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Thread: Trigonometry prove question

  1. #1
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    Trigonometry prove question

    I should add this is desperate. The bracket after m and c just shows that the value of m and c are assumed to be different for each straight line.* means multiply.

    Let a and b be the straight lines with equations y=m(1)x +c and y=m(2)x +c where m(1)*m(2) is not equal to 0. Use appropriate trigonometric formula to prove that a and b are perpendicular if and only if m(1)m(2) =1.


    Please note that by appropriate- keep means to keep it as simple as possible- note this is high school maths so please try not to get too advance.
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  2. #2
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    Quote Originally Posted by kingkaisai2
    I should add this is desperate. The bracket after m and c just shows that the value of m and c are assumed to be different for each straight line.* means multiply.

    Let a and b be the straight lines with equations y=m(1)x +c and y=m(2)x +c where m(1)*m(2) is not equal to 0. Use appropriate trigonometric formula to prove that a and b are perpendicular if and only if m(1)m(2) =1.


    Please note that by appropriate- keep means to keep it as simple as possible- note this is high school maths so please try not to get too advance.
    Actually you mean that,
    $\displaystyle m_1\cdot m_2=-1$
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  3. #3
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    Let $\displaystyle \theta $ represent the angle a line forms with the x-axis. Now, this is important, there are two possible angles the one on the right and left. We let $\displaystyle \theta$ be the one on the right.

    There is a theorem, given a line $\displaystyle y=mx+b$ then $\displaystyle \tan \theta =m$. This theorem is crucial in this proof.

    Let us have two lines. One $\displaystyle y=m_1x+b_1$ crosses the x-axis at $\displaystyle M_1$ and the other $\displaystyle y=m_2x+b_2$ crosses the x-axis at, $\displaystyle M_2$. Let $\displaystyle P$ be the point of intersection of these two lines. Let us assume that they are perpendicular, meaning that $\displaystyle \not < M_1PM_2=90^o$. Let $\displaystyle \theta_1$ we the angle formed by the first line, and $\displaystyle \theta_2$ be the angle formed by the second line. Then, because we have a right triangle, $\displaystyle \theta_2=\theta_1+90^o$. Therefore, $\displaystyle \tan \theta_1 \tan (90^o+\theta_1)$ can expressed as, $\displaystyle -\tan \theta_1 \cot \theta _1$ cuz, $\displaystyle \tan (90^o+x)=-\cot x$ but because tangent and cotanget are reciprocals functions there product gives one thus,
    $\displaystyle -\tan \theta_1 \cot \theta_1 =-1$
    This means that,
    $\displaystyle \tan \theta_1 \tan \theta_2=-1$
    But, $\displaystyle m_1=\tan \theta_1, m_2=\tan \theta_2$
    Thus,
    $\displaystyle m_1m_2=-1$
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  4. #4
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    For the converse,
    --------
    Let us have two lines. One $\displaystyle y=m_1x+b_1$ crosses the x-axis at $\displaystyle M_1$ and the other $\displaystyle y=m_2x+b_2$ crosses the x-axis at, $\displaystyle M_2$. Let $\displaystyle P$ be the point of intersection of these two lines. Let us assume that $\displaystyle m_1m_2=-1$. Let $\displaystyle \theta_1$ we the angle formed by the first line, and $\displaystyle \theta_2$ be the angle formed by the second line. Then, because we have a triangle, $\displaystyle \theta_2=\theta_1+P$. Where $\displaystyle P=\not < M_1PM_2$. Therefore, $\displaystyle m_1m_2=-1$ but, $\displaystyle m_1=\tan \theta_1$ and $\displaystyle m_2=\tan \theta_2$ thus,
    $\displaystyle \tan \theta_1\tan (\theta_1+P)=-1$
    Divide both sides by $\displaystyle \tan \theta_1$,
    $\displaystyle \tan(\theta_1+P)=-\cot \theta_1$ because you get a reciprocal. Now using the same reason as above you can write, $\displaystyle \tan(\theta_1+P)=\tan(\theta_1+90^o)$
    Thus,
    $\displaystyle \theta_1+P=\theta_1+90^o$
    Thus,
    $\displaystyle P=90^o$.
    Proof is complete
    -----------


    Note (not for you): I know I could have not divided both sides sides by tangent without assuming the line was non-vertical and I cannot conclude,
    $\displaystyle x=y$ from $\displaystyle \tan x=\tan y$ without denostrating this hold for angles $\displaystyle 0\leq x\leq \pi$, i.e. domain of triangle.
    So please do not bother me with these posts, I omitted that because that was high schools math.
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