This is Nathan with the week 2 Blog Prompt response.
To solve a quadratic trig equation you basically go back to Algebra 2, but these equations involve stuff like sin, cos, and tan. You can factor out the original equation and group the two equations and set them equal to zero. Here is an example:
2sin^2x+sinx-1=0 Here, you just factor out and group. (2sinx-1)(sinx+1)=0 Now you just, set each equation equal to zero and solve. a.)2sinx-1=0 sinx=1/2 x=π/6 or 5π/6 Degrees value=30 and 150. b.)sinx+1=0 sinx=-1 x=3π/2 x=270 degrees
If there is more than one trig function in the equation, identities are needed to reduce the equation to a single function for solving.
2cos^2x+3sinx-3=0 By using the sin^2x+cos^2x=1 formula you get: 2(1-sin^2x)+3sinx-3=0 2-2sin^2x+3sinx-3=0 -2sin^2x+3sinx-1=0 Now, multiply the equation by -1 to get: 2sin^2x-3sinx+1=0 Now, you factor and solve. (2sinx-1)(sinx-1)=0 a.)2sinx-1=0 sinx=1/2 x=π/6, 5π/6 b.)sinx-1=0 sinx=1 x=π/2
this lawrence doing the blog prompt. i love these things. they help me pass this class somewhat.
it is very simple. you have to go back to algebra 2 and set everything equal to zero, but the ones in advanced math have sin, cos, tan, and whatever else they want to put in there. you also have to know your identities for some of them. yes that means going back to the part where most of us failed.
example 1)
1 - 2sin2x + sin x = 1 Set one side equal to 0 0 = 2sin2x - sin x now you have to factor 0 = sin x(2sin x - 1) sin x = 0 or 2sin x - 1 = 0 sin x = 0 or sin x = 1/2 x = sin-10 or x = sin-1(1/2) x = 0, 180, 30, 150
example 2)
cos 2x - cos x = cos x - cos x
cos 2x - cos x = 0
Note: cos 2x = 2 cos ² x - 1
2 cos squared x - 1 - cos x = 0 2 cos squared x - cos x - 1 = 0
now you factor (2 cos x + 1)(cos x - 1) = 0
Set each factor equal to zero and solve for x.
2 cos x + 1 = 0
2 cos x + 1 - 1 = 0 - 1 2 cos x = -1 (2 cos x)/2 = -1/2 cos x = -1/2
If a trig equation can be solved analytically, these steps will do it: 1.Put the equation in terms of one function of one angle. 2.Write the equation as one trig function of an angle equals a constant. 3.Write down the possible value(s) for the angle. 4.If necessary, solve for the variable. 5.Apply any restrictions on the solution.
example 1 cos(4A) − sin(2A) = 0 ⇒
2sin²(2A) + sin(2A) − 1 = 0
You want to solve for sin(2A). You should recognize that the equation is really a quadratic,
2y² + y − 1 = 0, where y = sin(2A)
It can be factored in a straightforward way:
(sin(2A) + 1) (2sin(2A) − 1) = 0
From algebra you know that if a product is 0 then you solve by setting each factor to 0:
sin(2A) + 1 = 0 or 2sin(2A) − 1 = 0
sin(2A) = −1 or sin(2A) = 1/2
example 2 3tan²(B/2) − 1 = 0
To solve it, add 1 to both sides and divide by 3:
tan²(B/2) = 1/3
and then take square root of both sides:
tan(B/2) = ±√(1/3) or ±√(3)/3
It’s important to remember to use the plus-or-minus sign ± when taking the square root of both sides; otherwise you could overlook some solutions.
How do you solve a quadratic formula?? hmmmmm..let me see.... I have no idea, but wait let me look it up online. (lol)... 10 min later: ohhh... I remember this from Algebra 2. Answer: You solve a quadratic formula by factoring.
Example#1 x2 + 5x + 6 (x + 2)(x + 3) = 0 x + 2 = 0 or x + 3 = 0 x = –2 or x = – 3 The solution to x2 + 5x + 6 = 0 is x = –3, –2
Example#2 (x – 3)(x – 4) = 0 x – 3 = 0 or x – 4 = 0 x = 3 or x = 4
Thanks to www.purplemath.com I have a better understanding. :)
HEY IT IS TAYLOR HERE AND HERE IS MY COMMENT ON THIS WEEKS BLOG!!!
