On Friday, we started Chapter 8. Although we did not get much time in we learned 8-1 which is more about inverses.To refresh an inverse (-1) finds an angle where the function has a certain value.
Ex. Sin^-1 (#) or arcsin (#)
Cos^-1(#) or arcos(#)
In order to solve for trig in 8-1 you must first get the function by itself and then take the inverse. This takes you back to what you learned in Algebra to getting variables by themselves. With inverses there will be 2 answers, with a few exceptions. The next thing to do is find where the angle is based on trig function and if number is positive or negative.
Steps:
1. Take the inverside of the postive number to find the Q1 angle.
To get either Q2, Q3, or Q4 follow these rules--
Q2: make negative, add 180degrees
Q3: Add 180degrees
Q4: Make positive, add 360degrees
Ex- Solve for sinx=.8
Sin^-1=(.8) *Sin is postive in both quadrants 1 and 2.
In calculator once you plug in Sin^-1(.8) the answer you get is 53.130. Now you need to find the second quadrant. Your rules say to make the number negative then add 180 degrees. This gives you the answer 49.988.
x=53.130, 49.988
In the case of having to get a trig function alone for example:
5sin + 10 = 15 you would do your normal steps until getting to the form of theta=sin^-1(#)
Sunday, September 5, 2010
How to solve a simple trig equation
NOTES:
Solve for x.
4sin(theta)= -2
sin(theta)= (-1/2)
(theta)= sin^-1(1/2)
Now when you figure that out you put that in the quadrant 1.
you are going to end up with 30 degrees.
Now you need to find what is in quadrant 3 and quadrant 4 because that is where sin is negative.
so you are going to add 180 and you will get 210 degrees in quadrant 3.
then you take (-30) and add that to 360 degrees in quadrant 3.
and you will end up with 330 degrees in quadrant 4.
So (theta)= 210 degrees
330 degrees
- When solving a trig equation you are first going to want to solve for theta.
- Then you need to get the trig function by itself.
- After you need to take an inverse function.
- An inverse has 2 answers with only SOME exceptions.
- Find where the angle is based on the trig function.
- Last you need to determine weather its # is +ve or -ve.
- Take the inverse of the +ve number to find the quadrant 1 angle.
- When dealing with a quadrant 2 angle make -ve DEGREES and add 180 (degrees).
- When dealing with a quadrant 3 angle you are just going to want to add 180 (degrees).
- And when dealing with a quadrant 4 angle make -ve DEGREES and then add 360 (degrees).
Solve for x.
4sin(theta)= -2
sin(theta)= (-1/2)
(theta)= sin^-1(1/2)
Now when you figure that out you put that in the quadrant 1.
you are going to end up with 30 degrees.
Now you need to find what is in quadrant 3 and quadrant 4 because that is where sin is negative.
so you are going to add 180 and you will get 210 degrees in quadrant 3.
then you take (-30) and add that to 360 degrees in quadrant 3.
and you will end up with 330 degrees in quadrant 4.
So (theta)= 210 degrees
330 degrees
7-6 Inverse Angles
In section 7-6, we learned how to find the inverse of angles. It was one of the easiest things we did in chapter 7 because you could use your calculator for most of it. However, there are a few times where it cannot be done on a calulater, so you still need to know what you are doing.
When you are given a function with ^-1 at the end of it, that means you are going to find the inverse.
ex: tan^-1(#)
or arc tan (#)
The purpose of an inverse is to find where the angle function has a certain value. It is pretty much finding the opposite.
Putting it in your calculator is pretty simple.
ex: sec^-1 (a/b)= sin ^-1(b/a)
csc^-1 (a/b)= sin^-1 (b/a)
cot^-1 (a/b)= tan^-1 (b/a)
When you plug it in, the mode must be in degrees to get the correct answer. Once you get your answer in decimals, you round to the hundreds.
ex: tan^-1 (5) = 78.69
If you are given something that looks like this:
trig^-1 (trigx)= x
You need to draw it in the coordinate plane and use a formula to find trig 1.
ex: cos (tan^-1 (4/5))
- you need to graph the fraction on a coordinate plane and draw a triangle connecting it to the origin.
- once it is graphed you need to figure out the other side of the triangle, using the pythagorean triples, you can determine that the other side will be 3.
- the formula for tan is cos is y/r, so your answer will be 4/3
When you are given a function with ^-1 at the end of it, that means you are going to find the inverse.
ex: tan^-1(#)
or arc tan (#)
The purpose of an inverse is to find where the angle function has a certain value. It is pretty much finding the opposite.
