Scribe Post !

Trigonometric Identities

Okay everybody, Good evening ! How are you guys? Well, I hope you all are doing fine, this is my second scribe post, and a brief summary of how our math class went was, we had a quiz first thing, corrected it, showed us different ways on how to answer the questions, and also note that, ( those questions in the small quiz were in the old exams, which may come up in our final exams ). After doing our corrections on the quiz, then corrected our math homework, which was done the last night. Mr. K then corrected it, and he spoke fairly fast, it was probably because we were running out of time, because of the corrections made on the quiz.

Well, on the quiz, I felt I knew what I was doing, and when we corrected it, I, for one, had some minor mistakes, and errors, which caused me dearly, because I could of had a higher mark, but yes, I will learn from this mistake which is really the BEST part of learning. One of the questions that seemed tricky to me was number one, it was only tricky because I had not thought about the identities shown to us on the dictionary. The question on the quiz was like this:

Simplify: sin2θ/sinθ, now, by looking at that, do you think you know how to answer it? I bet you do, well here's the answer :

Simplify: sin2θ/sinθ, now from here, automatically you can go straight to 2cosθ, which is the answer. But just to show you how I got it, then here it is, because math, is, FUN!

= sinθcosθ+cosθsinθ/sinθ

= sinθcosθ/sinθ + cosθsinθ/sinθ

The sinθ's reduce, and you are left with 2cosθ, which was told earlier, is the answer.

Now, After we corrected those questions with Mr. K , we then went to our homework from last night, and corrected it with Mr. K. Here are some of the questions and answers from the hoemwork:

2. Simplify: tanθ/cscθ

3. Write an equivalent experession: cos(3x)(2x)+sin(3x)sin(2x)

5. Prove: 1/1-sinθ + 1/ 1+sinθ = 2tan^2θ+2

Question number 2 looks like this:

tanθ/cscθ

=(sinθ/cosθ )/(1/sinθ)

=(sinθ/cosθ)(sinθ/1)
=sin^2θ/cosθ

( now from here, this is the answer, but to make things more interesting it can turn out to this !

=sin^2θ/cosθ

=(sinθ/cosθ)(sinθ)

=tanθsinθ

For question number 3, this was how it was solve:

cos(3x)cos(2x)+sin(3x)sin(2x)

= cosx

(it's as easy as that !)

Finally but not least, the last question:

1/1-sinθ + 1/1+sinθ = 2tan^2+2

=1+sinθ+1-sinθ/ 1-sin^2θ

=2/cos^2θIV>

=2sec^2θ

Now, looking at that, you wonder, how did you get the " +2 ", well have no fear, the answer is here. The answer is, if you can't solve it like that, then go to the other side and try to simplify that and see what you get. This is what we got:

1/1-sinθ + 1/1+sinθ = 2tan^2+2

= 2(tan^2θ+1)

= 2sec^2θ

It's just like that, no hassles or nothing, just a small tad bit of thinking, well actually a lot of thinking, if you haven't seen these kinds of problems yet. Also note that, in those kinds of proving questions do not forget to (*NOT PASS THE WALL OF CHINA) , which is the big long line that seperates the two terms telling you that you cannot move anything from either side.

Now this is oliver_796 ABOUT to sign off, but wait, what's that? I here a .. I hear a .. I hear a .. MESSAGE. *DING-DONG. YES? The message is: PLEASE CONTINUE TO ASK QUESTIONS IN CLASS, BECAUSE IT'S A SURE GOOD WAY TO SHOW THE TEACHER THAT YOU WANT, AND ARE EAGER TO LEARN. BY SHOWING THIS, THE TEACHER WILL THERFORE MAKE YOU LEARN, WHICH IS VERY GREAT. =) , okay dokes? Continue to study with your study groups! Continue to perform random acts of kindness ( quote by Mr. K ) and go out there and learn ! Until next time, Take care everybody.

THE NEXT SCRIBE IS ANH! HAVE FUN!