Another Trigonometry Word Problem (rather difficults)

• Jan 28th 2009, 10:57 AM
violetice
Another Trigonometry Word Problem (rather difficults)
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?

http://img168.imageshack.us/img168/6434/abcdcp3.jpg
• Jan 28th 2009, 07:37 PM
Soroban
Hello, violetice!

Quote:

Code:

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

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

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

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

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

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

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

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

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

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

Quote:

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

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

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

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

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

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

• Jan 29th 2009, 10:28 AM
violetice
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?
• Feb 1st 2009, 09:59 PM
thebomb551
No, he took the derivitive of the function and found the zeros, this is different than just finding the zeros...
• Feb 2nd 2009, 08:30 AM
violetice
No, I meant to not find the zeros, but find the maximum point of the parabola.
• Feb 6th 2009, 09:34 PM
thebomb551
...To find a maximum, you take the derivative of your function and find the zeros of that funciton.