Perfect Sixes
Using only addition, subtraction, multiplication, division, factorials, and square roots inserted between digits, find a way to make each of the following statements true:
1 1 1 = 6
2 2 2 = 6
3 3 3 = 6
4 4 4 = 6
5 5 5 = 6
6 6 6 = 6
7 7 7 = 6
8 8 8 = 6
9 9 9 = 6
0 0 0 = 6
(ex: 2 + 2 + 2 = 6)
Answer at bottom of the page.
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We have some grasp now over the unit circle. By placing three different right triangles in the first quadrant of the unit circle, we can use our knowledge of their side length ratios to come up with the coordinates of where those triangles intersect the circle. After that, we can apply the idea of reference angles and how the coordinate plane works to generate coordinates for the other three quadrants.
The conceptual understanding of the Unit Circle at this level really is ~80% of the battle when it comes to trigonometry in this class. However, when we first started talking about angles, we discussed them in terms of degrees. We also discussed how one theory as to why there are 360 degrees is due to ancient calendars that tracked the sun (where one cycle was 360 data points (we know it is 365 now, obviously)) and how 360 has a lot of factors. Degrees are great because we've become accustomed to them over our academic careers and we have a pretty intuitive ability to quantify something like a 45° angle or even a 30° angle (if I asked you to hold out your thumb and index finger at a 30° angle, I'd bet you could do it within a few degrees of accuracy). However, this entire description of angles might be rooted in something that is very specific to our planet and how it interacts with the sun. A more elegant and universal mathematical way to quantify angles is something called a Radian. Person reading this right now, from this point in your math career onwards, radians will be the preferred way we describe angle measures. Here is a fantastic video explaining what a radian is from a guy with a lot of energy who holds his marker in a strange way.
A radian is the measure of an the central angle formed between two rays of a circle that subtend an arc with a length of exactly one radian. Like a degree, a radian is the same measure for any circle, regardless of radius. This seems counter-intuitive because a larger circle will have a longer radius, but proportionally the measure of a radian will remain constant.
Dividing a circle into quarters and sixths gives us the same subdivisions with radians as it does with degrees. For example, while 1/6 of a half circle (180°) is 30°, 1/6 of a half circle (π) is just π/6. In many ways, radian values are actually much more convenient and easier to use. It should be your goal as a maths student studying trigonometry to become just as comfortable using radians as you are with degrees. It is of course possible to convert between the two, but this step is largely unnecessary and can lead to wasting valuable time during assessments.
The last pieces of the puzzle for the basics of the unit circle involve negative angles and angles greater than 360° (or 2π). When we discussed reference angles we effectively concluded that instead of knowing the sine of an obtuse angle, like 150° for example, we could say that 150° is just a 30° reference angle in Quadrant II, and that sine (y-value) is positive in Quadrant II, so sin(150°) is the same as sin(30°), or 1/2 (there is more to this story, but we will get into that later). For negative angles and angles greater than 360°, we can apply a similar way of thinking. For example, a 390° is going to have the same y-value as a 30°, we've just completed one full rotation (360°) before going that extra 30°. Because of this, when we have angles greater than 360° (or 2π), we can just subtract 360° from that angle to get an angle with equivalent trigonometry values.
We briefly discussed negative angles when learning about co-terminal angles, but negative angles are simply angles that are drawn from the standard position but in a clockwise direction (as opposed to counterclockwise). So an angle of -30° would be the same thing as a 330° angle. An angle of -120° would be the same thing as 240°. Notice that there is a difference of 360° between these. For radians, there would be a difference of 2π between a negative angle and its corresponding positive angle (or a multiple of 2π if the negative angle is really big, like 25π/3 or something). Here's an example. If we wanted to know the sine of a 25π/3 angle [sin(25π/3)], we could just keep subtracting 2π until we were within the unit circle. 25π/3 - 6π/3 = 19π/3, which is still outside of the unit circle, so keep going. 19π/3 - 6π/3 = 13π/3 - 6π/3 = 8π/3 - 6π/3 = 2π/3. Finally, after subtracting 2π four times, we arrive at 2π/3. This means that for all intents and purposes, a 25π/3 angle is equivalent to a 2π/3 angle. We can then evaluate sin(2π/3), which is a 60° (π/3) reference angle in quadrant to, as √3/2.
Before moving on to get started on what is really the meat behind trigonometry, there are two important circle-related measurements that are useful to know: arc length and sector area. Introducing the measurement of radians makes these calculations extraordinarily simple, and they are also very useful in all sorts of different real-world applications. These calculations can also be accomplished when using degrees, but the procedure is just a bit more complicated.
The arc length on a circle is defined as the length of the arc subtended by two rays. For example, if there was a circle centered at the origin and one ray was extended along the positive x-axis and another was extended at a 135° in standard position, the arc length is the portion of the circumference of the circle between the two points of intersection. It should be clear that this is really going to depend primarily on the radius of the circle.
The sector area of a circle is the area of a section of the circle confined by two rays and an arc. A slice of pie or a slice of pizza would be a good example of what a sector is. Actually, pizza value is really a great application for sector area (and circle area in general)!
So for arc length we see that the length of an arc on a circle is just some portion of it's circumference. When working with radians, if we had a central angle of one radian, then the arc length would just be one radius long. This direct relationship is described by the equation s=rθ (when θ is in radians). When working with degrees, if a full 360° was one circumference of the circle (2πr), then some fraction of 360° is that same fraction of the circumference of the circle. In other words, s=(θ/360°)*2πr, or more commonly s=(θ/180°)πr.
For sector area we can also think about portions of a whole. In radians, the full 2π interior angle (the entire circle) has a radius of πr², so if we had some fraction of 2π, we would have that same fraction of the area. In other words A=(θ/2π)*πr². In this equation the π cancels out so we are left with the common way to represent sector area of A=(1/2)r²θ. Following a similar process gives us the formula for the area of a sector when working with degrees, A=(θ/360°)πr².
These two measures, arc length and sector area, come up surprisingly often in real world calculations and are important to know. When it comes to the study of Trigonometry (trig=triangles nometry=measurement) perhaps the most important thing to understand are trigonometric functions. We will start by looking at the first trigonometric function, sine.
The sine function is the the relation between the length of the opposite site of the angle and the angle. As we draw a triangle in different positions within the unit circle, the y-value of the intersection between the triangle and the unit circle repeats itself over a regular interval of 2π, which is known as a period. Because of this, we call sine (and all trig functions for that matter) a periodic function.
We will dive deeper into other trig functions and how to go about translating trig functions and what that looks like next week.
(note: there are other valid solutions!)
(1 + 1 + 1)! = 6
2 + 2 + 2 = 6
3×3 - 3 = 6
√4 + √4 + √4 = 6
5/5 + 5 = 6
6×6/6 = 6
7 - 7/7 = 6
8 - √(√(8 + 8) = 6
√(9×9) - √9 = 6
(0! + 0! + 0!)! = 6