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Math Help - Find sinά

  1. #1
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    Post Find sinά

    Hi, Can anyone give me an idea how can I solve this?

    Find sinά, if sin(45- ά)= - 2/3

    I'd appreciate your help.Thanks
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  2. #2
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    Have you have any ideas so far?
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  3. #3
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    I think it's about applying addition/reduction formulas, but nothing concrete
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  4. #4
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    How about applying the arcsin function to both sides? What would that give you?
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  5. #5
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    I'm sorry I dont know what arcsin function is:$:$
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  6. #6
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    The arcsin function is also written \sin^{-1}(x). It is the functional inverse of sin. Generally speaking, if you have \sin(x)=y, then x=\arcsin(y)=\sin^{-1}(y), modulo the symmetries of the sin function. You follow?
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  7. #7
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    Im sorry I havent learned inverse functions..there isnt any other way right..?
    Can u please tell me how arcsin function is applied to both sides?

    I really appreciate ur help. Thanks
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  8. #8
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    It's just like "undoing" any other function. To undo the exponential function, you might have this:

    e^{x}=y

    \ln(e^{x})=\ln(y) (applied logarithm to both sides)

    x=\ln(y). (simplified)

    Same thing here:

    \sin(x)=y

    \sin^{-1}(\sin(x))=\sin^{-1}(y)

    x=\sin^{-1}(y).

    So how could you use this in your situation?
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  9. #9
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    45- ά= arcsin-2/3 am I close..?:$
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  10. #10
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    Right. Now, with your notation, you should make sure you have ά in degrees. What do you get when you solve for it?
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  11. #11
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    Sorry for disturbing again, but can you solve that for me please?

    Thanks for your patience.I wish my maths techer was like you.
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  12. #12
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    If you're doing trig, you should know how to solve an equation like a-x=b, for x. That's what you have here. What do you get?
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  13. #13
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    [IMG][/IMG]

    I'd like to know if this procedure is allowed in Maths.

    Thanks a lot
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  14. #14
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    The idea is sound, but you've made a mistake or two in carrying it out. Going from

    \displaystyle\frac{\sqrt{2}}{2}[\cos(\alpha)-\sin(\alpha)]=-\frac{2}{3} to

    \displaystyle\cos(\alpha)-\sin(\alpha)=\frac{2\sqrt{2}}{3},

    you lost the minus sign. You should have gotten

    \displaystyle\cos(\alpha)-\sin(\alpha)=-\frac{2\sqrt{2}}{3}.

    This is a mistake that, due to another sign error, you later correct! This is a bit sloppy.

    So, I'd say I agree with your last step, but only as the result of two self-correcting sign errors.

    I should also point out that you have done one (what we call) irreversible step, when you squared the equation. Squaring an equation can sometimes introduce spurious results. I think in your case it does. So, when you get your answers, double-check them against your original equation, and make sure they satisfy the original equation. You should always double-check the answers to problems anyway, to make sure they make sense. Make sense?
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