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Homework Help: Evaluating sin (-pi/12) help

  1. Jan 11, 2005 #1
    How do u evaluate [tex] \sin \frac {-\pi} {12} [/tex] ?

    I converted the radian into degrees and got -15 degrees which is the same as 345 degrees. I made a circle with a cartesian plane and then sketched the angle it is in the 4th quadrant where sin is negative, but how do I evaluate? There is no special triangle that I can use.
  2. jcsd
  3. Jan 11, 2005 #2
    I assume by evaluate you mean get an answer without the sin and the pi???

    The Bob (2004 ©)
  4. Jan 11, 2005 #3
    Well the question just says Evaluate the following, give exact value when possible
  5. Jan 11, 2005 #4
    [tex]\sin\frac{-\pi}{12} = \sin(\frac{2\pi - 3\pi}{12}) = \sin(\frac{\pi}{6}-\frac{\pi}{4})[/tex]

    Can you take it from here???

    The Bob (2004 ©)
  6. Jan 11, 2005 #5
    Can't you use a calculator? :wink:
  7. Jan 11, 2005 #6
    I cant use a calculator already asked the teacher. Is it possible to do the question with degrees instead and with a diagram ? I dont understand what to do bob.
  8. Jan 11, 2005 #7
    Use the identity [tex]\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)[/tex].
  9. Jan 12, 2005 #8
    Wrong one Recon :wink:

    It should be [tex]\sin(A-B) = \sin A \cos B - \sin B \cos A[/tex]

    The Bob (2004 ©)
  10. Jan 12, 2005 #9
    There is no need to use degrees at all. So far you know that [tex]\sin \frac{-\pi}{12} = \sin(\frac{\pi}{6}-\frac{\pi}{4})[/tex]

    Then you apply one of the rules that you know from compound angles. In this case you have a sine function and the two different parts, in the brackets, are subtract.

    Therefore you apply: [tex]\sin(A-B) = \sin A \cos B - \sin B \cos A[/tex] That will then give you sine and cosine parts that you should know how to solve from there. If you are really stuck, look back at the last thread you started where I spend ages trying to understand what Dex and BobG were going on about.
  11. Jan 12, 2005 #10
    If you get stuck, aisha, then I can help some more. At the moment I am at college, trying not to think about my maths examination today, and I have the answer to the side of me. The answer I have has no trigonometry function or a sinlge Pi symbol. I have got rid of them all using the same functions that I have told you.

    The Bob (2004 ©)
  12. Jan 12, 2005 #11


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    Hint : do you know the double angle and half-angle formulae for sine ? Do you know how to solve a simple quadratic equation ? Then you can find the answer.

    That's one way. The other way is the one that has already been suggested by others.
  13. Jan 12, 2005 #12


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    Dump the degrees. They don't do you much good in trig. Instead, you have to memorize the cosine and sine of some of your key angles. The key angles are 0 rad, pi/6, pi/4, pi/3, and pi/2. If you use these as your reference angles and figure out how the cosine and sine change in each quadrant, you have all you need.

    After that, it's all using identities to find the angles in between the key angles.

    Try this thread https://www.physicsforums.com/showthread.php?t=58516&page=2&pp=15. This is the one that The Bob was talking about.
  14. Jan 12, 2005 #13
    I would be interested to see this method.

    I am going to post the next bit of the answer, it is obvious but it means that I can copy and paste it later :wink::

    [tex]\sin \frac{-\pi}{12} = \sin (\frac{2\pi - 3\pi}{12}) = \sin (\frac{\pi}{6} - \frac{\pi}{4})[/tex]

    Then you apply [tex]\sin (A - B) = \sin A \cos B - \sin B \cos A[/tex]

    Therefore: [tex] \sin (\frac{\pi}{6} - \frac{\pi}{4}) = \sin \frac{\pi}{6} \cos \frac{\pi}{4} - \sin \frac{\pi}{4} \cos \frac{\pi}{6}[/tex]

    The Bob (2004 ©)
  15. Jan 12, 2005 #14
    Hey thanks everyone I think i got it

  16. Jan 12, 2005 #15


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    Perhaps aisha should disregard what I wrote. You see, it can be done, it's just that it's very tedious.

    The half-angle sine formula is worse, the half-angle cosine formula is slightly better.

    Let the required value be [itex]s[/tex]

    Then :

    [tex]\cos{\frac{\pi}{6}} = 1 - 2s^2[/tex]

    [tex](2s)^2 = 2 - \sqrt{3}[/tex]

    Getting the square root of that surd is quite icky, and involves a simultaneous quadratic form. But after discarding inadmissible roots, it comes out nicely to :

    [tex]s = \frac{1}{4}(\sqrt{2}-\sqrt{6})[/tex]

    Your method should be far less tedious and neater.
    Last edited: Jan 12, 2005
  17. Jan 12, 2005 #16


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    You slipped up somewhere. Aisha, could you please use a calculator to confirm your answers first ? Even we do it, it's easy to make mistakes in these problems.

    We're not asking you to cheat or anything, but it's good to determine for yourself if your answer is correct or not before posting it.
  18. Jan 13, 2005 #17
    [tex]\sin\frac{-\pi}{12} = \frac{\sqrt{2} - \sqrt{6}}{4}[/tex]

    This is what I got because:

    [tex]\sin\frac{-\pi}{12} = \sin(\frac{\pi}{6} - \frac{\pi}{4})[/tex]

    [tex]\sin(\frac{\pi}{6} - \frac{\pi}{4}) = \sin\frac{\pi}{6} \cos\frac{\pi}{4} - \sin\frac{\pi}{4} \cos\frac{\pi}{6} = (\frac{1}{2} \times \frac{\sqrt{2}}{2}) - (\frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2})[/tex]

    [tex](\frac{1}{2} \times \frac{\sqrt{2}}{2}) - (\frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2}) = \frac{\sqrt{2}}{4} - \frac{\sqrt{2} \sqrt{3}}{4} = \frac{\sqrt{2}}{4} - \frac{\sqrt{2 \times 3}}{4} = \frac{\sqrt{2} - \sqrt{6}}{4}[/tex] or [tex]\frac{\sqrt{2}}{4}(1 - \sqrt{3})[/tex]

    The Bob (2004 ©)
  19. Jan 13, 2005 #18
    A thought:

    Perhaps what your teacher is trying to get you to consider is the difference between 'even' and 'odd' functions. Which are defined as follows:

    If F(x) is an even function then:

    F(-x) = F(x)

    If F(x) is an odd function then:

    F(-x)= -F(x)

    For the case of trigometric functions cos(x) is even and sin(x) is odd. This should not be too difficult to figure out if you play with it for a while plus as you can see from the solution that The Bob has given you demonstrates that sin(x) is odd.
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