Trig university question

**smplease** · March 8th 2008, 06:56 PM

Hey I'm currently doing a math bridging course in university and cant seem to remember a lot of the math i learnt in school

the question is

prove that tan^2 theta - sin^2 theta = tan^2 theta*sin^2 theta

sorry i dont have math equation writing program..it looks a bit weird in words.. any help would be greatly appreciated

thanks again

**mr fantastic** · March 8th 2008, 07:45 PM

Originally Posted by smplease

Hey I'm currently doing a math bridging course in university and cant seem to remember a lot of the math i learnt in school

the question is

prove that tan^2 theta - sin^2 theta = tan^2 theta*sin^2 theta

sorry i dont have math equation writing program..it looks a bit weird in words.. any help would be greatly appreciated

thanks again

Left hand side = $\tan^2 \theta - \sin^2 \theta$

$= \frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta$

$=\frac{\sin^2 \theta}{\cos^2 \theta} - \frac{\sin^2 \theta \, \cos^2 \theta}{\cos^2 \theta}$

$=\frac{\sin^2 \theta - \sin^2 \theta \, \cos^2 \theta}{\cos^2 \theta}$

$=\frac{\sin^2 \theta (1 - \cos^2 \theta)}{\cos^2 \theta}$

$=\frac{\sin^2 \theta (\sin^2 \theta)}{\cos^2 \theta}$

$=\frac{\sin^2 \theta}{\cos^2 \theta} \times \sin^2 \theta$

$\tan^2 \theta \times \sin^2 \theta$

= Right Hand Side.

Q.E.D.

**smplease** · March 9th 2008, 06:13 PM

Originally Posted by mr fantastic

Left hand side = $\tan^2 \theta - \sin^2 \theta$

$= \frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta$

$=\frac{\sin^2 \theta}{\cos^2 \theta} - \frac{\sin^2 \cos^2 \theta}{\cos^2 \theta}$

$=\frac{\sin^2 \theta - sin^2 \cos^2 \theta}{\cos^2 \theta}$

$=\frac{\sin^2 \theta (1 - \cos^2 \theta)}{\cos^2 \theta}$

$=\frac{\sin^2 \theta (\sin^2 \theta)}{\cos^2 \theta}$

$=\frac{\sin^2 \theta}{\cos^2 \theta} \times \sin^2 \theta$

$\tan^2 \theta \times \sin^2 \theta$

= Right Hand Side.

Q.E.D.

Thanks so much for the reply and the answer I understand most of it except for the 4th line down..

where does the 1 in (1- cos) come from?

thanks again

**mr fantastic** · March 9th 2008, 09:16 PM

Originally Posted by smplease

Thanks so much for the reply and the answer I understand most of it except for the 4th line down..

where does the 1 in (1- cos) come from?

thanks again

$\sin^2 \theta$ gets taken out as a common factor:

$ =\frac{{\color{red}1} \sin^2 \theta - \sin^2 \theta \, {\color{red}\cos^2 \theta}}{\cos^2 \theta} $

$ =\frac{\sin^2 \theta ({\color{red}1 - \cos^2 \theta})}{\cos^2 \theta} $

**DivideBy0** · March 9th 2008, 09:19 PM

Originally Posted by mr fantastic

$\sin^2 \theta$ gets taken out as a common factor:

$ =\frac{{\color{red}1} \sin^2 \theta - \sin^2 {\color{red}\cos^2 \theta}}{\cos^2 \theta} $

$ =\frac{\sin^2 \theta ({\color{red}1 - \cos^2 \theta})}{\cos^2 \theta} $

Maybe he was just a bit confused because the second $\sin^2{\theta}$ didn't have a $\theta$ in it.

**mr fantastic** · March 9th 2008, 09:26 PM

Originally Posted by DivideBy0

Maybe he was just a bit confused because the second $\sin^2{\theta}$ didn't have a $\theta$ in it.

Thanks for that. I only just noticed then when you pointed it out. I hope it was obvious a theta was meant to be there.

**smplease** · March 10th 2008, 03:38 PM

Originally Posted by mr fantastic

Thanks for that. I only just noticed then when you pointed it out. I hope it was obvious a theta was meant to be there.

my eyes are now open

thanks so much for the help i understand it now

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