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Math Help - Solving equations of inverse functions

  1. #1
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    Solving equations of inverse functions

    The original problem is : arcsin4x + arccosx = \pi/6 and is also attached as an image.

    I have it to "X=-(cosU * (3^1/2)/2 - sinU * 1/2 ) but that might be wrong, I don't know.

    I suppose starting at 2 and then asking for help at 4:51 means I'm going to have to review this chapter, as I've had to ask for help in every section.
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  2. #2
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    Hello Wolvenmoon
    Quote Originally Posted by Wolvenmoon View Post
    The original problem is : arcsin4x + arccosx = \pi/6 and is also attached as an image.

    I have it to "X=-(cosU * (3^1/2)/2 - sinU * 1/2 ) but that might be wrong, I don't know.

    I suppose starting at 2 and then asking for help at 4:51 means I'm going to have to review this chapter, as I've had to ask for help in every section.
    Let y = \arccos x and z = \arcsin 4x

    Then y+z=\tfrac{\pi}{6}, x = \cos y and 4x=\sin z

    So \sin z = 4\cos y = 4\cos(\tfrac{\pi}{6}-z)

    =4\cos(\tfrac{\pi}{6})\cos z + 4\sin(\tfrac{\pi}{6})\sin z

    =2\sqrt3\cos z + 2\sin z

    \Rightarrow \sin z = -2\sqrt3\cos z

    \Rightarrow \tan z = -2\sqrt 3

    \Rightarrow \sin z = -\frac{2\sqrt3}{\sqrt{1+(2\sqrt3)^2}}=-\frac{2\sqrt39}{13}

    \Rightarrow x = -\frac{\sqrt39}{26}

    Grandad
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  3. #3
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    Hi

    I have an interesting method

    Consider a cyclic quadralateral  ABCD and

    set  AC is the unit diameter of the circle

    Therefore ,  AD \perp CD and  AB \perp BC

    and let  AD = 4x , BC = x

    This circle is constructed for this problem

     \angle ACD = \arcsin(4x) , \angle ACB = \arccos(x)

    there sum is  \frac{\pi}{6}

    we can using the fact that  BD / \sin(\frac{\pi}{6}) = 2R = 1 to find the length of  BD ,  BD = \frac{1}{2}

    Finally , we end up the geometry part by applying Ptolemy's Theorem :

     AD \cdot BC + AB \cdot CD = AC \cdot BD

     4x^2 + \sqrt{1-x^2} \sqrt{1-16x^2} = \frac{1}{2}

     1 - 17x^2 + 16x^4 = \frac{1}{4} - 4x^2 + 16x^4

     x^2 = \frac{3}{4(13)}
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  4. #4
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    Hello simplependulum
    Quote Originally Posted by simplependulum View Post
    Hi

    I have an interesting method

    Consider a cyclic quadralateral  ABCD and

    set  AC is the unit diameter of the circle

    Therefore ,  AD \perp CD and  AB \perp BC

    and let  AD = 4x , BC = x

    This circle is constructed for this problem

     \angle ACD = \arcsin(4x) , \angle ACB = \arccos(x)

    there sum is  \frac{\pi}{6}

    we can using the fact that  BD / \sin(\frac{\pi}{6}) = 2R = 1 to find the length of  BD ,  BD = \frac{1}{2}

    Finally , we end up the geometry part by applying Ptolemy's Theorem :

     AD \cdot BC + AB \cdot CD = AC \cdot BD

     4x^2 + \sqrt{1-x^2} \sqrt{1-16x^2} = \frac{1}{2}

     1 - 17x^2 + 16x^4 = \frac{1}{4} - 4x^2 + 16x^4

     x^2 = \frac{3}{4(13)}
    Interesting indeed!

    But just slightly flawed, I'm afraid, because it's impossible to draw so that \angle BCD = \frac{\pi}{6}.