Putting it in your calculator is pretty simple.
ex: sec^-1 (a/b)= sin ^-1(b/a)
csc^-1 (a/b)= sin^-1 (b/a)
cot^-1 (a/b)= tan^-1 (b/a)
When you plug it in, the mode must be in degrees to get the correct answer. Once you get your answer in decimals, you round to the hundreds.
ex: tan^-1 (5) = 78.69
If you are given something that looks like this:
trig^-1 (trigx)= x
You need to draw it in the coordinate plane and use a formula to find trig 1.
ex: cos (tan^-1 (4/5))
- you need to graph the fraction on a coordinate plane and draw a triangle connecting it to the origin.
- once it is graphed you need to figure out the other side of the triangle, using the pythagorean triples, you can determine that the other side will be 3.
- the formula for tan is cos is y/r, so your answer will be 4/3
Review of 7-6; Inverses
An inverse,-1 finds an angle.
Use your calculator and make sure its in degrees.
An inverse finds the ANGLE where the function has a certain VALUE.
*if you look for key words like this it will be easier to determine what to do with the problem.
If you are given trig1(trig2^-1(a/b)) then draw the points in a coordinate plane, and use formulas to find trig1 **trig1 is what we are solving for, but trig2 gives us the information we need to plug into the formulas.
Example: Tan^-1(0)= y/x= 0 + 180(degrees)n
- sin^-1(#) or arcsin(#) **Both arcsin and sin are the same thing just written out differently.
- cos^-1(#) or arcos(#) **Same here, both arcos and cos are the same thing just written out differently.
Use your calculator and make sure its in degrees.
An inverse finds the ANGLE where the function has a certain VALUE.
*if you look for key words like this it will be easier to determine what to do with the problem.
If you are given trig1(trig2^-1(a/b)) then draw the points in a coordinate plane, and use formulas to find trig1 **trig1 is what we are solving for, but trig2 gives us the information we need to plug into the formulas.
- sin^-1(sinx)=x
- cos^-1(cosx)=x
Example: Tan^-1(0)= y/x= 0 + 180(degrees)n
Review of 7-5
This week in Advanced Math we took two chapter tests – a multiple choice and free response. In chapter seven we learned how to convert units of measurement, find coterminal angles, use formulas in word problems, find sine and cosine, find reference angles, and also find inverses. We were even introduced to the trig chart. In 7-5 we had to use all the functions. Given one function, we were taught how to find the other five. For example:
EXAMPLE 1: Given tan theta = -5/12 for 0 <> pi. Find the other five trig functions.
A). Draw your coordinate plane. (Draw it in the quadrant according to 0 <> pi, which would be quadrant II).
B). Since tan is equaled to y/x, 5 will be y and 12 will be x.
C). Now that we know we’re looking for r, you would use the Pythagorean Theorem to find it.
But since we know that (5, 12, 13) is a Pythagorean Triple, we don’t have to do that step.
D). Now, we can find the other five trig functions. Just plug the numbers into each function.
*** y=5, x=12, r=13 (But remember, it was in quadrant II so x had to be negative in order for that to work, so 12 will be negative when used).
Sin theta = y/r = 5/13
Cos theta = x/r = -12/13
Cot theta = x/y = -12/5
Sec theta = r/x = -13/12
Csc theta = r/y = 13/5
We know that these functions are easy to remember if you know sin, cosine, and tangent.
Cosecant is the reciprocal of sine, secant is the reciprocal of cosine, and cotangent is the reciprocal of tangent. All you do is plug your numbers into the function!
Also, we were taught how to find reference angles. It is the same as section 7-4. You still follow the same steps and if it is a trig chart angle, plug it in.
EXAMPLE 1: Given tan theta = -5/12 for 0 <> pi. Find the other five trig functions.
A). Draw your coordinate plane. (Draw it in the quadrant according to 0 <> pi, which would be quadrant II).
B). Since tan is equaled to y/x, 5 will be y and 12 will be x.
C). Now that we know we’re looking for r, you would use the Pythagorean Theorem to find it.
But since we know that (5, 12, 13) is a Pythagorean Triple, we don’t have to do that step.
D). Now, we can find the other five trig functions. Just plug the numbers into each function.
*** y=5, x=12, r=13 (But remember, it was in quadrant II so x had to be negative in order for that to work, so 12 will be negative when used).
Sin theta = y/r = 5/13
Cos theta = x/r = -12/13
Cot theta = x/y = -12/5
Sec theta = r/x = -13/12
Csc theta = r/y = 13/5
We know that these functions are easy to remember if you know sin, cosine, and tangent.
