Advanced Functions
Chapter 7: Trigonometric Identities and Equations
Quadratic Trigonometry Equations:
A question given in this sub-chapter would look like:
(2cscx-1)^2 = 9 [-pi,0]
First start off by square rooting both sides which would give 2cscx-1 = +-3. Now solve for cscx of both 3 and -3.
Starting off with 2cscx-1 = 3 if you solve this algebraically you'll get cscx = 2 but we can simplify this more by finding sinx as cscx = 1/sinx. So sinx= 1/2. Now you want to solve for x and to do this you want to look at the sin graph and determine where sin x = 1/2 on the graph. In this case there is no solution to this one as 1/2 is out of the given domain for the sin graph. Meaning this one has no solution so now we want to look at 2cscx-1 = -3
If we do the same thing we did with the first equation to 2cscx-1=-3 it would turn out as sin x = -1/1 = -1. Looking at the sin graph again you will see that at -1 x equals -pi/2 which is inside our given domain of [-pi,0]. Therefore the answer would be -pi/2
Another question that could be given is:
cos^2x - 2sin^2x = 1 within the domain of -2pi
Proving Identities:
Proving identities is all about using common sense and testing out different ways to make the left side equal the right side. A question that might be asked is:
sinxtanx = secx-cosx
Your goal is to make it so the left side is exactly the same as the right side. To do this you must use the plenty of identities given to you.
First off to solve this question separate the left side from the right side and solve them individually.
Starting with the left side which is sinxtanx we know that tanx = sinx/cosx then if we put sinx into this it will become sin^2/cosx. This is all we can do to the left side so now you move to the right side as you know what you need to get the right side to look like.
The right side is secx-cosx. We know that secx = 1/cosx. Now it looks like 1/cosx - cosx. We want to get cosx to have the same denominator as 1/cosx so we can subtract them. To do this multiply cosx by cosx/cosx. This will give you 1/cosx - cosxcosx/cosx. Combine the two fractions and you will get 1-cos^2x/cosx. Using one of the identities that we know 1-cos^2x = sin^2x meaning the right side looks like sin^2x/cosx
Therefore our left side = right side
Another example of a proving identities question could be:
csc2x + cot2x = cotx
First look at the left side which equals csc2x + cot2x. We know that csc2x = 1/sin2x and cot2x = cos2x/sin2x. We also know that sin2x can be written as 2sinxcosx so for both of the fractions that will replace the denominator. Also cos2x can be written as 2cos^2x - 1 so that will replace the numerator of cos2x/2sinxcosx. Now add the two fractions together giving you 2cos^2x/2sinxcosx. Now cancel out terms that are similar in the numerator and denominator. Turning the fraction into cosx/sinx. We can't do anything else to the left side so we'll look at the right side.
Now looking at the right side the only thing we need to do is change cotx into cosx/sinx
Therefore our left side = right side
Double Angle Formulas:
In this sub-chapter you'll be given a question like:
Given sin theta = 3/5. pi/2 < theta < pi, find sin2theta + cos2theta.
Just looking at the question you can see what quadrant theta is in as it is greater than 90 degrees and less than 180 degrees. This means it is in the second quadrant.
Recall some formulas such as:
sin2x = 2sinxcosx
cos2x = cos^2x - sin^2x
sin^2x + cos^2x = 1
First we can find cos^2theta by substituting 3/5 into the third formula I listed. If you do this and solve for cos^2theta you will get 16/25. Then square root that to find costheta which is +-4/5 but since we know this is in the second quadrant cos is negative so it's just -4/5.
Now you can find sin2theta using the first formula I listed by substituting the values we found and the values given into the equation. If you solve for sin2theta you will find that it equals -24/25
Using the second formula I listed you can find cos2theta by plugging in the values we found. If you solve for that cos2theta will equal 7/25
Now just add -24/25 + 7/25 and that will give you the answer of -17/25
Compound Angle Formulas
When solving compound angle formulas questions will ask you to find exact values like in the question:
Find the exact value of cos (13pi/12)
To solve this you must first turn the radians into degrees. You can do this by remembering that pi = 180 degrees. So 13x180 = 2340/12 = 195 degrees.
Now use common exact values to find two angles that add or subtract to 195 degrees. For example 5pi/6 = 150 degrees and pi/4 = 45 degrees. These both add up into 195 degrees.
Then the equation will turn into:
cos 13pi/12 = cos (5pi/6 +pi/4)
Our new equation looks like one of the identities which is cos (a + b). This identity turns into:
cosacosb - sinasinb.
The equation will turn into:
cos (5pi/6) cos (pi/4) - sin (5pi/6) sin (pi/4)
The next step is to write cos (5pi/6) and sin (5pi/6) in terms of related acute. To do that you must first determine what quadrant 5pi/6 is in. 5pi/6 = 150 degrees which means its in the second quadrant because it is more than 90 degrees and less than 180 degrees. Then you have to find the reference angle of 5pi/6 in radians. To do this you know that pi = 180 degrees. So 6pi/6 = 180 degrees as well. so thinking about it to get 5pi/6 to 6pi/6 all you have to do is add pi/6 to it. Meaning pi/6 is our reference angle. But since cos is negative in the second quadrant cos (5pi/6) will turn into -cos (pi/6). sin (5pi/6) will just turn into sin (pi/6) as sin is positive in the second quadrant. We know sin is positive and cos is negative due to the CAST rule.
