   Chapter 11, Problem 7P

Chapter
Section
Textbook Problem

# (a) Show that for xy ≠ −1, arctan   x − arctan   y = arctan x − y 1 + x y if the left side lies between −π/2 and π/2. (b) Show that arctan 120 119 − arctan 1 239 = π / 4 . (c) Deduce the following formula of John Machin (1680–1751): 4 arctan 1 5 − arctan 1 239 = π 4 (d) Use the Maclaurin series for arctan to show that 0.1973955597 < arctan 1 5 < 0.1973955616 (e) Show that 0.004184075 < arctan 1 239 < 0.004184077 (f) Deduce that, correct to seven decimal places, π ≈ 3.1415927. Machin used this method in 1706 to find π correct to 100 decimal places. Recently, with the aid of computers, the value of π has been computed to increasingly greater accuracy. In 2013 Shigeru Kondo and Alexander Yee computed the value of π to more than 12 trillion decimal places!

(a)

To determine

To show: arctanxarctany=arctanxy1+xy for xy1 .

Explanation

Given:

arctanx and arctany lies between π2 and π2 .

Result used:

The formula of tan(ab) is tanatanb1+tanatanb (1)

Proof:

Consider a=arctanx and b=arctany .

Substitute a and b values in equation (1),

tan(ab)=tan(arctanx)tan(arctany)1+tan(arctanx)tan(arctany)

tan(ab)=xy1+xy

(b)

To determine

To show: arctan120119arctan1239=π4 .

(c)

To determine

To deduce: 4arctan15arctan1239=π4 .

(d)

To determine

To show: 0.197395597<arctan15<0.1973955616 .

(e)

To determine

To show: 0.004184075<arctan1239<0.004184077 .

(f)

To determine

To deduce: Correct to seven decimal places π3.1415927

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