Starting off, we were supposed to go onto some new stuff today but since none of us did our homework from yesterday, we ended up going through those homework questions instead.

Having already done (a) yesterday, we continued on with (b) and (c).

- Since the arctrig function is the argument of the outer trig function, we can make this equal to x
- We then took the argument of tanx (Opposite/Adjacent) and formed our triangle - Quick Note: To get a better visual of which Quadrant you're in, draw your triangle as if it were in that Quadrant. Most of the triangles drawn on today's slides didn't follow this but Mr. K said it was better to do it this way
- After making the triangle from the argument of tanx, you can now solve for secx
- Follow the same procedure to get (c)

These questions are a bit trickier, considering the fact that you have to remember your sine/cosine/tan(?!) dances.

- (a) is easy enough to do in our heads since they are easy angles that can be taken off the unit circle
- After finding the arctrig values within the brackets, take the sine of them and from there it should be smooth sailing

- (b) is not as simple as (a), since you don't have nice pretty unit circle values this time
- Like the previous question, you take the arguments of the outer function and make them equal to, in this case, since there are two arguments inside, we shall make one equal to x and the other equal to y
- Make your triangles for x and y
- You should now notice that you have enough information to complete your cosine dance! All that's left is to input the values and then do basic algebra. Tada!

- The part in the box was just some small explanation on the different between an arctrig and its inverse.
- As you can see, if the whole expression is to the exponent negative 1, then that is the reciprocal
- Do not confuse the reciprocal with the arctrig

- Same deal as (b) with this question.
- No need to make a triangle with arctan since taking the tan of an arctan will just leave you with the tan of that value anyways!
- After making your triangle for y and finding the value for tany, you can now use the tangent law thing (I've forgotten the name for these things...addition law?) and solve. The part on the left side was just Mr. K building his tan law thing.

- This question basically gives away to you what quadrant x is in. Since it says that x > 0, that means that it will be in quadrant 1 for both questions
- From then on, its mostly the same procedure as the last few questions. Take the inside argument and make it equal to y (x is being used this time so use y instead...or whatever variable you want)
- Make your triangles
- Solve
- ???
- Profit!!

^{2}as my example:

- Take the nearest multiple of 10 of the number, n, you are trying to square, in this case its 10
- Find the value needed to get back to 12, in this case its 2.
- We then add 2 to 12 to get 14.
- Multiply your (n+value) by (n-value). So from our example we should have (10)(14) = 140.
- Now go back to the value we needed and square that. It should be easy since its only a one-digit number. From our example, we get 4.
- Add the number you just got to the product we made and voila! We got our answer!

That was all for today's class. Homework is to graph the three arctrig functions from memory. Don't cheat! As for the last scribe for this cycle, that shall be Yinan! Now have a nice day as I go back to rest on my back and maybe kill some more brain cells with my super long naps.

## 3 comments:

I love the trick about squaring numbers!

I know, right? We were all amazed in class too.

There's also a trick for squaring 3-digit numbers and I believe benofschool posted a video showing how to do it. Just click on the OnMyMind tag on the side and look for it. :D

I found it at http://www.youtube.com/watch?v=I9t-gYnPNaw. It seems to me to be a fairly straightforward application of the form of a perfect square of a binomial. Using Difference of Squares backwards seemed smarter to me.

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