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Math Help - 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) \sin x =  - \frac{1}<br />
{2}.

    2) \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; January 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) sin x = - \frac{1}{2}

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

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

    cos x( 2 sin x - 1) = 0

    Cos x = 0

    OR

    2 Sin x - 1 = 0

    3) sin^{2} x - sin x = 0

    sin x(sin x - 1) = 0

    Sin x = 0

    OR

    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 \cos (\alpha  + \beta ).

    Consider the following sketch:



    Construction:

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

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

    We have

    \cos (\alpha  + \beta ) = \frac{{\overline {MN} }}<br />
{{\overline {QM} }} = \frac{{\overline {OM}  - \overline {ON} }}<br />
{{\overline {QM} }} = \frac{{\overline {OM} }}<br />
{{\overline {QM} }} \cdot \frac{{\overline {MP} }}<br />
{{\overline {MP} }} - \frac{{\overline {RP} }}<br />
{{\overline {QM} }} \cdot \frac{{\overline {QP} }}<br />
{{\overline {QP} }}.

    Rearrange

    \cos (\alpha  + \beta ) = \frac{{\overline {MP} }}<br />
{{\overline {QM} }} \cdot \frac{{\overline {OM} }}<br />
{{\overline {MP} }} - \frac{{\overline {QP} }}<br />
{{\overline {QM} }} \cdot \frac{{\overline {RP} }}<br />
{{\overline {QP} }} = \cos \beta \cos \alpha  - \sin \beta \sin \alpha \quad\blacksquare

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

    ---

    Now, a proof with complex numbers.

    \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

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

    \sqrt{(\cos b-\cos a)^2 + (\sin b - \sin a)^2} = \sqrt{2 - 2\cos (a-b)}
    (\cos b - \cos a)^2 + (\sin b - \sin a)^2 = 2 - 2 \cos (a-b)
    \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)
    \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:


    \frac{{7\pi }}<br />
{6} + 2n\pi


    Does that look correct to you guys?

    For #2 I got:

    x = \frac{\pi }<br />
{2} + 2n\pi

    For #3:

    x = 2n\pi ;\frac{\pi }<br />
{2} + 2n\pi


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