Cosecant is the reciprocal of sine, secant is the reciprocal of cosine, and cotangent is the reciprocal of tangent. All you do is plug your numbers into the function!
Also, we were taught how to find reference angles. It is the same as section 7-4. You still follow the same steps and if it is a trig chart angle, plug it in.
This week in advanced math we review Chapter Seven. We also had two test this week which made us put what we learned to the test. We really had to study and know the Trig Chart, as well as all of the formulas we have learned. On Monday through Wednesday evening we had to do a blog on SIN, COS, TAN, COT, SEC, and CSC. Then on Friday we went on to the next section, section 8-1. 8-1 has to do with finding the angles within quadrants. Here are the steps to finding the quadrant angle.
1) First you must solve for theta, to do this you get the trig function by itself and then take an inverse.
2) Then you find if the angle is based on trig function and if the number is tve or -ve.
3)
a)Take the inverse of the tve number to find the quadrant 1 angle.
b)Make -ve and add 180 to find quadrant 2 angle.
c) Add 180 to find quadrant 3 angle.
d)Make -ve add 360 for quadrant 4 angle.
Example: 2Cos(theta) +7=5
1) Subtract 7 from 5 to get 2Cos(theta)= -2
2) Divide -2 by 2 to get Cos(theta)= -1
3)Move cos to the other side to get theta= cos^-1 (1)
4) Cos is negative in quadrants 2 and 3 and cos6-1 (1) = 0.
5)so in quadrant 2 theta=180 and in quadrant 3 theta=180.
1) First you must solve for theta, to do this you get the trig function by itself and then take an inverse.
2) Then you find if the angle is based on trig function and if the number is tve or -ve.
3)
a)Take the inverse of the tve number to find the quadrant 1 angle.
b)Make -ve and add 180 to find quadrant 2 angle.
c) Add 180 to find quadrant 3 angle.
d)Make -ve add 360 for quadrant 4 angle.
Example: 2Cos(theta) +7=5
1) Subtract 7 from 5 to get 2Cos(theta)= -2
2) Divide -2 by 2 to get Cos(theta)= -1
3)Move cos to the other side to get theta= cos^-1 (1)
4) Cos is negative in quadrants 2 and 3 and cos6-1 (1) = 0.
5)so in quadrant 2 theta=180 and in quadrant 3 theta=180.
Saturday, September 4, 2010
Trigonometric Equations and Applications
In section 8-1 we learned how to solve for theta while using inverses. Basically , you are solving an equation with a trig function. An inverse almost always has two answers. You must also remember that your calculator must be in degree mode. If the answer is asked for in radians, wait until the end and use the formula pi/180 to convert to radians.
FIRST;
Get the trig function by itself.
2sinex +6=0
*Subtract six .
Divide both sides by two; sin therefore is equal to 3.
NEXT;
Make variable equal to the inverse of the trig function.
X= sin^-1 (3)
LASTLY;
Solve for x while using coordinate plane.
Ex. 1
Sinx=(-4)
X=sin^-1 (4)
Because sine can never be 4,or greater than 1 for the matter, the answer therefore is NO SOLUTION.
Ex.2.
Csc x= (2)
X= csc^-1 (2) * Csc is related to the sine function therefore you use the inverse of sine.
X=sin^-1 (1/2) * You must also find the reciprocal in order to make the statement true.
X=30
Sine is positive in both quadrant 1 and 2. We already know that sin^-1 is 30 in quadrant 1. In order to find sin^-1 in quadrant 2, we make the number (30) negative and add 180.
-(30)+ 180 = 150
The answer is x=30
x= 150
FIRST;
Get the trig function by itself.
2sinex +6=0
*Subtract six .
Divide both sides by two; sin therefore is equal to 3.
NEXT;
Make variable equal to the inverse of the trig function.
X= sin^-1 (3)
LASTLY;
Solve for x while using coordinate plane.
Ex. 1
Sinx=(-4)
X=sin^-1 (4)
Because sine can never be 4,or greater than 1 for the matter, the answer therefore is NO SOLUTION.
Ex.2.
Csc x= (2)
X= csc^-1 (2) * Csc is related to the sine function therefore you use the inverse of sine.
X=sin^-1 (1/2) * You must also find the reciprocal in order to make the statement true.
X=30
Sine is positive in both quadrant 1 and 2. We already know that sin^-1 is 30 in quadrant 1. In order to find sin^-1 in quadrant 2, we make the number (30) negative and add 180.
-(30)+ 180 = 150
The answer is x=30
x= 150
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