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Math Help - Find a in terms of b

  1. #1
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    Find a in terms of b

    For \pi<a<\frac{3\pi}{2} with cos(a) = -sin(b) where 0<b<\frac{\pi}{2}, find a in terms of b.

    I am terrible at maths so please show any simple easy to understand working out so that I can understand better.

    Any help would be appreciated!
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  2. #2
    Member HappyJoe's Avatar
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    Quote Originally Posted by Joker37 View Post
    For \pi<a<\frac{3\pi}{2} with cos(a) = -sin(b) where 0<b<\frac{\pi}{2}, find a in terms of b.

    I am terrible at maths so please show any simple easy to understand working out so that I can understand better.

    Any help would be appreciated!
    Recall the following "transition formula" for going from cosine to sine, where x is arbitrary:

    \cos(x) = \sin(x+\frac{\pi}{2}).

    Since the numbers \frac{\pi}{2} and -\frac{3\pi}{2} are different by 2\pi, and since sine is periodic with period 2\pi, we have also that

    \sin(x+\frac{\pi}{2})=\sin(x-\frac{3\pi}{2}),

    whence

    \cos(x) = \sin(x-\frac{3\pi}{2}).

    Since \pi<a<\frac{3\pi}{2}, we have that a-\frac{3\pi}{2} lies in the interval (-\frac{\pi}{2},0).

    Since you want to solve \cos(a) = -\sin(b) for a, we have by the transition formula above that we might just as well solve

    \sin(a-\frac{3\pi}{2}) = -\sin(b)=\sin(-b).

    for a, where the last equality follows from sine being an odd function.

    Since 0<b<\frac{\pi}{2}, we have that -b lies in the interval (-\frac{\pi}{2},0). Hence both a-\frac{3\pi}{2} and -b are numbers in the interval (-\frac{\pi}{2},0), on which sine is injective, so in solving

    \sin(a-\frac{3\pi}{2}) =\sin(-b)

    for a, we can apply \sin^{-1} to both sides to obtain

    a-\frac{3\pi}{2} = -b,

    or

    a = -b+\frac{3\pi}{2}.
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  3. #3
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    I don't understand all of the above.

    Below was my attempt:
    cos(a)=-sin(b)
    cos(\pi+a)=-cos(\frac{\pi}{2}-b)
    \pi+a=\frac{\pi}{2}-b
    a=\frac{-\pi}{2}-b

    However the answer is:
    \frac{3\pi}{2}-b

    Where did I go wrong?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Joker37 View Post
    I don't understand all of the above.

    Below was my attempt:
    cos(a)=-sin(b)
    cos(\pi+a)=-cos(\frac{\pi}{2}-b)
    How is
    cos(a) = cos( \pi + a)

    -Dan
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  5. #5
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    Quote Originally Posted by topsquark View Post
    How is
    cos(a) = cos( \pi + a)

    -Dan
    cos(a)=-sin(b)
    -cos(\pi+a)=-cos(\frac{\pi}{2}-b)

    ?

    But still, how would this change the result?
    edit: -\pi -a = \frac{\pi}{2}-b

    ?

    edit2: a=\frac{3\pi}{2}-b

    Is this the right methodology?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Joker37 View Post
    cos(a)=-sin(b)
    -cos(\pi+a)=-cos(\frac{\pi}{2}-b)

    ?

    But still, how would this change the result?
    edit: -\pi -a = \frac{\pi}{2}-b

    ?

    edit2: a=\frac{3\pi}{2}-b

    Is this the right methodology?
    Looks good to me. But as HappyJoe's post indicates there are a few landmines we need to step over to say that we can do this.

    -Dan
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  7. #7
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    Quote Originally Posted by topsquark View Post
    Looks good to me. But as HappyJoe's post indicates there are a few landmines we need to step over to say that we can do this.

    -Dan
    What sort of things?
    I don't really understand HappyJoe's working out.
    I'm terrible at maths.
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Joker37 View Post
    What sort of things?
    I don't really understand HappyJoe's working out.
    I'm terrible at maths.
    I think it's best for you to go through HappyJoe's post and let us know what line or lines you don't understand. We can help you better that way.

    -Dan
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