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Math Help - Trigonometry functions

  1. #1
    Junior Member
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    Jan 2009
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    Trigonometry functions

    Could someone please explain these 2 problems for me?

    #1: If 0 <x < {\pi} and sinx + cosx = \frac{1}{5}, then find tan x.

    #2: What are the solutions for sec[arctan(\frac{-3}{5})] and sin[arcsin(\frac{5}{13}) - arccos(\frac{-3}{5})]?

    Thank you!!!
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  2. #2
    Newbie
    Joined
    May 2009
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    #1: If 0 <x < {\pi} and sinx + cosx = \frac{1}{5}, then find tan x.


    Not sure about this, but I reckon you have to change sinx + cosx into the form Rsin(x + theta)

    \theta)=Rsinxcos\theta+Rcosxsin\theta" alt="Rsin(x+\theta)=Rsinxcos\theta+Rcosxsin\theta" />
    so Rsinxcos\theta+Rcosxsin\theta=sinx+cosx
    Equating terms Rsinxcos\theta=sinx
    and Rcosxsin\theta=cosx

    hence Rcos\theta=1 and Rsin\theta=1
    so tan\theta=1,\theta=\frac{\pi}{4}

    then find R=\frac{1}{\sqrt{2}}

    so you will have \frac{1}{\sqrt{2}}sin(x+\frac{\pi}{4})=\frac{1}{5}

    hope this helps, it should be easy to find x now and therefore find tan x.... there is probably an easier way though!!
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