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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: beezy4eva on February 28, 2008, 09:35:33 pm

Title: reciprocal circular functions/identities qus
Post by: beezy4eva on February 28, 2008, 09:35:33 pm: simplify the folowing expressions:
1.(sec θ -cos θ)(cosec θ -sin θ)

2.(1-cos²θ)(1+cot²θ)

3.sec²θ-cosec²θ
tan²θ-cot²θ

help is appreciated :) thanks
Title: Re: reciprocal circular functions/identities qus
Post by: Glockmeister on February 28, 2008, 09:49:44 pm: First One

This took longer than expected
Note, I'm letting the $\theta = x$ , cause I can't be bother typing it out

$(sec{x} - cos{x})(cosec{x} - sin{x})$
$= \frac{1}{cos{x}sin{x}} - \frac{sin{x}}{cos{x}} - \frac{cos{x}}{sin{x}} + sin{x}cos{x}$

Using $1 = sin^2{x} + cos^2{x}$ , we can substitute the $1$ on the numerator of the first fraction for $sin^2{x} + cos^2{x}$

$= \frac{sin^2{x} + cos^2{x}}{cos{x}sin{x}} - \frac{sin{x}}{cos{x}} - \frac{cos{x}}{sin{x}} + sin{x}cos{x}$

$= \frac{sin{x}}{cos{x}} + \frac{cos{x}}{sin{x}} - \frac{sin{x}}{cos{x}} - \frac{cos{x}}{sin{x}} + sin{x}cos{x}$

$= sin{x}cos{x}$

$= \frac{1}{2}sin{2x}$

Second one

$(1-\cos^2{x})(1+\cot^2{x})$
$= (sin^2{x})(cosec^2{x})$
$= (sin^2{x})(\frac{1}{sin^2{x}})$
$= 1$

Third One

This one was an interesting one. My initial thought was to use the definition of the reciprocals, but that got me nowhere.

$\frac{sec^2{\theta}-cosec^2{\theta}}{tan^2{\theta}-cot^2{\theta}}$

$Let sec^2{\theta} = tan^2{\theta} + 1 and Let cosec^2{\theta} = cot^2{\theta} + 1 (this is your identities)$

$= \frac{tan^2{\theta} + 1-(cot^2{\theta} + 1)}{tan^2{\theta}-cot^2{\theta}}$

$= \frac{tan^2{\theta}-(cot^2{\theta} )}{tan^2{\theta}-cot^2{\theta}}$

$= 1$