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Trigonometry homework problems

  1. Feb 26, 2012 #1
    1. The problem statement, all variables and given/known data
    I need to prove that

    [itex]\frac{\frac{1}{2} cot20^{o}-cos10^{o}}{\frac{1}{2}+sin10^{o}}=\frac{\sqrt{3}}{3}[/itex]

    3. The attempt at a solution

    I try to do it by this way

    [itex]\frac{\frac{1}{2} cot10^{o}-cos10^{o}}{\frac{1}{2}+sin10^{o}}=\frac{cos20^{o}-2cos10^{o}sin20^{o}}{(1+2sin10^{o})sin20^{o}}= \frac{ cos20^{o}-\frac{1}{2}-sin10^{o}}{(1+2sin10^{o})sin20^{o}}[/itex]

    Please help, it is very urgent.
    Last edited: Feb 26, 2012
  2. jcsd
  3. Feb 26, 2012 #2


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    Re: Trigonometry

    After taking some time, I figured out how to get that [itex]\displaystyle 2\cos10^{o}\sin20^{o}=\frac{1}{2}+\sin10^{o}\,,[/itex] using a product to sum identity.

    Do a similar thing to your denominator:
    [itex]\displaystyle (1+2\sin10^{o})\sin20^{o}=\sin20^{o}+\sin10^{o}+ \sin30^{o}=\sin10^{o}+\sin20^{o}+\sin30^{o}[/itex]​
    That's also the same as cos60° + cos70° + cos80° .

    Your numerator can be written as -cos60° + sin70° - cos80° or equivalently, -sin10° + cos20° - sin30° .

    I don't know if any of that helps.

    Also remember that [itex]\displaystyle \tan30^{o}=\cot60^{o}=\frac{\sqrt{3}}{3}[/itex]
    Last edited: Feb 26, 2012
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