At the start of the week, we learned all about using the formulas we got last week. The formulas are used for many things, but primarily for solving trig equations. For example, to prove an equation like sin(8x) = 2sin(4x)cos(4x), you change sin(8x) to sin(2(8x)) and then you can use the double angle formula of sin(2x) = 2sin(x)cos(x) to get the other side of the equation. These formulas also assisted us in multiple ways later in the week. Next, we learned about inverse trig functions and how to solve them, such as sin(sin-1(π/6)) ends up equaling π/6 because the sine and the inverse sine cancel each other out. However, it's less easy with other types of equations, such as sin(cos-1(1/2)), in which case we have to realize that the inverse cosine represents the theta of the sine. We think about what angle would produce a cosine of 1/2, and we get π/6, and we know that the sine of that is √(3)/2, so that is our answer. This can also be used to solve simpler equations, such as one where the inverse function is the only function.
We also learned about graphing inverse trig functions, where the original non inverse graph must satisfy both the horizontal and the vertical line tests. Therefore, the domain and the range switch positions, with several changes to make sure the graph passes both tests(such as in the case of cosine, where we must examine a period of half of what it normally would be). Essentially, the period and the amplitude switch places in the function. Then, we learned about rewriting inverse trig functions as algebraic expressions, such as in the case of tan(cos-1x), where we draw a triangle using the information we have(cos θ = (x/1)), and then find the missing side and solve the outer trig function. This results in an answer of √(1-x2)/x.
Then we learned about solving trig equations, in a two step process: 1. Isolate the trig function, 2. Solve and apply to the period. For example, in the case of 2sin(x) -1 = 0, we can easily isolate sin(x) = 1/2, so we know that x is either equal to π/6 or 5π/6. However, we also have to consider that these two points will repeat along the sine wave, so our final answer is x = π/6 + 2kπ, x = 5π/6 + 2kπ where k represents the number of periods gone through. We also learned to make sure that, in solving it, we keep the same period that it originally had, because otherwise we might accidentally remove some answers from what we get.
Finally, we learned about Polar Coordinates, which determine a point using a radius and an angle as opposed to its vertical and horizontal position. For example, polar coordinates would determine a point in the form of (5, π/2) as opposed to (0, 5). We also learned about some of the relationships between polar and rectangular coordinates, such as x = rcos(θ), y = rsin(θ), r2 = x2 + y2, and tan(θ) = y/x.
Tuesday, April 13, 2010
Tuesday, March 30, 2010
Formulas for Success
The week began with a short talk about stretching trig graphs. For sine and cosine, lessening the period presses the graph closer against the zero, but for tangent, if you make the period smaller, the graph becomes smaller from both ends. After this, we begun on Chapter 8. We learned about the trig identities, some of which we already knew: the reciprocal properties(csc = 1/sin, etc), the pythagorean identities(sin2 + cos2 = 1 and such), and the even and odd properties(sin(-x) = -sin(x)). We also learned about some new identities, known as cofunctions. Cofunctions state that sin((π/2) - u) = cos u. This works both ways, and additionally, cotangent and tangent are linked in cofunctions, as are cosecant and secant.
Using these identities, we practiced verifying equations including the trig functions. We learned two general rules for doing this: to always start with the most complex side of the equation, and to try and convert to sine and cosine wherever possible. For example, in the case of the equation (sin x + cos x)2 = 1 + 2(sin x)(cos x), we break the more complex left side down into (sin x + cos x)(sin x + cos x), which we then multiply out into sin2 x + cos2 x + 2(sin x)(cos x), and since sin2 x + cos2 x is equal to one, we get the right side of our equation.
Next we learned about the addition and subtraction formulas, which allow us to find the exact value of any angle, not just multiples of 30 or 45. For example, to find the sine of 15 degrees, we would use the subtraction formula sin(x-y) = (sin x)(cos y) - (cos x)(sin y) and since fifteen is equivalent to 45-30, we can express sin(15) = (sin 45)(cos 30) - (cos 45)(sin 30), which comes out to being √(6)-√(2)/4.
