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Math Help - Sin(y)=Cos(y/3)

  1. #1
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    Sin(y)=Cos(y/3)

    Hello everyone, I came across an old trigonometry problem which I'm not able to solve again and it's driving me crazy, here it is:

    find the value of tan(x+y) knowing that:
    tan(x)=-sqrt(2)/2
    sin(y)=cos(y/3) with 19pi/4<=y<=5pi ( y in [19pi/4;5pi] ).

    The final answer should be tan (x+y)= sqrt(2)-3.

    My only problem is that I've a really hard time to find y. I tried to add pi/2 on both side to have cos or sin on both sides but it didn't help then I tried with a variable, pi/3=a => sin(3a)=cos(a) => sin(a)(4cosˆ2(a)-1)=cos(a) but then again it led me to nothing...
    Can you please explain to me how to find y.

    ( I did the problem backwards and you should fin tan(y)=1-sqrt(2)).

    Thanks in advance!!
    Last edited by sunmalus; March 20th 2010 at 01:52 AM.
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  2. #2
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    Hello, sunmalus!

    Find the value of \tan(x+y)

    knowing that:
    . . \tan x \:=\:-\frac{\sqrt{2}}{2}

    . . \sin y\:=\:\cos\tfrac{y}{2}\;\;\text{ where }\tfrac{19\pi}{4}\:\leq\: y\:\leq\: 5\pi . . {\color{blue}y}
    is in Quadrant 2

    The final answer should be: \tan (x+y)\:=\: \sqrt{2}-3 . .
    I'm not getting this

    We have: . \sin y \;=\;\cos\tfrac{y}{2}

    Then:. . . . \sin y \;=\;\pm\sqrt{\frac{1 + \cos y}{2}}

    Square: . \sin^2\!y \;=\;\frac{1 + \cos y}{2}

    Then: . 1-\cos^2\!y \;=\;\frac{1+\cos y}{2} \quad\Rightarrow\quad 2\cos^2\!y + \cos y - 1 \;=\;0

    Factor: . (2\cos y - 1)(\cos y + 1) \:=\:0


    Then: . \begin{array}{ccccccccc}\cos y \;=\; \frac{1}{2} & \Rightarrow & y \;=\; \pm\frac{\pi}{3} & \text{not in Quadrant 2} \\ \\[-3mm] \cos y \;=\; \text{-}1 & \Rightarrow & \boxed{y \;=\; \pi} \end{array}


    And we have: . \tan(x + \pi) \;=\;\tan x \;=\;-\frac{\sqrt{2}}{2}


    What's going on?

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  3. #3
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    Quote Originally Posted by sunmalus View Post
    Hello everyone, I came across an old trigonometry problem which I'm not able to solve again and it's driving me crazy, here it is:

    find the value of tan(x+y) knowing that:
    tan(x)=-sqrt(2)/2
    sin(y)=cos(y/2) with 19pi/4<=y<=5pi ( y in [19pi/4;5pi] ).

    The final answer should be tan (x+y)= sqrt(2)-3.

    My only problem is that I've a really hard time to find y. I tried to add pi/2 on both side to have cos or sin on both sides but it didn't help then I tried with a variable, pi/3=a => sin(3a)=cos(a) => sin(a)(4cosą2(a)-1)=cos(a) but then again it led me to nothing...
    Can you please explain to me how to find y.

    ( I did the problem backwards and you should fin tan(y)=1-sqrt(2)).

    Thanks in advance!!
    Does \sin{y} = \cos{\frac{y}{2}} or does \sin{y} = \cos{\frac{y}{3}}?
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  4. #4
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    correction

    I am very very sorry, it's cos(y/3), I was a bit tired when I posted this. But thanks to Soroban I think I got it now.

    Again a big sorry and many thanks!!
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  5. #5
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    I spoke too fast, I'm still stuck...
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  6. #6
    RJH
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    Whew! that was a little lengthy!
    Here's how I found y >>>>

    <br />
sin(y) = cos(y/3) [1]

    sin(y) = 3sin(y/3) - 4sin^3(y/3)<br />
    and..
    cos(y/3) = sqrt(1-sin^2(y/3))
    so sub into [1]..
    (3sin(y/3) - 4sin^3(y/3))^2 = 1-sin(y/3)
    9sin^2(y/3) + 16sin^6(y/3) - 24sin^4(y/3) = 1-sin^2(y/3)<br />
    same as..

    (1-sin^2(y/3))(8sin^2(y/3))(1-2sin^2(y/3)) + sin^2(y/3) = 1 -sin^2(y/3)<br />
    so..
    8sin^4(y/3)-8sin^2(y/3)+1 = 0
    <br />
sin^2(y/3) = 1/2 + sqrt(32)/16 (quadratic formula)

    solution we want is -9pi/8
    move y in positive direction to between 19pi/4 and 5pi there's your y

    -9pi/8 + 3(2pi) = 39pi/8

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  7. #7
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    Thanks

    Thanks a lot to all of you!
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  8. #8
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    RJH, I've a last question for you. How did you proceed to find that:

    <br />
9sin^2(y/3) + 16sin^6(y/3) - 24sin^4(y/3) = 1-sin^2(y/3)<br />

    =>

    <br />
(1-sin^2(y/3))(8sin^2(y/3))(1-2sin^2(y/3)) + sin^2(y/3) = 1 -sin^2(y/3)<br />

    because I looked how you started and I wanted to try to do the rest by myself but I got stuck at that step...

    Do you have a methode or something, I know that you want to find something with the same power and I was looking for this, but by just "randomly" trying to find what it could be it would have taken me a while...

    So any advice if I come across something like this again?
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  9. #9
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    Hi

    I propose you an alternative way

    \sin y = \cos \left(\frac{y}{3}\right) is equivalent to

    \cos \left(\frac{\pi}{2}-y\right) = \cos \left(\frac{y}{3}\right)

    \cos a = \cos b leads to a = b + 2k \pi or a = -b + 2k \pi

    Therefore
    \frac{\pi}{2}-y = \frac{y}{3} + 2k \pi or \frac{\pi}{2}-y = -\frac{y}{3} + 2k \pi

    y = \frac{3\pi}{8} + 3k \frac{\pi}{2} or y = \frac{3\pi}{4} + 3k \pi
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  10. #10
    RJH
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    Thank you to running-gag!
    I knew there must be a simpler way to do it...
    sunmalus, I highly recommend that you follow running-gag's reasoning, my way is far too long winded
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