What is the sin of 90 degrees, Tutoring Precalculus

We looked at coordinates and their rays from the origin and the six trigonometric functions. To do that, we used SOHCAHTOA and to remember the other three functions, you can use the word ‘chocolate’ since CHO, cscx=H/O is the reciprocal of sinx.
To visualize the ‘triangles’ for the two directions on the x-ax and the two directions on the y-axis, I’ll draw something like 85 degrees instead of 90 since it’s fairly close and then label one side of the triangle as zero and the other two as equal values (1 usually works well).


Tutoring Precalculus, Focusing on the Unit Circle and 30-60-90 and 45-45-90 Triangles

We started by focusing on the unit circle. There are two main triangles that form the basis for a lot of what they are doing in the class. 30-60-90 and 45-45-90. Knowing the ratios of the sides for those triangles allows you to more easily do much of the work. So we looked at those triangles in various positions.


Knowing what sinx and cosx look like graphed can be useful as well.

We also looked at how triangles with 85 degrees and 5 degrees are similar to 90 and 0 degrees and how that can be useful.

Tutoring Precalculus, Trigonometric Functions and Properties

We started by looking at a problem with triangles. Two smaller triangles made up a larger triangle. The larger triangle was a right triangle. Basically, we started finding different angles, and use the law of sines at one point. Then we could use a trigonometric identity.

For another problem, it helped to create a new variable.

We spent a little time on graphs of trigonometric identities and also understanding the domain and range. That included looking at the amplitude and expansion and compression of functions.

Tutoring Precalculus, Why Inverse Trigonometric Functions have Restricted Domains

The first thing we saw was some points taken off for answers on a test because they were not in the right range.

It would seem like there should be multiple answers for something like arccos(1/2), but in order for the inverse trigonometric functions to actually be functions (which has a few requirements) the domain is restricted.

sin(x) looks like this, it keeps going in both the positive and negative directions along the x-axis


arcsin(x) looks like this,


If it kept going farther either above the top or below the bottom, it would fail the vertical line test and not be a function. I think that may be the simplest way to think about why the domain is restricted.

The main change to his approach should be to put the triangles for trigonometric functions on a set of axes properly. Should make things easier to figure out and more accurate. Often he correctly used a 30 60 90 triangle for example, but sometimes not in the correct orientation which caused problems.

Knowing what the oscillating functions look like on a graph would be good too as well as having a few more of the values memorized, but also understanding them in several ways.

Correcting a Problem with Bases and Exponents

Thanks for posting the photo of your work George, that makes it much easier to follow. Especially because the syntax in your typed version of the equations is not quote right.

If you followed exactly what you typed, it would be like this:

literal equations

Because of missing parentheses. I thought that was probably not what you meant.

However, with the photo, it’s clear what the original problem is and what to fix:

george work

And it’s also good to see your work.

Essentially, in the problem you have the same base for all the terms (2/5) and similar exponents.

For the first variation, you have an additional negative sign in the exponent of the left term.


The blue is the original, orange is simplified.

For the second variation, you have an additional negative sign in the exponent of the right term.

two fifths varitaion 2


Left Side:

Your work is mostly right, but here’s the first mistake I see on the left side:

You state that the exponent on the left term should be greater than the exponent on the right term. That would be true if the base was greater than 1, but it’s (2/5). If for example, you square (2/5) you get 4/25 which is a smaller number than 2/5.

Therefore you want the base on the left term to be less than the base on the right term. So switch the direction of the inequality sign.

-4x – 2 < 3x – 18

7x > 16

x > 16/7

Right Side:

I would recommend keeping the base the same for all the terms, 2/5 for this problem. At least for me, it’s easier to work through that way.

It seems like on the right side of the board, you turned the 2/5 into 5/2 which you would do if you had the exponent of -1 like you did on the left side of the board, but you don’t have that exponent of -1 in this variation. So it should be 2/5 as the base, not 5/2 for the left term.

The other difference for the right side of the board is the negative sign in exponent of the right term (in front of the 3).

You basically turned the base of the right term into 5/2, which is okay.

But you added a negative sign to the exponent on the left term, which isn’t there, to turn that base into 5/2 as well.

Let’s say we do go with 2/5 as the base for both terms.

Again, if the base is less than 1, a smaller exponent will result in a larger number.

4x + 2 < -3x + 18


7x < 16

x < 16/17

Obediah Jones was correct in one of his comments to you, although

“The fact that they stemmed from the same inequality is why the value of x is either less than or greater than the exact same number, 16/7 16/7.”