Then we learned about how to make a graph involving two trigonometric functions by expressing them as only one, through the equation Asin(x) + Bcos(x) = ksin(x+ϕ) where ϕ is an angle and all of the following are true: k = √(A2 + B2), cos ϕ = A/k, and sin ϕ = B/k. By using this function, we can graph two separate functions as one graph. Lastly, we learned a lot of formulas: double angle formulas, going from power 2 to power none, half angle formulas, product to sum, and sum to product. We briefly touched on how to use these with the promise that we'd learn more about them next week.
Using these identities, we practiced verifying equations including the trig functions. We learned two general rules for doing this: to always start with the most complex side of the equation, and to try and convert to sine and cosine wherever possible. For example, in the case of the equation (sin x + cos x)2 = 1 + 2(sin x)(cos x), we break the more complex left side down into (sin x + cos x)(sin x + cos x), which we then multiply out into sin2 x + cos2 x + 2(sin x)(cos x), and since sin2 x + cos2 x is equal to one, we get the right side of our equation.
Next we learned about the addition and subtraction formulas, which allow us to find the exact value of any angle, not just multiples of 30 or 45. For example, to find the sine of 15 degrees, we would use the subtraction formula sin(x-y) = (sin x)(cos y) - (cos x)(sin y) and since fifteen is equivalent to 45-30, we can express sin(15) = (sin 45)(cos 30) - (cos 45)(sin 30), which comes out to being √(6)-√(2)/4.
Then we learned about how to make a graph involving two trigonometric functions by expressing them as only one, through the equation Asin(x) + Bcos(x) = ksin(x+ϕ) where ϕ is an angle and all of the following are true: k = √(A2 + B2), cos ϕ = A/k, and sin ϕ = B/k. By using this function, we can graph two separate functions as one graph. Lastly, we learned a lot of formulas: double angle formulas, going from power 2 to power none, half angle formulas, product to sum, and sum to product. We briefly touched on how to use these with the promise that we'd learn more about them next week.
Tuesday, March 23, 2010
Harmony
The week started with a discussion on 'harmonic motion', which describes "the displacement of an object at a given time". This equation is given as y(t) = asin(wt) or y(t) = acos(wt), where w represents a variable, and is used in finding the period(given as 2π/w), t represents time, and a represents amplitude. We also learned about a new facet of graphs of harmonic motion- frequency. Frequency is the inverse of period, and represents how many periods occur per unit time.
We learned this as being represented by the 'ferris wheel' example- a person's height while riding on a ferris wheel can be described through harmonic motion. If the bottom of the ferris wheel is 2 feet off the ground, and it has a radius of 8 feet, and we know it rotates twice every minute, an equation could be set up like: h(t) = 10 + 8sin(4πt) where t is in minutes. 4π is derived from the frequency equation, which would be w/2π = 2, so w = 4π.
Another, similar scenario is a mass on a spring. The mass is pulled down or pressed up by a certain amount, and then released. This is almost exactly like the ferris wheel in the nature of setting up an equation, but in real life scenarios, we must consider dampening motion, which means that each time the mass goes up and down, it will lose amplitude. A dampened spring has an equation in the form of y(t) = ke-ctcos(wt) Where c is the spring dampening constant, and k is the initial amplitude.
At this point in the week, we learned about what was going to be on the test, and later on, took the test. This made it so that what we learned in class was much shorter than in weeks where we had both days, and so my blog post is shorter than usual.
We learned this as being represented by the 'ferris wheel' example- a person's height while riding on a ferris wheel can be described through harmonic motion. If the bottom of the ferris wheel is 2 feet off the ground, and it has a radius of 8 feet, and we know it rotates twice every minute, an equation could be set up like: h(t) = 10 + 8sin(4πt) where t is in minutes. 4π is derived from the frequency equation, which would be w/2π = 2, so w = 4π.
