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Homework Help: Trigonometry Identities.

  1. Aug 16, 2006 #1
    hey i have a problem which im puzzled over. i im given that:
    [tex]\sin A = \frac{3}{5}[/tex]
    and i am asked to find exact values for:
    [tex]\sin 2A[/tex]
    [tex]\cos 2A[/tex]
    [tex]\tan 2A[/tex]
    where [tex]90^0 \leq A \leq 180^0[/tex] (the power to zero is the degree sign)
    i have gone about solving this by the use of a pythagorien (?) triple, therefore
    [tex]\cos A = \frac{4}{5}[/tex] and
    [tex]\tan A = \frac{3}{4}[/tex].
    however i do not understand how A can be greator than 90 since this is in a right angle triangle. thanks in advance for those who help,
  2. jcsd
  3. Aug 16, 2006 #2


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    Homework Helper

    The Pythagorean Identity states that: sin2x + cos2x = 1, where x is some angle.
    So in this case, we have:
    sin2A + cos2A = 1
    [tex]\Leftrightarrow \cos ^ 2 A = 1 - \sin ^ 2 A = 1 - \left( \frac{3}{5} \right) ^ 2 = \frac{16}{25}[/tex]
    So there will be 2 possible value for cosA:
    [tex]\Leftrightarrow \cos A = \pm \frac{4}{5}[/tex], right?
    Since (4 / 5)2, and (-4 / 5)2 both return 16 / 25.
    So how can we know what the value of cos A is? The problem gives us more information, that the angle A is between 90o, and 180o, i.e in the second quadrant. By looking at the unit circle, can you see what sign cos A takes? Is it positive or negative?
    After having cos A, one can use the Double-Angle Formulae to finish the problem:
    sin(2A) = 2 sin(A) cos(A)
    cos(2A) = cos2(A) - sin2(A) = 1 - 2sin2(A) = 2cos2(A) - 1.
    Ok, can you go from here? :)
  4. Aug 17, 2006 #3
    hmmmm.....i see. i think i understand that now, thanks,
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