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Thread: a cos x + b sinx

  1. #1
    Member Furyan's Avatar
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    a cos x + b sinx

    Hello

    I would appreciate some help with the following question, which is, prove that:

    $\displaystyle \cos\theta - \sin\theta = -\sqrt{2}\sin(\theta - 45^o)$

    If I do it like this:

    $\displaystyle \cos\theta - \sin\theta \equiv R\sin(\theta + \alpha)$

    $\displaystyle \equiv R(\sin\theta\cos\alpha + \cos\theta\sin\alpha)$

    $\displaystyle 1 = R\sin\alpha$

    $\displaystyle -1 = R\cos\alpha$

    $\displaystyle R = \sqrt{2}$

    $\displaystyle \alpha = arctan (-1) = - 45^o$

    I get:

    $\displaystyle \sqrt{2}\sin(\theta - 45^o)$

    I thought of doing it like this:

    $\displaystyle \cos\theta - \sin\theta \equiv\sin(\theta - \alpha)$

    $\displaystyle \equiv R(\sin\theta\cos\alpha - \cos\theta\sin\alpha)$

    $\displaystyle \equiv -R(-\sin\theta\cos\alpha +\cos\theta\sin\alpha)$

    but then $\displaystyle (-R)^2 = R^2$

    So I can't seem to do it.

    I did first have to show that:

    $\displaystyle \cos\theta - \sin\theta = \sqrt{2}\cos(\theta + 45^o)$, which I was able to do.

    Any help would be very much appreciated,

    Thank you
    Last edited by Furyan; Jul 19th 2012 at 04:33 PM.
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  2. #2
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    Re: a cos x + b sinx

    Quote Originally Posted by Furyan View Post
    prove that:

    $\displaystyle \cos\theta - \sin\theta = -\sqrt{2}\sin(\theta - 45^o)$
    Rewrite the right-hand side using the formula $\displaystyle \sin(\theta-\alpha)=\sin\theta\cos\alpha - \cos\theta\sin\alpha$.
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  3. #3
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    Re: a cos x + b sinx

    Why you do not try this,

    $\displaystyle \cos \theta -\sin \theta=\sqrt{2}(\cos \theta \frac{1}{\sqrt{2}}-\sin\theta\frac{1}{\sqrt{2}})$
    $\displaystyle =-\sqrt{2}(\sin\theta\cos 45 -\cos\theta\sin45)=-\sqrt{2}\sin{(\theta-45)}$
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  4. #4
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    Re: a cos x + b sinx

    Quote Originally Posted by Furyan View Post
    If I do it like this:

    $\displaystyle \cos\theta - \sin\theta \equiv R\sin(\theta + \alpha)$

    $\displaystyle \equiv R(\sin\theta\cos\alpha + \cos\theta\sin\alpha)$

    $\displaystyle 1 = R\sin\alpha$

    $\displaystyle -1 = R\cos\alpha$

    $\displaystyle R = \sqrt{2}$

    $\displaystyle \alpha = arctan (-1) = - 45^o$

    I get:

    $\displaystyle \sqrt{2}\sin(\theta - 45^o)$
    The error here is that $\displaystyle \tan(\alpha)=-1$ does not imply that $\displaystyle \alpha=-45^\circ$. In this case, $\displaystyle \sin\alpha=1/\sqrt{2}$ and $\displaystyle \cos\alpha=-1/\sqrt{2}$, so $\displaystyle \alpha=135^\circ$. Thus,

    $\displaystyle \cos\theta-\sin\theta=\sqrt{2}\sin(\theta+135)=\sqrt{2}\sin(1 80-\theta-135)=\sqrt{2}\sin(45-\theta)$ $\displaystyle =-\sqrt{2}\sin(\theta-45)$.
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  5. #5
    Member Furyan's Avatar
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    Re: a cos x + b sinx

    Thank you both for taking the time to reply.

    (There's nothing in the chapter I'm working through about either of the methods you have shown me, but I will study your posts in detail to see what I can learn from them.

    Thank you very much for you efforts.)

    Edit: Ah, actually there is an example showing how to determine which quadrant the angle is in. I never learnt that, I'm not sure how I got by without it. It seems pretty important, in fact, looking back it's one of the first things I should have learnt in trigonometry.
    Last edited by Furyan; Jul 20th 2012 at 05:08 AM. Reason: I was mistaken.
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  6. #6
    Member Furyan's Avatar
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    Re: a cos x + b sinx

    Thank you,

    (I will see if I can get:

    $\displaystyle \sqrt{2}\sin(45 - \theta)$

    just now seeing that $\displaystyle \alpha= 135$ is beyond me, but I guess I need to know that.)

    Thank you very much indeed for your help.

    Edit: I think I understand now. Since $\displaystyle \sin\alpha > 0$ and both $\displaystyle \cos\alpha$ and $\displaystyle \tan\alpha < 0, \alpha$ is in the second quadrant, $\displaystyle 90^o \leq \alpha \leq 180^o$. Thanks for pointing that out. I'm glad I know that now.
    Last edited by Furyan; Jul 20th 2012 at 05:21 AM.
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