Another, similar scenario is a mass on a spring. The mass is pulled down or pressed up by a certain amount, and then released. This is almost exactly like the ferris wheel in the nature of setting up an equation, but in real life scenarios, we must consider dampening motion, which means that each time the mass goes up and down, it will lose amplitude. A dampened spring has an equation in the form of y(t) = ke-ctcos(wt) Where c is the spring dampening constant, and k is the initial amplitude.
At this point in the week, we learned about what was going to be on the test, and later on, took the test. This made it so that what we learned in class was much shorter than in weeks where we had both days, and so my blog post is shorter than usual.
Tuesday, March 16, 2010
Sketchers
The week started with more on finding the terminal point- by taking the sin and cosine of the angle, which is simple if it is a multiple of 30 or 45 and you have memorized the 30-60-90 right triangle and the 45-45-90 right triangle. For example, an angle with a measure of 120 would have a reference angle of 60, so its terminal point would be at (-1/2, √(3)/2) since cosine is negative and sine is positive in the second quadrant. We then learned about even and odd functions, cosine and secant being even and all the rest being odd. For an even function, the negative angle yields the same result as the positive one, so sec(-x) = sec(x). For the rest, if the angle is negative, the result is also negative: tan(-x) = -tan(x).
Next, we started learning about trigonometric graphs, which are the graphs of the trig functions. sine and cosine are periodic graphs, which means they always come back to the same point and then repeat, after one full cycle. We briefly touched back over translations(sin (x) + 3 moves the graph up three, for example) and then we started learning about the general equation for graphing a trig function. This is y = a sin(kx) where a is the amplitude, (how high the sine or cosine wave goes, and how low it goes below zero) and k is a constant which is used to find the period. For sine and cosine, the period is 2π/k, and is essentially how long one cycle of the graph is. So an equation like 3sin(2x) would indicate a wave that starts at zero, goes up to 3 and then drops back to zero when it reaches π/2 on the x axis, then drops to -3, and rises back up to zero once it reaches π on the x-axis, then repeats. The bigger k is, the more the graph is horizontally compressed. If k is below one but above zero, it stretches the graph horizontally.
Then we learned about the phase shift, represented by b, and present in the equation for graphing a trig function. The equation, when you consider the phase shift, is y = a sin (k(x-b)). The phase shift essentially just shifts the entire graph left or right. If it is positive, the equation shifts to the right, and if it is negative, the graph shifts to the left(the variable itself- because there is a minus sign before it, a negative phase shift would be indicated as (x+b).) So if I had an equation such as y = 4 sin (2x + 1/2), my graph would start at -1/4, and go to (-1/4 + π), going vertically up to 4 and then down to -4. As shown in the example just given, oftentimes it is necessary to factor out the k from the x and the phase shift. Finally, we learned about some of the differences among graphing the trigonometric functions. For example, tangent and cotangent use π/k for their period, rather than 2π/k, and start at -π/2k, instead of zero. Cosecant and secant are simply the opposites of sine and cosine, with the difference that while cosecant never actually touches the sine graph, secant briefly touches the cosine graph.
Next, we started learning about trigonometric graphs, which are the graphs of the trig functions. sine and cosine are periodic graphs, which means they always come back to the same point and then repeat, after one full cycle. We briefly touched back over translations(sin (x) + 3 moves the graph up three, for example) and then we started learning about the general equation for graphing a trig function. This is y = a sin(kx) where a is the amplitude, (how high the sine or cosine wave goes, and how low it goes below zero) and k is a constant which is used to find the period. For sine and cosine, the period is 2π/k, and is essentially how long one cycle of the graph is. So an equation like 3sin(2x) would indicate a wave that starts at zero, goes up to 3 and then drops back to zero when it reaches π/2 on the x axis, then drops to -3, and rises back up to zero once it reaches π on the x-axis, then repeats. The bigger k is, the more the graph is horizontally compressed. If k is below one but above zero, it stretches the graph horizontally.
