Results 1 to 7 of 7

Thread: Solving equations of inverse functions

  1. #1
    Member
    Joined
    Aug 2009
    Posts
    101

    Solving equations of inverse functions

    The original problem is : $\displaystyle arcsin4x + arccosx = \pi/6$ 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.
    Attached Thumbnails Attached Thumbnails Solving equations of inverse functions-flash0011.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello Wolvenmoon
    Quote Originally Posted by Wolvenmoon View Post
    The original problem is : $\displaystyle arcsin4x + arccosx = \pi/6$ 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 $\displaystyle y = \arccos x$ and $\displaystyle z = \arcsin 4x$

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

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

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

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

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

    $\displaystyle \Rightarrow \tan z = -2\sqrt 3$

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

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

    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jan 2009
    Posts
    715
    Hi

    I have an interesting method

    Consider a cyclic quadralateral $\displaystyle ABCD $ and

    set $\displaystyle AC $ is the unit diameter of the circle

    Therefore , $\displaystyle AD \perp CD$ and $\displaystyle AB \perp BC $

    and let $\displaystyle AD = 4x , BC = x $

    This circle is constructed for this problem

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

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

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

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

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

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

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

    $\displaystyle x^2 = \frac{3}{4(13)} $
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    Hello simplependulum
    Quote Originally Posted by simplependulum View Post
    Hi

    I have an interesting method

    Consider a cyclic quadralateral $\displaystyle ABCD $ and

    set $\displaystyle AC $ is the unit diameter of the circle

    Therefore , $\displaystyle AD \perp CD$ and $\displaystyle AB \perp BC $

    and let $\displaystyle AD = 4x , BC = x $

    This circle is constructed for this problem

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

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

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

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

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

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

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

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

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

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

    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 $\displaystyle x$.

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

    Grandad
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Jan 2009
    Posts
    715
    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 $\displaystyle \angle BCD = \frac{\pi}{6}$.

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

    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 $\displaystyle x$.

    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 !
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Aug 2009
    Posts
    101
    *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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, Wolvenmoon!

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


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

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


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

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

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


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


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

    . . . . . . . . . . . . . . $\displaystyle 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: .$\displaystyle x \;=\;-\frac{\sqrt{39}}{26} $



    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: Nov 30th 2011, 01:41 AM
  2. Need help solving this problem. (Inverse Functions)
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: Aug 13th 2011, 05:41 AM
  3. Replies: 3
    Last Post: Jun 10th 2011, 12:25 AM
  4. Solving inverse trig equations
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Sep 24th 2010, 06:44 PM
  5. Inverse functions, solving equations.
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Oct 27th 2009, 08:02 AM

Search Tags


/mathhelpforum @mathhelpforum