October 2021 Paper 3 Q14

OCR MEICurrent spec5 marksTrigonometry

14

The questions in this section refer to the article on the Insert. You should read the article before attempting the questions.

The relevant parts of the article “Adding arctangents” are reproduced below; the line numbers are those printed on the Insert.

Line 7
It can be shown that \(\arctan\left(\frac{1}{2}\right) + \arctan\left(\frac{1}{3}\right) = \arctan 1\).

Lines 22–23
The arctangent addition formula is a further generalization:
\(\arctan x + \arctan y = \arctan\left(\dfrac{x + y}{1 - xy}\right)\), as long as \(xy < 1\).

Lines 39–40
• For \(n\) a positive integer, \(\arctan\left(\dfrac{1}{n+1}\right) + \arctan\left(\dfrac{1}{n^2+n+1}\right) = \arctan\left(\dfrac{1}{n}\right)\); this follows directly from the arctan addition formula in line 23.

(a) Show that
\(\arctan\left(\dfrac{1}{n+1}\right) + \arctan\left(\dfrac{1}{n^2+n+1}\right) = \arctan\left(\dfrac{1}{n}\right) \Rightarrow \arctan\left(\dfrac{1}{2}\right) + \arctan\left(\dfrac{1}{3}\right) = \arctan 1.\) [1]
(b) Use the arctan addition formula in line 23 to show that
\(\arctan\left(\dfrac{1}{n+1}\right) + \arctan\left(\dfrac{1}{n^2+n+1}\right) = \arctan\left(\dfrac{1}{n}\right)\), as given in line 39. [4]