/(6) \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/menbjm5743w7i1q140a4t9ivmlklppc05u.png)
Simplifying the fraction yields:
![\[ \sin(100) \approx 100 - 166666.67 \]](https://img.qammunity.org/2024/formulas/mathematics/high-school/i61fs9cs73ac4chal17u6h0w2nrnoj2fxv.png)
The result is approximately
. However, given that the range of
is between -1 and 1, this approximation is indeed ridiculous. It violates the bounds imposed by the sine function, emphasizing the limitations of polynomial approximations, especially when extrapolating far beyond the region where the polynomial is derived.
In conclusion, the approximation
is absurd, showcasing the inaccuracy of polynomial approximations beyond their valid ranges.
The question probable maybe:
Approximate sin(100) using the degree 3 Maclaurin polynomial for sin(x) Notice how ridiculous this approximation is, given that - 1 <= sin(x) <= 1
sin(100)≈ ?/6 (your answer ? is an integer).