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Thread: Find a in terms of b

  1. #1
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    Find a in terms of b

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

    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
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    Quote Originally Posted by Joker37 View Post
    For $\displaystyle \pi<a<\frac{3\pi}{2}$ with $\displaystyle cos(a) = -sin(b)$ where $\displaystyle 0<b<\frac{\pi}{2}$, find $\displaystyle a$ in terms of $\displaystyle b$.

    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 $\displaystyle x$ is arbitrary:

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

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

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

    whence

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

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

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

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

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

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

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

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

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

    or

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

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

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

    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:
    $\displaystyle cos(a)=-sin(b)$
    $\displaystyle cos(\pi+a)=-cos(\frac{\pi}{2}-b)$
    How is
    $\displaystyle cos(a) = cos( \pi + a)$

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

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

    ?

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

    ?

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

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

    ?

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

    ?

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

    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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