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<p>Let&nbsp;\(a\) and \(b\) be distinct positive real numbers.</p>
<ol class="abc-list"><li data-list-item-id="reciprocal-a" data-list-item-path="reciprocal-a"><div class="list-item-paragraph">Prove by contradiction that&nbsp;\(\frac{1}{a}+\frac{1}{b}&gt;\frac{4}{a+b}\).<span class="inline-marks">[5]</span></div></li><li data-list-item-id="reciprocal-b" data-list-item-path="reciprocal-b"><div class="list-item-paragraph">Give a counterexample to show that this inequality need not hold when&nbsp;\(a\) and \(b\) are nonzero but not both positive.<span class="inline-marks">[2]</span></div></li></ol>

Let \(a\) and \(b\) be distinct positive real numbers.

Prove by contradiction that \(\frac{1}{a}+\frac{1}{b}>\frac{4}{a+b}\).

Give a counterexample to show that this inequality need not hold when \(a\) and \(b\) are nonzero but not both positive.

Proof by Contradiction

Disproof by Counterexample

Reciprocal sum bounded by the arithmetic mean

Mathematical Induction

<p>Let&nbsp;\(x\) and \(y\) be positive real numbers such that&nbsp;\(x\ne y\).</p>
<ol class="abc-list"><li data-list-item-id="source-a" data-list-item-path="source-a"><div class="list-item-paragraph">Prove by contradiction that&nbsp;\(\frac{x}{y}+\frac{y}{x}&gt;2\).<span class="inline-marks">[5]</span></div></li><li data-list-item-id="source-b" data-list-item-path="source-b"><div class="list-item-paragraph">Give a counterexample to show that this inequality may not hold if&nbsp;\(x\) and \(y\) are not both positive.<span class="inline-marks">[2]</span></div></li></ol>

Let \(x\) and \(y\) be positive real numbers such that \(x\ne y\).

Prove by contradiction that \(\frac{x}{y}+\frac{y}{x}>2\).

Give a counterexample to show that this inequality may not hold if \(x\) and \(y\) are not both positive.

Reciprocal inequality for distinct positive numbers

<p>Prove by induction that&nbsp;\(n! \le n^n\) for&nbsp;\(n \in \mathbb{Z}^+\).<span class="inline-marks">[7]</span></p>

Prove by induction that \(n! \le n^n\) for \(n \in \mathbb{Z}^+\).

<ol class="abc-list"><li data-list-item-id="t06-p1-q06-a" data-list-item-path="t06-p1-q06-a"><div class="list-item-paragraph">Show that&nbsp;\(\cos2nx=\cos((2n+1)x)\cos x+\sin((2n+1)x)\sin x\).<span class="inline-marks">[2]</span></div></li><li data-list-item-id="t06-p1-q06-b" data-list-item-path="t06-p1-q06-b"><div class="list-item-paragraph">Hence prove by mathematical induction that&nbsp;\(\sin x+\sin3x+\sin5x+\cdots+\sin((2n-1)x)=\frac{1-\cos2nx}{2\sin x}\) for all \(n\in\mathbb Z^+\), \(\sin x\ne0\).<span class="inline-marks">[7]</span></div></li><li data-list-item-id="t06-p1-q06-c" data-list-item-path="t06-p1-q06-c"><div class="list-item-paragraph">Hence solve \(\sin x+\sin3x=\cos x\) for&nbsp;\(0&lt;x&lt;\pi\).<span class="inline-marks">[5]</span></div></li></ol>

Show that \(\cos2nx=\cos((2n+1)x)\cos x+\sin((2n+1)x)\sin x\).

Hence prove by mathematical induction that \(\sin x+\sin3x+\sin5x+\cdots+\sin((2n-1)x)=\frac{1-\cos2nx}{2\sin x}\) for all \(n\in\mathbb Z^+\), \(\sin x\ne0\).

Hence solve \(\sin x+\sin3x=\cos x\) for \(0<x<\pi\).

Compound and Double-Angle Identities

Trigonometric Equations and Identities - General

Proof by Mathematical Induction

Trigonometry

<ol class="abc-list"><li data-list-item-id="imported-item-block-0-item-0" data-list-item-path="imported-item-block-0-item-0"><div class="list-item-paragraph">Show that&nbsp;\(\sin 2 nx = \sin((2n + 1)x) \cos x - \cos((2n + 1)x) \sin x\).<span class="inline-marks">[2]</span></div></li><li data-list-item-id="imported-item-block-0-item-1" data-list-item-path="imported-item-block-0-item-1"><div class="list-item-paragraph"><p><strong>Hence</strong> prove, by induction, that \[\cos x + \cos 3x + \cos 5x + ... + \cos((2n - 1)x) = \frac{\sin 2nx}{2 \sin x},\] for all \(n \in \mathbb{Z}^+\), \(\sin x \neq 0\).</p><span class="inline-marks">[12]</span></div></li></ol>

Show that \(\sin 2 nx = \sin((2n + 1)x) \cos x - \cos((2n + 1)x) \sin x\).

<p>Prove, by induction, \((1+a)^n \ge 1+na\), where&nbsp;\(a \in \mathbb{R}^+\), for all positive integers of \(n\).<span class="inline-marks">[5]</span></p>

Prove, by induction, \((1+a)^n \ge 1+na\), where \(a \in \mathbb{R}^+\), for all positive integers of \(n\).

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