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Compound Angles

  1. Nov 1, 2012 #1
    In this question, my answer to part (a) is correct, & leads into (b). With (b) my answer is also correct, except for the sign. Can anyone help me figure what I need to do?

    Many thanks.

    1. The problem statement, all variables and given/known data

    Q. [itex]90^{\circ}<A<180^{\circ}[/itex] such that [itex]\sin(A+\frac{\pi}{6})+\sin(A-\frac{\pi}{6})=\frac{4\sqrt{3}}{5}[/itex]. Find (a) [itex]\sin A[/itex] & (b) [itex]\tan A[/itex]

    2. Relevant equations

    3. The attempt at a solution

    (a) [itex]\sin A\cos \frac{\pi}{6}+\cos A\sin{\pi}{6}+\sin A\cos\frac{\pi}{6}-\cos A\sin\frac{\pi}{6}=\frac{4\sqrt{3}}{5}[/itex]
    [itex]2\sin A\cos\frac{\pi}{6}=\frac{4\sqrt{3}}{5}[/itex]
    [itex]\frac{2\sqrt{3}\sin A}{2}=\frac{4\sqrt{3}}{5}[/itex]
    [itex]2\sqrt{3}\sin A=\frac{8\sqrt{3}}{10\sqrt{3}}[/itex]
    [itex]\sin A=\frac{4}{5}[/itex]

    (b) [itex]\tan A=\frac{\sin A}{\cos A}[/itex]
    [itex]\sin^2A+cos^2A=1[/itex]
    [itex](\frac{4}{5})^2+cos^2A=1[/itex]
    [itex]\cos^2A=1-\frac{16}{25}[/itex]
    [itex]\cos A=\sqrt{\frac{9}{25}}[/itex]
    [itex]\cos A=\frac{3}{5}[/itex]
    [itex]\tan A=\frac{4}{5}/\frac{3}{5}[/itex]
    [itex]\tan A=\frac{4}{3}[/itex]

    Answer: (From text book): (b) [itex]\frac{-4}{3}[/itex]
     
  2. jcsd
  3. Nov 1, 2012 #2

    CAF123

    User Avatar
    Gold Member

    Use the condition given in the question: π/2 ≤ A ≤ π. if you like, draw the CAST diagram or sketch the cosine graph - what is the sign of cos here?
    Note also [itex] \sqrt{9/25} = +/- 3/5 [/itex]
     
  4. Nov 1, 2012 #3
    Great. Thank you.
     
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