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Thread: Need some help with few trig equations...

  1. #1
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    Need some help with few trig equations...

    Find all solutions of the given equation:

    1.) 2 sin x + 1 = 0

    2.) sin 2x - cos x = 0

    3.) sin² x - sin x = 0

    **Not looking for the entire answer, just how to start.

    Prove that the given trig identity is true:

    cos (a - b) = (cos a) (cos b) + (sin a) (sin b)

    **I know that it's true because it states the formula in the book but I just don't know how to prove it, lol.
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  2. #2
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    1) $\displaystyle \sin x = - \frac{1}
    {2}.$

    2) $\displaystyle \sin 2x = 2\sin x\cos x$ & factorise.

    3) Factorise and proceed as 2).

    ---

    About last formula, it requires a geometric reasoning to prove it, or you can also use complex numbers.
    Last edited by Krizalid; Jan 20th 2008 at 03:49 PM.
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  3. #3
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    Could you explain a little further on...all of them?

    For #1, -1/2 is what I got on my own, but I don't know how to put the answer in terms of (ex. from another problem: x= pi/3 + 2n(pi) for any int n).

    As for #2 and #3, can you give me the first step to factorizing? I'm clueless on where to start.

    And #4...I still have no clue.
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  4. #4
    Bar0n janvdl's Avatar
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    Quote Originally Posted by toop View Post
    Could you explain a little further on...all of them?

    For #1, -1/2 is what I got on my own, but I don't know how to put the answer in terms of (ex. from another problem: x= pi/3 + 2n(pi) for any int n).

    As for #2 and #3, can you give me the first step to factorizing? I'm clueless on where to start.

    And #4...I still have no clue.
    Krizalid seems to be away for the moment so I'll jump in...

    1) $\displaystyle sin x = - \frac{1}{2}$

    $\displaystyle x = sin^{-1} ( \frac{1}{2} )$ (The value of x must be in quadrants 3 and 4.

    2) $\displaystyle 2 \ sin x \ cos x - cos x= 0$

    $\displaystyle cos x( 2 sin x - 1) = 0$

    $\displaystyle Cos x = 0$

    OR

    $\displaystyle 2 Sin x - 1 = 0$

    3) $\displaystyle sin^{2} x - sin x = 0$

    $\displaystyle sin x(sin x - 1) = 0$

    $\displaystyle Sin x = 0$

    OR

    $\displaystyle sin x - 1 = 0$

    4) This is a long process of proving. Look it up on the net, or it is most probably in your textbook.
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  5. #5
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    Quote Originally Posted by toop View Post
    cos (a - b) = (cos a) (cos b) + (sin a) (sin b)
    First of all, we're gonna find a formula for $\displaystyle \cos (\alpha + \beta ).$

    Consider the following sketch:



    Construction:

    • Consider a rectangle $\displaystyle NOPR.$
    • Extend $\displaystyle \overline{ON}.$ We get point $\displaystyle M.$
    • Join $\displaystyle M$ & $\displaystyle P.$
    • Extend $\displaystyle \overline{NR}.$ We get point $\displaystyle Q.$
    • Draw $\displaystyle \overline {QP} \perp\overline {MP} ,$ now join $\displaystyle M$ & $\displaystyle Q.$
    • Let $\displaystyle \measuredangle \,OMP = \alpha$ & $\displaystyle \measuredangle \,QMP = \beta .$

    Start by showing that $\displaystyle \measuredangle \,OMP = \measuredangle \,PQR = \alpha .$

    We have

    $\displaystyle \cos (\alpha + \beta ) = \frac{{\overline {MN} }}
    {{\overline {QM} }} = \frac{{\overline {OM} - \overline {ON} }}
    {{\overline {QM} }} = \frac{{\overline {OM} }}
    {{\overline {QM} }} \cdot \frac{{\overline {MP} }}
    {{\overline {MP} }} - \frac{{\overline {RP} }}
    {{\overline {QM} }} \cdot \frac{{\overline {QP} }}
    {{\overline {QP} }}.$

