## Monday, February 11, 2008

### Trigonometric Equations

Alrightey I finally got home. Sorry for the late post had some technical difficulties with the computer also. Anyways here's the recap of what we did on todays slides.

Slide 2:

Here we see the reciprocal trig functions we learned about yesterday. Just for review, the new trig functions we learned were cosecant, secant, and cotangent. Mr. K emphasized that most kids go wrong when they think that cosecant is with cos and secant is with sin. So make sure to not do that! Anyways as it says on the slide, we get cosecant by 1/sin(theta), secant by 1/cos(theta), and cotangent by 1/tan(theta). Or we can just change the numerator with the denominator to save time. As I learned today, don't switch any signs. If it's negative in the denominator it will still be negative when it becomes your numerator when you flip the fraction.

Slide 3:

Ok the question here was a bit of a review from the questions we've been doing for the past few days. First we must find the hypotenuse of the big triangle. We do that by plotting the point (9,-40), which are your x and y values. Then we must find the cos and sin values. We do this by finding the angles adjacent to theta over the hypotenuse (for the cos), and for sin we put the angle opposite of theta over the hypotenuse. The cos and sin value you get will be the coordinates for theta! As for the second question on the slide, just list the cos, sin, tan, secant, cosecant, and cotangent values. Remember for the reciprocal trig functions, just flip the fraction.

Slide 5:

We've been practicing how to find the exact values with questions like these on this slide. Doing these questions are a lot easier if you know the values around the unit circle. K i'll make this brief, all you have to do is substitue the exact values in and simplify!

Slide 6:

Oh yes I remember this slide from this morning. Mr. K challenged our mathematical minds to do these questions. But like he said there's a method to his madness or something like that. I think we all know how to do these kinds of questions. We just isolate x or factor.

Slide 7:

We now see what Mr. K was trying to do by starting us off easy. We do exactly what we did on the last slide. We simply isolate sinx and get the value of 1/2. Now what on the unit circle do we know has a y value of 1/2? pi/6, and 5pi/6 of course.

Slide 8:

1+2cosx = 5cosx

The question gets a bit tougher but we still do the same things. Isolate cosx and it will equal 1/3. Now we take our trusty calculators and press 2nd fnc cos then 1/3. Make sure to be on radians. We then get the value 1.2309. But there is another value because cos is also positive in quadrant 4. How do we get the value in quadrant 4? Just put in your calculator 2pi - 1.2309 to get the other value of x, which is 5.0522 the related angle. K a quick summary, if looking for related angle in q2 you take the reference angle and subtract it from pi. If looking for related angle in q3 you take the reference angle and add it to pi. And that's where we ended!

Hope you guys understood my rambling. Oh yeah next scriiibe is...........................JAMIE since I took your turn today. KK goodniight everyone *faints*.