    The fact is that x < 0, and your solution of  x^2 = \frac{3}{4(13)} stops short of solving for x. If you take the positive square root and substitute back, you'll find that \angle BCD \approx 76.1^o and \angle ACD  \approx 73.9^o. This gives \angle BCD = 150^o, which represents the solution to the equation \arcsin 4x + \arccos x = \frac{5\pi}{6}.

    The negative solution to your equation is, of course, the correct one - and it's the same as the one I obtained by a more conventional approach. Sadly, it's not possible to draw your diagram with a negative value of x.

    Unless, of course, you can think of a modification ...?

    Grandad
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  5. #5
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    Quote Originally Posted by Grandad View Post
    Hello simplependulumInteresting indeed!

    But just slightly flawed, I'm afraid, because it's impossible to draw so that \angle BCD = \frac{\pi}{6}.

    The fact is that x < 0, and your solution of  x^2 = \frac{3}{4(13)} stops short of solving for x. If you take the positive square root and substitute back, you'll find that \angle BCD \approx 76.1^o and \angle ACD \approx 73.9^o. This gives \angle BCD = 150^o, which represents the solution to the equation \arcsin 4x + \arccos x = \frac{5\pi}{6}.

    The negative solution to your equation is, of course, the correct one - and it's the same as the one I obtained by a more conventional approach. Sadly, it's not possible to draw your diagram with a negative value of x.

    Unless, of course, you can think of a modification ...?

    Grandad
    Yes . we cannot draw a line segment with length -x

    Actually . I didn't realise of the mistake until you told me the impossibility of drawing this ...

    Now , i have no confidence to modify my solution because Ptolemy's theorem

    requires positve lengths but i am suprised that I still can find out the

    exact answer !


    Luckily , this method still can be applied to solve this kind of problems if the answer is positive !
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  6. #6
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    *Whoosh* What just happened!? I think something might have flown over my head *sonic boom!*

    Yep, something flew over my head so quickly it broke the sound barrier.
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  7. #7
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    Hello, Wolvenmoon!

    I got Grandad's answer . . . and simplependulum's.


    Solve for x\!:\;\;\arcsin(4x) + \arccos(x) \:=\:\frac{\pi}{6}
    Let: . \alpha \:=\:\arcsin(4x) \quad\Rightarrow\quad \sin\alpha \:=\:4x \quad\Rightarrow\quad\cos\alpha\:=\:\sqrt{1-16x^2} .[1]

    Let: . \beta \:=\:\arccos(x) \quad\Rightarrow\quad \cos\beta \:=\:x \quad\Rightarrow\quad \sin\beta \:=\:\sqrt{1-x^2} .[2]


    We have: . \alpha + \beta \;=\;\frac{\pi}{6}

    Then: . \sin(\alpha + \beta) \;=\;\sin\frac{\pi}{6} \quad\Rightarrow\quad \sin\alpha\cos\beta + \cos\alpha\sin\beta \;=\;\frac{1}{2}

    Substitute [1] and [2]: . 4x\cdot x + \sqrt{1-16x^2}\cdot\sqrt{1-x^2} \;=\;\frac{1}{2}


    Multiply by 2: . 8x^2 + 2\sqrt{(1-16x^2)(1-x^2)} \;=\;1 \quad\Rightarrow\quad 2\sqrt{(1-16x^2)(1-x^2)} \;=\;1 - 8x^2


    Square both sides: . 4(1-16x^2)(1-x^2) \;=\;1 - 16x^2 + 64x^4 \quad\Rightarrow\quad 4 - 68x^2 + 64x^4 \;=\;1 - 16x^2 + 64x^4

    . . . . . . . . . . . . . . 52x^2 \:=\:3 \quad\Rightarrow\quad x^2 \:=\:\frac{3}{52} \quad\Rightarrow\quad x \;=\;\pm\sqrt{\frac{3}{52}} \;=\;\pm\frac{\sqrt{39}}{26}


    The only answer that checks is: . x \;=\;-\frac{\sqrt{39}}{26}



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