Then we learned about the phase shift, represented by b, and present in the equation for graphing a trig function. The equation, when you consider the phase shift, is y = a sin (k(x-b)). The phase shift essentially just shifts the entire graph left or right. If it is positive, the equation shifts to the right, and if it is negative, the graph shifts to the left(the variable itself- because there is a minus sign before it, a negative phase shift would be indicated as (x+b).) So if I had an equation such as y = 4 sin (2x + 1/2), my graph would start at -1/4, and go to (-1/4 + π), going vertically up to 4 and then down to -4. As shown in the example just given, oftentimes it is necessary to factor out the k from the x and the phase shift. Finally, we learned about some of the differences among graphing the trigonometric functions. For example, tangent and cotangent use π/k for their period, rather than 2π/k, and start at -π/2k, instead of zero. Cosecant and secant are simply the opposites of sine and cosine, with the difference that while cosecant never actually touches the sine graph, secant briefly touches the cosine graph.
Monday, March 8, 2010
Those are the laws
We started the week by learning about how to find one trig function based on another, using the trig substitutions we learned last week. For example, if you know that sin θ = 3/4, then from tan θ = sin θ/cos θ and cos2θ = 1 - sin2θ you can determine that tan θ is equal to approximately 1.134. For determining whether answers were positive or negative, we learned about using the quadrant, or figuring out what quadrant, the value given is in. As we learned last week, all the functions are positive in Quadrant I, sin and csc are in quadrant II, tan and cot are in quadrant III, and cos and sec are in quadrant IV.
Next, we learned about the law of sines, used for solving triangles. The law is that sin A/a = sin B/b, so if you know an angle and the side opposite it, and any other measurement in the triangle, you can solve the triangle. For example, if I knew that one of the angles in a triangle was 50 degrees and the side opposite it was 20 ft long, and I also knew that another angle in the triangle is 20 degrees, I could use the law of sines: sin 50/20 = sin 20/b, 20(sin 20)/sin 50 = b. However, we learned that when you use the law of sines to find an angle, there is a chance you will get an 'ambiguous case', in which there are multiple possible answers. The general rule that we were given in this case is if solving for an angle using the law of sines, if you get a number greater than one for sine, there is no solution, and if you get a number smaller than one for sine, then there are two possible solutions: the angle you get, and its supplement. The law of sines is usable only if you know two angles and a side, or two sides and an angle opposite one of them(ASA, SAA, SSA).
The other two possibilities if you know three components of a triangle are SAS and SSS, which can be solved using the law of cosines. The law of cosines states, by the definition from class: The square of any side of a triangle is equal to the sum of the squares of the other sides, minus double the product of these sides and the cosine of the included angle. That is to say, a2 = b2 + c2 - 2bc(cos A). This can be used if you know the internal angle to two sides, or if you know all sides of the triangle and no angles. The law of cosines, unlike the law of sines, has only one answer. We also briefly discussed Heron's Formula, a way for determining the area of a triangle if you know all of its sides. This formula is A = √(s(s-a)(s-b)(s-c)) where s is half the perimeter of the triangle.
Finally, we learned about the terminal point- the point which lies on the terminal side of an angle on the unit circle and meets the perimeter of the circle. To find the terminal point, you simply take the cosine of the angle for the x coordinate, and the sine for the y coordinate.
And now for my first question: Why is it that the law of sines produces two possible answers and the law of cosines does not? It seems like the angle and its supplement would still be valid if using the law of cosines to determine the angle. Thanks very much!
Next, we learned about the law of sines, used for solving triangles. The law is that sin A/a = sin B/b, so if you know an angle and the side opposite it, and any other measurement in the triangle, you can solve the triangle. For example, if I knew that one of the angles in a triangle was 50 degrees and the side opposite it was 20 ft long, and I also knew that another angle in the triangle is 20 degrees, I could use the law of sines: sin 50/20 = sin 20/b, 20(sin 20)/sin 50 = b. However, we learned that when you use the law of sines to find an angle, there is a chance you will get an 'ambiguous case', in which there are multiple possible answers. The general rule that we were given in this case is if solving for an angle using the law of sines, if you get a number greater than one for sine, there is no solution, and if you get a number smaller than one for sine, then there are two possible solutions: the angle you get, and its supplement. The law of sines is usable only if you know two angles and a side, or two sides and an angle opposite one of them(ASA, SAA, SSA).
