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“Inverse Trigonometric Functions: Telescoping Series and Parameter Problems” is available to Standard subscribers.

Beyond the textbook. Add up long strings of arctangents by letting the terms cancel in pairs. Find the range of expressions built from inverse functions. Then decide for which values of a parameter an equation has a solution. It is at the level of JEE Advanced, A-Level Further Maths and IB Higher Level, with a full timed paper.

3Lessons12Problems2Python Labs1Question Set1Mock Test

Read the first lesson freeSums of arctangents and telescoping series

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Sums of arctangents and telescoping series: A long sum of inverse tangents can often be done in a line. Write each term as a difference $\tan^{-1}a_{r+1} - \tan^{-1}a_r$

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Range and graph questions: To find the range of an expression built from inverse functions, work out what it equals on each piece of its domain, because the branches often change the formula.

\tan^{-1}x + \tan^{-1}\dfrac{1}{x}

takes only the two values

\pm\dfrac{\pi}{2}

, and

\sin^{-1}\dfrac{2x}{1 + x^2}

rises to

\dfrac{\pi}{2}

x = 1

and then falls again. Count solutions by sliding a horizontal line across the graph. For a periodic composite, the period of a sum is the least common multiple of the periods.

Equations with a parameter: In an equation such as

\left(\sin^{-1}x\right)^2 + \left(\cos^{-1}x\right)^2 = a

, write

t = \sin^{-1}x

and

\cos^{-1}x = \dfrac{\pi}{2} - t

. The equation becomes a quadratic in

t

, where

t

must stay in

\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]

. Different

x

give different

t

, so the number of solutions

x

is the number of roots

t

in that interval. The vertex and the end values of the parabola give the range of the left side and the counts for each value of the parameter.