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Thread: trigo problem no. 11

  1. #1
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    trigo problem no. 11

    p180 ex8b q5
    prove the following identities:
    Attached Thumbnails Attached Thumbnails trigo problem no. 11-trigo11.gif  
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  2. #2
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    Just solve for $\displaystyle \frac x2$. So $\displaystyle y:=\frac x2$ then $\displaystyle \frac{\sin{2y}-\cos{y}}{1-\cos{2y}-\sin{y}}=\frac{\cos{y}}{\sin{y}}$. And solving that is quite straight forward...
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  3. #3
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    Quote Originally Posted by james_bond View Post
    Just solve for $\displaystyle \frac x2$. So $\displaystyle y:=\frac x2$ then $\displaystyle \frac{\sin{2y}-\cos{y}}{1-\cos{2y}-\sin{y}}=\frac{\cos{y}}{\sin{y}}$. And solving that is quite straight forward...
    Just to clarify:

    Replace $\displaystyle \frac{x}{2}$ with y. Note that this means you replace x with 2y.

    Then you're required to prove the identity $\displaystyle \frac{\sin{2y}-\cos{y}}{1-\cos{2y}-\sin{y}}=\frac{\cos{y}}{\sin{y}}$.

    And this is, if not quite straight forward, at least more straight forward.
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  4. #4
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    OK, so step by step:
    $\displaystyle \frac{\sin{2y}-\cos{y}}{1-\cos{2y}-\sin{y}}=\frac{\cos{y}}{\sin{y}}\Longleftrightarro w\frac{2\sin y\cos y-\cos y}{1-(\cos^2y-\sin^2y)-\sin y}=\frac{\cos y}{\sin y}\Longleftrightarrow$
    $\displaystyle \Longleftrightarrow\frac{2\sin y\cos y-\cos y}{1-\cos^2y+\sin^2y-\sin y}=\frac{\cos y}{\sin y}\Longleftrightarrow\frac{2\sin y\cos y-\cos y}{\sin^2y+\sin^2y-\sin y}=\frac{\cos y}{\sin y}\Longleftrightarrow$$\displaystyle \frac{2\sin y\cos y-\cos y}{2\sin^2y-\sin y}=\frac{\cos y}{\sin y}\Longleftrightarrow\frac{\cos y \left(2\sin y-1\right)}{\sin y \left(2\sin y-1\right)}=\frac{\cos y}{\sin y}\Longleftrightarrow 0=0$
    which is true.
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  5. #5
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    i think of another method, but seems slow
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  6. #6
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    Hello, afeasfaerw23231233!


    Your textbook has an error.
    . . Incredibly, it produces the desired result!


    Let $\displaystyle t = \tan\frac{x}{2}$

    Then we have: .$\displaystyle \frac{\dfrac{2t}{1+t^2} - \dfrac{1}{1+t^2} }{1 - \dfrac{1-t^2}{1+t^2} - \dfrac{t}{1+t^2}} $ .
    . . . This is not correct!

    Do you know the basis for this strange substitution?
    In case you don't, I'll explain it.

    We let: .$\displaystyle t \:=\:\tan\frac{x}{2}$

    So we have: .$\displaystyle \tan\frac{x}{2} \:=\:\frac{t}{1} \:=\:\frac{opp}{adj}$

    That is, angle $\displaystyle \frac{x}{2}$ is in a right triangle with: $\displaystyle opp = t,\;adj = 1$
    . . Using Pythagorus, we have: .$\displaystyle hyp = \sqrt{1+t^2}$

    Then: .$\displaystyle \boxed{\sin\frac{x}{2} \:=\:\frac{t}{\sqrt{1+t^2}}},\quad\boxed{\cos\frac {x}{2} \:=\:\frac{1}{\sqrt{1+t^2}}}$


    We also have: .$\displaystyle \sin x \;=\;2\sin\frac{x}{2}\cos\frac{x}{2}\;=\;2\frac{t} {\sqrt{1+t^2}}\frac{1}{\sqrt{1+t^2}}\quad\Rightarr ow\quad\boxed{ \sin x \;=\;\frac{2t}{1+t^2}} $

    and: .$\displaystyle \cos x \;=\;\cos^2\!\frac{x}{2} \,- \,\sin^2\!\frac{x}{2} \;=\;\left(\frac{1}{\sqrt{1+t^2}}\right)^2 - \left(\frac{t}{\sqrt{1+t^2}}\right)^2\quad\Rightar row\quad\boxed{ \cos x \;=\;\frac{1-t^2}{1+t^2}}$


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Back to the problem . . .

    We have: .$\displaystyle \frac{\sin x - \cos\frac{x}{2}}{1 - \cos x - \sin\frac{x}{2}} $

    Substitute: .$\displaystyle \frac{\dfrac{2t}{1+t^2} - {\color{red}\dfrac{1}{\sqrt{1+t^2}}}} {1 - \dfrac{1-t^2}{1+t^2} - {\color{red}\dfrac{t}{\sqrt{1+t^2}}}} $ . .
    Your book left out the radicals.


    Multiply top and bottom by $\displaystyle (1+t^2)\sqrt{1+t^2}$

    . . $\displaystyle \frac{2t\sqrt{1+t^2} - (1 + t^2)}{(1+t^2)\sqrt{1+t^2} - (1-t^2)\sqrt{1+t^2} - t(1+t^2)} \;=\;\frac{2t\sqrt{1+t^2} - 1 - t^2}{2t^2\sqrt{1+t^2} - t - t^3}$


    Factor and reduce: .$\displaystyle \frac{(2t\sqrt{1+t^2} - 1 - t^2)}{t(2t\sqrt{1+t^2} - 1 - t^2)} \;=\;\frac{1}{t}$


    Therefore, we have: .$\displaystyle \frac{1}{t} \;=\;\frac{1}{\tan\frac{x}{2}} \;=\;\cot\frac{x}{2}$ . . . . There!

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