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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Matt The Rat on April 20, 2008, 05:52:11 pm

Title: Trig Question
Post by: Matt The Rat on April 20, 2008, 05:52:11 pm: I was wondering if someone would be able to help out with this question. Thanks in advance! :)

$Given that \sin(\theta + \alpha) = \lambda cos(\theta - \alpha) , show that \tan(\theta) = \frac{\lambda - \tan(\alpha)} { 1 - \lambda \tan(\alpha)}$
Title: Re: Trig Question
Post by: ice_blockie on April 20, 2008, 05:59:18 pm: Use compound angle formula, rearrange & simplify!
Title: Re: Trig Question
Post by: Mao on April 20, 2008, 06:01:16 pm: (looks like dinner got delayed)

*typing*

$\sin(\theta+\alpha) = \lambda \cos(\theta-\alpha)$

using compound angle formula:

$\sin(\theta) \cdot \cos (\alpha) + \sin (\alpha) \cdot \cos (\theta) = \lambda \cdot (\cos (\theta) \cdot \cos (\alpha) + \sin (\theta) \cdot \sin (\alpha))$

$\sin (\theta) \cdot \cos (\alpha) - \lambda \cdot \sin (\theta) \cdot \sin (\alpha) = \lambda \cdot \cos (\theta) \cdot \cos (\alpha )- \sin (\alpha) \cdot \cos (\theta)$

$\sin(\theta) \cdot (\cos (\alpha) - \lambda \cdot \sin (\alpha)) = \cos (\theta) \cdot (\lambda \cdot \cos (\alpha) - \sin (\alpha))$

$\frac{\sin(\theta)}{\cos (\theta)} = \frac{\lambda \cdot \cos (\alpha) - \sin (\alpha)}{\cos (\alpha) - \lambda \cdot \sin (\alpha)}$

$\tan (\theta) = \frac{\lambda \cdot \cos (\alpha) - \sin (\alpha)}{\cos (\alpha) - \lambda \cdot \sin (\alpha)} \cdot \frac{\sec (\alpha)}{\sec (\alpha)}$

$\tan (\theta) = \frac{\lambda - \tan (\alpha)}{1-\lambda \cdot \tan (\alpha)}$

I tried a graphical approach first and got absolutely nowhere, i suppose there's times when you just ought to trust algebra.
Title: Re: Trig Question
Post by: ice_blockie on April 20, 2008, 06:10:26 pm: Just one question...where did you get this question from? I haven't seen it before!
Title: Re: Trig Question
Post by: Matt The Rat on April 20, 2008, 06:15:13 pm: Thanks Mao and ice_blockie!