The other two possibilities if you know three components of a triangle are SAS and SSS, which can be solved using the law of cosines. The law of cosines states, by the definition from class: The square of any side of a triangle is equal to the sum of the squares of the other sides, minus double the product of these sides and the cosine of the included angle. That is to say, a2 = b2 + c2 - 2bc(cos A). This can be used if you know the internal angle to two sides, or if you know all sides of the triangle and no angles. The law of cosines, unlike the law of sines, has only one answer. We also briefly discussed Heron's Formula, a way for determining the area of a triangle if you know all of its sides. This formula is A = √(s(s-a)(s-b)(s-c)) where s is half the perimeter of the triangle.
Finally, we learned about the terminal point- the point which lies on the terminal side of an angle on the unit circle and meets the perimeter of the circle. To find the terminal point, you simply take the cosine of the angle for the x coordinate, and the sine for the y coordinate.
And now for my first question: Why is it that the law of sines produces two possible answers and the law of cosines does not? It seems like the angle and its supplement would still be valid if using the law of cosines to determine the angle. Thanks very much!
Monday, March 1, 2010
Going 'round in circles.
This week, we started with learning a bit more about arcs and sectors, starting with using the formula from last week to solve for various parts of a sector. One important thing about using this formula is that the angle that creates the sector(the subtending angle) must be in radians for the formula to be true, so we learned about converting degree measures to radians- Simply multiply by π/180. We also learned about another formula used to finding the area of a sector: A = 1/2r2θ, used in the same method as the arc length formula. After that, we moved into circular motion, consisting of linear speed and angular speed. Linear speed is the actual speed of a point on a spinning object, and is defined by v = s/t, whereas angular speed is simply how fast something rotates in a circle, without basis on how big the circle is. Angular speed is defined by ω = θ/t.
Next we learned about the trigonometric functions: sine, cosine, and tangent, and their inverses: cosecant, secant, and cotangent. These functions define angles in a right triangle, with the sine of an angle being the opposite side of the right triangle over the hypotenuse, cosine being the adjacent side over the hypotenuse, and tangent being the opposite side over the adjacent. These functions can be used to solve any right triangle. For example, if you know how long the shadow of a building is, and what the angle of elevation to the sun is, you can determine the height of the building: tan θ = x/b, so x = b tan θ.
Finally, we learned about using the trigonometric functions on an angle in the unit circle. On an angle, you can name a point along the terminal side of the angle, called the terminal point. The x and y coordinates of this point can give you the radius of the circle by using the pythagorean theorem(x2 + y2 = r2, where r is the radius of the circle). When we are using radians, the radius of the circle is equal to one unit, so because of the definitions of sine, cosine, and tangent, the sine of an angle is equal to the y-component of the terminal side, and the cosine is equal to the x-component. This led us into a discussion about quadrantal angles, angles for which any one of the trigonometric functions are undefined, such as 90 degrees and 180 degrees. We then learned about the trigonometric functions being positive or negative in the four quadrants: All of them produce a positive answer in quadrant I, sine is positive ins quadrant II, tangent is positive in quadrant III, and cosine is positive in quadrant IV.
Next we learned about the trigonometric functions: sine, cosine, and tangent, and their inverses: cosecant, secant, and cotangent. These functions define angles in a right triangle, with the sine of an angle being the opposite side of the right triangle over the hypotenuse, cosine being the adjacent side over the hypotenuse, and tangent being the opposite side over the adjacent. These functions can be used to solve any right triangle. For example, if you know how long the shadow of a building is, and what the angle of elevation to the sun is, you can determine the height of the building: tan θ = x/b, so x = b tan θ.
