Use the double-angle identity before solving · source reference

Complete question

Show that $\tan x+\cot x>4$ for $0<x<\pi/2$ simplifies to $0<\sin2x<1/2$. [4] Hence solve it exactly in terms of pi. [2]

[2018/DHS/Prelim/I/4]

Printed working

$$\begin{aligned} \tan x+\cot x&>4\\ \tan x+\cot x-4&>0\\ \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}-4&>0\\ \frac{\sin^2x+\cos^2x-4\sin x\cos x}{\sin x\cos x}&>0\\ \frac{1-2\sin2x}{\tfrac12\sin2x}&>0\\ \frac{1-2\sin2x}{\sin2x}&>0.\qquad(*) \end{aligned}$$

The printed sign line has critical values 0 and $\tfrac12$ in the variable $\sin2x$ and signs negative, positive, negative. Therefore $0<\sin2x<\tfrac12$ (shown).

Alternative method for (*)

Let $y=\sin2x$:

$$\frac{1-2\sin2x}{\sin2x}>0\Rightarrow\frac{1-2y}{y}>0\Rightarrow(1-2y)y>0.$$

The downward quadratic has roots 0 and $\tfrac12$, so $0<y<\tfrac12$, hence $0<\sin2x<\tfrac12$ (shown).

From the sine sketch on $0<x<\pi/2$, the required intervals are

$$0<x<\frac{\pi}{12}\quad\text{or}\quad\frac{5\pi}{12}<x<\frac\pi2.$$

Teacher margin notes

Sketch the graph of $y=\sin2x$, $0<x<\tfrac12\pi$, and the lines $y=0$ and $y=\tfrac12$. Find the ranges of values of $x$ for which the curve $y=\sin2x$ lies between the two lines.

Source diagrams

Provenance

Question page references: 1. Solution page references: 7. The original compilation header and printed page numbers are preserved in the edition notes. Original PDFs and page images remain only in the private QA archive.