Another Trigonometry Word Problem (rather difficults)

violetice · #1 $Old$ January 28th, 2009, 12:57 PM

In the diagram in the left below, AD is a diameter of the circle and is tangent to line m at D. If AD=1 AND BC is perpendicular to m, show that:

a) AB + BC = 1 +cos(x) - cos^2(x)

b) What value of x makes the sum AB + BC a maximum?

Soroban · #2 $Old$ January 28th, 2009, 09:37 PM

Hello, violetice!

Quote:

Code:

                A
              * * *
          *     |x *  *
        *       |     * *
       *      E + - - - -* B
                |     *  |
      *         |θ *     |*
      *        O*        |*
      *         |        |*
                |        |
       *        |        *
        *       |       *|
          *     |     *  |
      - - - - * * * - - -+- - - - m
                D        C

$AD$ is a diameter of circle $O$ and is tangent to line $m$ at $D.$
If $AD=1$ and $BC \perp m$ ,
. . a) show that: . $AB + BC \:= \:1 +\cos x - \cos^2\!x$

$\Delta OAB$ is isosceles with $OA = OB = \tfrac{1}{2}$

Then: . $x \,=\, \angle OAB \,=\, \angle OBA \quad\Rightarrow\quad \angle AOB \,=\,\theta \,=\,180^o - 2x$

Law of Cosines: . $AB^2 \:=\:OA^2 + OB^2 - 2(OA)(OB)\cos\theta$

We have: . $AB^2 \:=\:\left(\tfrac{1}{2}\right)^2 + \left(\tfrac{1}{2}\right)^2 - 2\left(\tfrac{1}{2}\right)\left(\tfrac{1}{2}\right)\cos(180^o-2x) \:=\:\tfrac{1}{4} + \tfrac{1}{4} - \tfrac{1}{2}[-\cos2x]$

. . $AB^2 \:=\:\frac{1}{2}+\frac{1}{2}\cos2x \:=\;\frac{1+\cos2x}{2} \;=\;\cos^2\!x \quad\Rightarrow\quad\boxed{ AB \:=\:\cos x}$

We see that: . $BC \:=\:AD - AE$

In right triangle $AEB\!:\;\;AE \:=\:AB\cos x \:=\:\cos^2\!x$

Hence: . $\boxed{BC \:=\:1 - \cos^2\!x}$

Therefore: . $AB + BC \:=\:\cos x + 1 - \cos^2\!x$

Quote:

b) What value of $x$ makes the sum $AB + BC$ a maximum?

We have: . $f(x) \:=\:1 + \cos x - \cos^2\!x$

Differentiate and equate to zero: . $-\sin x + 2\cos x\sin x \:=\:0$

. . Factor: . $\sin x(2\cos x - 1) \:=\:0$

We have: . $\sin x \:=\:0\quad\Rightarrow\quad x \:=\:0$ . This gives a minimum sum.

And: . $2\cos x-1\:=\:0\quad\Rightarrow\quad \cos x \:=\:\tfrac{1}{2} \quad\Rightarrow\quad \boxed{x \:=\:\frac{\pi}{3}}$

violetice · #3 $Old$ January 29th, 2009, 12:28 PM

A very pretty solution! thank you very much!

I also just noticed in part b, you could you the quadratic formula for 1-cosx-cosx^2 (and it'd basically be just -x^2-cosx+1) and find the maximum that way.

...or am i wrong?

thebomb551 · #4 $Old$ February 1st, 2009, 11:59 PM

No, he took the derivitive of the function and found the zeros, this is different than just finding the zeros...

violetice · #5 $Old$ February 2nd, 2009, 10:30 AM

No, I meant to not find the zeros, but find the maximum point of the parabola.

thebomb551 · #6 $Old$ February 6th, 2009, 11:34 PM

...To find a maximum, you take the derivative of your function and find the zeros of that funciton.