Section+1.4+Solving+Trig+Equations

=Section 1.4 Solving Trig Equations= 8/24/11

A review of Even and Odd Functions
math {}$Yesterday we looked at the graphs of $y=\sin\left( x \right)$ and $y=\cos\left (x \right )$, so today we reviewed the de{}finitions of even and odd functions. We looked at both graphical and algebraic de{}finitions. A function is even, if the graph of the functions has reflection symmetry through the y-axis. A function is odd, if the graph of the function has rotational symmetry about the origin. Algebraically, a function is even if $f(-x) = f(x) \forall x\epsilon\mathbb{R}$. Similarly, a function is odd if $f(-x)=-f(x) \forall x \epsilon \mathbb{R}.

From the graphs that we examined yesterday, we are able to conclude that $y=\sin\left ( x \right )$ is an odd function and that $y=\cos\left (x \right )$ is an even function. Therefore we can also conclude that $\sin(-x)=-\sin(x)$ and $\cos(-x)=\cos(x)$

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Minds in Action Dialogue
math {}$ In Class we read the Minds in Action Dialogue on page 18. In this dialogue Sasha and Tony are attempting to find solutions to the equation $\sin(x) = \frac{1}{2}$. They use calculators, unit circles, and memorized angles to find the two solutions between $[0, 2\pi)$ As they worked through their solution they brought up some important points that we need to keep in mind as we are solving equations on our own.

Sometimes it is helpful to substitute and alternate variable for the expression $\sin(x)$ or $\cos(x)$.

Sometimes it is appropriate to use the $\sin^{-1}$ button on the calculator and to have a "big honking" decimal as the answer; however, sometimes it is possible to give an exact answer and in those cases we want to give the exact answers.

We always need to be careful about what kinds of answers we are being asked for. The equation $\sin(x)=\frac{1}{2}$ has two solutions on $[0,2\pi)$, but the graphing calculator would supply only one of them (in an inexact form). We need to be able to be smarter than our calculator and know to look for the second solution. If we are asked for all solutions we need to remember to write them as $\frac{\pi}{6}+2n\pi, n\epsilon\mathbb{Z}$ and $\frac{5\pi}{6}+2n\pi, n\epsilon\mathbb{Z}$. It is even more complicated if we are asked to solve for all solutions between $-2\pi$ and $4\pi$.

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Sample Problems worked in Groups in Class.
math {}$ The first problem we looked at was $-8\cos(x)+5=11$. The first part of the algebra was easy.

$$-8\cos(x)+5=11$$ $$-8\cos(x)=6$$ $$\cos(x)=-\frac{3}{4}$$

Then the tricky part came. First we had to decide with $\frac{3}{4}$ was a value that we were supposed to have memorized (it wasn't). Next we had to look at $\cos^{-1}(-\frac{3}{4}) = 2.419$. This clearly gave us an answer to the equation. Now we had to find the second answer. Using the symmetry of the unit circle or the symmetry of the graph of $f(x)=\cos(x)$, we see that the second answer between $0$ and $2\pi$ is $2\pi-2.419$.

Next we looked at the equation $10\sin^2(x)-3\sin(x)=4$. This equation seemed easier if we subsituted $H=\sin(x)$. Once we substituted we had a simple quadratic $10H^2-3H=4$. Moving everything to one side of the equation $10H^2-3H-4=0$ allowed us to use the quadratic formula to find that $H=\sin(x)=\frac{4}{5}$ or $H=\sin(x)=-\frac{1}{2}$. With the second equation we have to use our brain (not our calculator) to find the solution. I know that the reference angle is $\frac{\pi}{6}$ and that the answers should be in the third and fourth quadrant since the $\sin(x)$ was negative. Therefore the two solutions in $[0,2\pi)$ are $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$. I have to use my calculator for $\sin(x)=\frac{4}{5}$. My calculator gives me that $x=.927$ and the symmetry of the unit circle or the symmetry of the graph of $f(x)=\sin(x)$ gives me that the second solution is $\pi-.927$.

One question that came up is how do you know whether to subract the calculator answer from $\pi$ or $2\pi$ to find the second solution between $0$ and $2\pi$.

A final question we looked at in class dealt with using the Pythagorean identity.

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