    Rearrange

    $\displaystyle \cos (\alpha + \beta ) = \frac{{\overline {MP} }}
    {{\overline {QM} }} \cdot \frac{{\overline {OM} }}
    {{\overline {MP} }} - \frac{{\overline {QP} }}
    {{\overline {QM} }} \cdot \frac{{\overline {RP} }}
    {{\overline {QP} }} = \cos \beta \cos \alpha - \sin \beta \sin \alpha \quad\blacksquare$

    Finally, write $\displaystyle \cos (\alpha - \beta ) = \cos \left[ {\alpha + ( - \beta )} \right]$ and you'll get the desired formula.

    ---

    Now, a proof with complex numbers.

    $\displaystyle \cos (\alpha - \beta ) = \text{Re} \Big[ {e^{i(\alpha - \beta )} } \Big] = \text{Re} \Big[ {(\cos \alpha + i\sin \alpha )(\cos \beta - i\sin \beta )} \Big].$

    Finally

    $\displaystyle \cos (\alpha - \beta ) = \text{Re}\, (\cos \alpha \cos \beta - i\cos \alpha \sin \beta + i\sin \alpha \cos \beta + \sin \alpha \sin \beta ).$

    And we happily get $\displaystyle \cos (\alpha - \beta ) = \cos \alpha \cos \beta + \sin \alpha \sin \beta\quad\blacksquare$

    As desired.
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  6. #6
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    Quote Originally Posted by toop View Post

    Prove that the given trig identity is true:

    cos (a - b) = (cos a) (cos b) + (sin a) (sin b)
    The best way is to use $\displaystyle \cos z = = \frac{1}{2}(e^{iz} + e^{-iz})$ --- but that is unfair.

    Here is another geometric proof that I remember from 10th grade.

    1)Construct the circle $\displaystyle x^2+y^2 = 1$.
    2)Construct those red lines touching at $\displaystyle P_1,P_2$.
    3)$\displaystyle a$ is the bigger angle from the x-axis.
    4)$\displaystyle b$ is the smaller angle from the [tex]x-axis.
    5)Coordinates of $\displaystyle P_1$ is $\displaystyle (\cos b,\sin b)$.
    6)Coordinates of $\displaystyle P_2$ is $\displaystyle (\cos a,\sin a)$
    7)The distance from $\displaystyle P_1 \to P_2$ is $\displaystyle \sqrt{(\cos b-\cos a)^2 + (\sin b - \sin a)^2}$.
    8)The angle between the two red lines is $\displaystyle a-b$.
    9)The length of $\displaystyle P_1P_2$ using law of cosines is $\displaystyle \sqrt{1^2+1^2 - 2 \cos (a-b)}$.
    10)Equality of #7 and #9 leads to ...

    $\displaystyle \sqrt{(\cos b-\cos a)^2 + (\sin b - \sin a)^2} = \sqrt{2 - 2\cos (a-b)}$
    $\displaystyle (\cos b - \cos a)^2 + (\sin b - \sin a)^2 = 2 - 2 \cos (a-b)$
    $\displaystyle \cos^2 b - 2\cos b\cos a + \cos^2 a + \sin^2 b - 2\sin a \sin b + \sin^2 a = 2 - 2 \cos (a-b)$
    $\displaystyle \cos (a-b) = \cos a \cos b + \sin a \sin b$
    Attached Thumbnails Attached Thumbnails Need some help with few trig equations...-picture.jpg  
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  7. #7
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    Wow, thanks for the help on #4.

    For #1 I got:


    $\displaystyle \frac{{7\pi }}
    {6} + 2n\pi $


    Does that look correct to you guys?

    For #2 I got:

    $\displaystyle x = \frac{\pi }
    {2} + 2n\pi $

    For #3:

    $\displaystyle x = 2n\pi ;\frac{\pi }
    {2} + 2n\pi $


    Thanks a lot guys!
    Last edited by toop; Jan 21st 2008 at 03:11 PM.
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