ATAR Notes: Forum

HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Extension 1 => Topic started by: hinakamishiro on July 23, 2016, 03:34:20 pm

Title: Proving triple angle formula for tan
Post by: hinakamishiro on July 23, 2016, 03:34:20 pm: Hi there! I was wondering how to prove the triple angle formula for tan using the triple angle formulas for sin and cos? Thanks!
Title: Re: Proving triple angle formula for tan
Post by: RuiAce on July 23, 2016, 05:02:25 pm: Quote from: hinakamishiro on July 23, 2016, 03:34:20 pm
Hi there! I was wondering how to prove the triple angle formula for tan using the triple angle formulas for sin and cos? Thanks!
$\begin{align*}\tan 3x&=\frac{\sin 3x}{\cos 3x}\\ &= \frac{3\sin x - 4\sin^3 x}{4\cos^3 x - 3\cos x}\\ &= \frac{3-4\sin^2 x}{4\cos^2x - 3}\times \frac{\sin x}{\cos x}\\ &= \frac{3\sec^2 x - 4\tan^2 x}{4 - 3\sec^2 x} \times \tan x \tag{!}\\ &= \tan x \times \frac{3(1+\tan^2x)-4\tan^2 x}{4-3(1+\tan^2 x)}\\ &= \tan x \times \frac{3 - \tan^2 x}{1 - 3\tan^2 x}\\ &= \frac{3\tan x -\tan^3x}{1-3\tan^2 x}\end{align*}$
$\text{We made use of the ratio identity }\tan\theta=\frac{\sin \theta}{ \cos \theta}\\ \text{as well as the Pythagorean identity }1+\tan^2 \theta =\sec^2 \theta\\ \text{The trick in line (!) was to divide top and bottom by } \cos^2 x$
Title: Re: Proving triple angle formula for tan
Post by: hinakamishiro on July 24, 2016, 12:18:16 pm: Quote from: RuiAce on July 23, 2016, 05:02:25 pm
$\begin{align*}\tan 3x&=\frac{\sin 3x}{\cos 3x}\\ &= \frac{3\sin x - 4\sin^3 x}{4\cos^3 x - 3\cos x}\\ &= \frac{3-4\sin^2 x}{4\cos^2x - 3}\times \frac{\sin x}{\cos x}\\ &= \frac{3\sec^2 x - 4\tan^2 x}{4 - 3\sec^2 x} \times \tan x \tag{!}\\ &= \tan x \times \frac{3(1+\tan^2x)-4\tan^2 x}{4-3(1+\tan^2 x)}\\ &= \tan x \times \frac{3 - \tan^2 x}{1 - 3\tan^2 x}\\ &= \frac{3\tan x -\tan^3x}{1-3\tan^2 x}\end{align*}$
$\text{We made use of the ratio identity }\tan\theta=\frac{\sin \theta}{ \cos \theta}\\ \text{as well as the Pythagorean identity }1+\tan^2 \theta =\sec^2 \theta\\ \text{The trick in line (!) was to divide top and bottom by } \cos^2 x$

Thanks! I would have never thought of that line.