Finally, we learned about using the trigonometric functions on an angle in the unit circle. On an angle, you can name a point along the terminal side of the angle, called the terminal point. The x and y coordinates of this point can give you the radius of the circle by using the pythagorean theorem(x2 + y2 = r2, where r is the radius of the circle). When we are using radians, the radius of the circle is equal to one unit, so because of the definitions of sine, cosine, and tangent, the sine of an angle is equal to the y-component of the terminal side, and the cosine is equal to the x-component. This led us into a discussion about quadrantal angles, angles for which any one of the trigonometric functions are undefined, such as 90 degrees and 180 degrees. We then learned about the trigonometric functions being positive or negative in the four quadrants: All of them produce a positive answer in quadrant I, sine is positive ins quadrant II, tangent is positive in quadrant III, and cosine is positive in quadrant IV.
Next, we were told about the reference angle, which is the angle between the terminal side of theta and the nearest x-axis. Taking any of the trig functions of this angle produces the same as angle theta! Lastly, we learned about trig identities:tan θ = (sin θ)/(cos θ), sin2θ + cos2θ = 1, tan2θ + 1 = sec2θ, and 1 + cot2θ = csc2θ. These formulas are used to determine all the trig functions based on just one or just a few, or to describe one in terms of another.
Friday, February 19, 2010
Okay, but how do you SOLVE one?
This week, we started off with learning to solve problems using the logarithms that we learned about last week. This begins with taking an equation that is either exponential or logarithmic, like 2x = 8 and taking the log of both sides, usually with a base that will make the equation easier to solve. For example, choosing base e when there is an e involved in the equation(to cancel it out), or choosing base 10 to eliminate a 10 or related number. At this point, you can use the laws of log to manipulate the equation and solve for the variable that you don't know. We did many problems based on this.
Two important applications of this are population growth and radioactive decay, which we learned about in class. Population growth is based on an identical formula to that of continuous interest: n(t) = n0ert and can be solved using logs. For example, if you wanted to find out how long it would take a population to reach 100 from 5, at a rate of 2%, you could use 100 = 5e.02t, then divide by 5 and take the natural log of both sides, ln 20 = ln e.02t, and then the laws of log allow you to simply get ln 20/.02 = t. Radioactive decay is similar, and can be solved in a way that is very much the same, except that it relies on the formula m(t) = m0e-rt.
After this we learned about angles, and the initial and terminal sides of one- the initial side being the side that starts the angle, and the terminal side being the one that ends it. We learned that the measure of an angle is essentially how open one is, measured in degrees or radians, and that when an angle's vertex is at the origin of the unit circle, and its initial side is on the positive x-axis, this is called standard position. We also learned that when the initial side of an angle goes counterclockwise to reach the terminal side, the angle is known as a "positive angle", and that when it goes clockwise, it is known as a "negative angle". These two types of angles can still have the same initial and terminal sides, but simply go different ways to reach there. Angles which are not the same, but reach the same terminal side are known as "coterminal" angles. Finally, we learned about arcs: segments of a circle defined by the terminal and initial sides of an angle. These arcs are found by the formula s = θR.
Two important applications of this are population growth and radioactive decay, which we learned about in class. Population growth is based on an identical formula to that of continuous interest: n(t) = n0ert and can be solved using logs. For example, if you wanted to find out how long it would take a population to reach 100 from 5, at a rate of 2%, you could use 100 = 5e.02t, then divide by 5 and take the natural log of both sides, ln 20 = ln e.02t, and then the laws of log allow you to simply get ln 20/.02 = t. Radioactive decay is similar, and can be solved in a way that is very much the same, except that it relies on the formula m(t) = m0e-rt.
After this we learned about angles, and the initial and terminal sides of one- the initial side being the side that starts the angle, and the terminal side being the one that ends it. We learned that the measure of an angle is essentially how open one is, measured in degrees or radians, and that when an angle's vertex is at the origin of the unit circle, and its initial side is on the positive x-axis, this is called standard position. We also learned that when the initial side of an angle goes counterclockwise to reach the terminal side, the angle is known as a "positive angle", and that when it goes clockwise, it is known as a "negative angle". These two types of angles can still have the same initial and terminal sides, but simply go different ways to reach there. Angles which are not the same, but reach the same terminal side are known as "coterminal" angles. Finally, we learned about arcs: segments of a circle defined by the terminal and initial sides of an angle. These arcs are found by the formula s = θR.
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