Question

Sin(x)=x- x3/3!+x5/5!-x7/7!+....The percentage relative approximate error in using three terms to find sin(4.1) is ______percent.

Answer 1

Answer:

The answer is "239.62%".

Step-by-step explanation:

Given value:

$\bold{\sin(x)=x- \frac{x^3}{3!}+\frac{x^5}{5!} - \frac{x^7}{7!}+....}$

Solve the first three values:

$\sin(x)=x- \frac{x^3}{3!}+\frac{x^5}{5!}$

put the value of sin(4.1):

$\to \sin(4.1)=(4.1)- \frac{(4.1)^3}{3!}+\frac{(4.1)^5}{5!}$

                 $=(4.1)- \frac{(4.1)^3}{3!}+\frac{(4.1)^5}{5!} \\\\=(4.1)- \frac{68.92}{3 \times 2\times 1}+\frac{1158.56}{5\times 4 \times 3 \times 2 \times 1}$

                 $=(4.1)- 11.48+9.65\\\\= 2.27$

The actual value of $\sin(4.1) = -1.59$

Calculating the percentage relative approximate error value:

Formula:

$=\frac{actual \ value - \ approx \ value }{actual \ value} \times 100$

$=\frac{-1.59 - 2.27 }{-1.59} \times 100\\\\= \frac{-3.81}{-1.59} \times 100\\\\= \frac{3.81}{1.59} \times 100\\\\=239.62 \ \%$