The equation will now look like:
-cos (pi/6) cos(pi/4) - sin (pi/6) sin (pi/4)
Next use the special triangles to find values of each angle in the equation. This would turn the equation into:
- square root of 3/2 x 1/square root of 2 - 1/2 x 1/square root of 2
Then solve it from there by doing the multiplication step first. Making the equation become:
-square root of 3/2square root of 2 - 1/2square root of 2
Since both fractions have the same denominator you can combine them into one fraction.
-square root of three - 1/2square root of 2
Now to get the square root of 2 out of the denominator multiply the whole fraction by square root of 2/square root of 2.
The equation will look like:
-square root of 6 - square root of 2/2 x 2
Then the equation would equal:
-square root of 6 - square root of 2/4
Chapter 5: Rational Functions, Equations, and Inequalities
Solving Rational Inequalities
This sub-chapter is about solving rational inequalities. An example of a question given in this section would be x+3/x+1 > or equal to x-2/x-3
To solve and graph this equation treat it like a normal inequalities question at the start. So move the whole right side of the equation to the left side making the equation be x+3/x+1 - x-2/x-3 > or equal to 0. Now you want to get a common denominator to subtract both of the equations. To do this multiple x+3/x+1 by x-3 and multiply x-2/x-3 by x+1. This will give (x+3)(x-3) - (x-2)(x+1)/(x+1)(x-3). Now expand the numerator to give a smaller term. expanding the numerator will give x^2 - 9 - x^2 - x + 2x + 2 in the numerator. Now adding and subtracting like terms in that numerator will give x - 7. So now the equation looks like x-7/(x+1)(x-3). You don't need to expand the denominator as having it already factored helps with finding the vertical asymptotes. With this you can now create the graph with the same steps you would use for a regular rational function. The vertical asymptotes are -1 and 3. The y-intercept is 2.3. The x-intercept is 7. Finally there's a horizontal asymptote at zero which is known using the rules from earlier. Then test values to figure out what quadrant each line in the graph will be in meaning testing values after 3, before -1, and in between -1 and 3.
Solving Rational Equations
This sub-chapter is about how to solve rational equations. An example of a rational equation is
x-2/x-3 = 0. So step by step this equation is going to get solved.
First off you want to find the zero of the denominator to find out the restriction on x, which in this case would be 3. Then multiple both sides of the equation by the lowest common denominator which is (x - 3). This would cancel out the denominator and keep the right side equal to zero. Leaving the equation to look like x - 2 = 0. Then use algebra to solve for x. This will give you 2. This would also be your answer for this question
Rational Functions
At the beginning of this chapter you will be given equations such as y = 1/x+2 and be asked to graph this equation.
This equation is pretty simple so there aren't many steps to it. First using the rules you've learned we know that the horizontal asymptote is at 0 because there is no degree on the numerator and there is a degree of 1 at the denominator. Now you must find the vertical asymptotes of the graph. To do this take a look at the denominator and find the zeros of it. This means our vertical asymptote would be at -2. We get -2 because you just make the denominator equal zero and solve for x algebraically. The last thing you must do for this equation is find the y-intercept. This is done by making x = 0 and solve for y. This will give a y-intercept of 0.5. With all this information on how to graph this equation all you must do is put in a graph. The next step to making this graph is finding out what quadrants our equation is in. To do this take values that are before and after the vertical asymptote and plug them into the equation. So you would take values -3 and 2. It doesn't have to be these values just choose values after the vertical asymptote. Now if you plug -3 into the equation you will get a negative value, this means on the left of the asymptote this part of the graph will be in the third quadrant making it negative. If you plug the value 2 into the equation you will get a positive value which means on the right of the asymptote the graph will be in the first quadrant and positive. With all this information you can now create an estimate of the graph this will create.
Another equation you could be given is 2x^2 + 3x/3x^2 - 48.
To graph this use the rules from before and since we know that the degree of N and D are the same, its the ratio of the lead coefficients. So 2/3 which equals 0.66. So that's the horizontal asymptote of this graph. Now recall how to find the vertical asymptote of rational equations. For this equation there are two vertical asymptotes. You will find that these two asymptotes are 4 and -4 by using the same method from the previous one. Again also find the y-intercept of this equation. But there's an extra step you must take for this equation and that is finding the x-intercepts. To find the x-intercepts you must first factor the numerator if possible, then make both factors equal zero. For this equation you just factor x out of the numerator then you x-intercepts would be x = 0 and x = -1.5. Again with all this information gathered put these important points onto the graph and then test values that are less than -4, greater than 4, and values in between -4 and 4. This will give you the locations of the lines you must draw to make an estimate of what the graph will look like.
Most questions in both 5.1 and 5.2 will ask the domain and range of these graphs which is pretty simple to solve as it's mainly common sense. So for the first example question the domain would be x is all real number but it can't equal -2. The range would be y = all real numbers except for 0.
When looking at rational functions and equations there are a couple of rules that you must keep in mind to graph these functions and equations.
These rules will help you find the horizontal asymptotes of the problems provided in this chapter:
If N is the degree of the numerator and D is the degree of the denominator
N < D means that the horizontal asymptote is at y = 0
Example: 1/x+1
N = D means the horizontal asymptote is y = ratio of leading coefficients of numerator and denominator
Example: 3x^2/2x^2+1
N > D means there is no horizontal asymptote
Example: x^3/x+1
The formula for a rational function is
f(x) = p(x)/q(x)
If you take a look at the simplest version of a rational function which would be 1/x. It would show you two curved lines that have asymptotes that they can't touch. Both the vertical and horizontal asymptote of this graph would be 0. These lines are pretty much opposite to each other and one is in the first quadrant, while the other is in the third quadrant. In this chapter you learn how to graph equations similar to this but much more complex.