Ok how to solve a quadratic trig equation. Well since the internet and my classmates gave me a clue as to what this was I will inform you.
First, you need to identify if the equation or any of its parts are trigonometric identities. If so replace them in the formula, if not try to solve it . You need to look out for factoring and remembering that dividing can not be used for canceling. Also make sure to note if it is positive or negative and where it is in the quadrant plane if you want to get the correct answer. You basically see what you can do to find your answer.
EXAMPLES
4 sin^2 + 2 sin=0 Since there are no identities in this formula
2 sin( 2 sin + 1)=0 After you write the problem done you factor out a sin and a 2 and this is how your answer should look like.
2 sin=0 2sin+1=0 Then you set both equations to equal zero.
Sin^-1 (½) Sin^-(-½) Then you divide by 2 for the first equation and set it to the inverse. Then you subtract 1 and divide by 2 for the 2nd equation and set it to inverse.
30 degrees Is the answer that you will get for both of them but since the first one is positive and the next one is negative you need to find 4 different answers. You just need to add 180 to 30, subtract 180 from 30, and subtract 30 from 360 to get your other answers.
Your final answers should be 30 degrees, 150 degrees, 210 degrees, and 330 degrees.
Cos^2- 1/2 cos = 0 Since there are no identities you will solve the problem normally.
Cos( cos -1/2 ) =0 After you write the problem done you factor out a cos and this is how your answer should look like.
Cos= 0 cos - 1/2 =0 Then you set both equations to equal zero.
Since one of our equations is zero we will use the unit circle. Cos is only zero at 90 degrees and 270 degrees. So we just need to work out the last equation.
Cos^1(½) This is on your trig chart and you should know that it is 60 degrees and since cos is positive we just need to find the 1st and second quadrant.
360 degrees - 60 degrees = 300 degrees You just need to subtract 60 degrees from 360 degrees to get your final answer.
Your answers should be 90 degrees, 270 degrees, 60 degrees, and 300 degrees.
That is how you solve quadratic trig equation. At least I think so lol.
http://www.regentsprep.org/Regents/math/algtrig/ATT10/trigequations2.htm That site helped me a bit but I have no clue if I copied their equations manly because I was just looking at the equation formation.
WELL UNTIL NEXT TIME JA NE ( Japanese word for goodbye)
Mary'ss blogggg how do you solve a quadratic trig equation? good question... hahhaah. You remember algebra 2 perhaps? well you are gonna need it! Its basically the same thing, you can sometimes pretend the trig function isn't there to solve, but most of the time, you will need identities to finish the equation.
Here's just a normal quadratic example 1
6x^2+7x+2 i'm pretty sure you take the product of the coefficient and the last number, and find factors that add or subtract to give you the middle coefficient. factors of 12 that add to give you 1 are -4 and 3. so you expand the equation like this 6x^2+3x+4x+2 then you group and factor out what you can 3x(2x+1)+2(2x+1) (3x+2)=0 2x+1=0 set them both equal to zero and solve x = -2/3 x=-1/2
Example with trig in it 2sin^2+sinx-1=0 factor and you get (2sin^2-1)(sinx+1) then you set them equal to 0 so you get sinx=1/2 and sinx=-1 so you take the inverses, and i'm sure you can handle that on your own. its almost the same. kbyyyeee!
to solve these equations, you have to apply the quadratic formula to find solutions to these equations. This is a multi-step process that starts with simplifying the equation. After the equation is simplified, you will be able to solve the quadratic formula and then enter that answer in to solve for X.
example 1: Solve. 0° ≤ x ≤ 360° tan2x – 3tanx – 4 = 0 let x = tan x x2 – 3x – 4 = 0 (x – 4)(x + 1) = 0 x – 4 = 0 x + 1 = 0 x = 4 x = -1 tan x = 4 tan x = -1 x = 76° x = 135° x = 256° x = 315
Example 2: Solve. 0° ≤ x ≤ 360° 2cos2x – sinx =1 Use the identity, cos2x = 1 – sin2x 2(1 – sin2x) - sinx = 1 2 – 2sin2x – sinx = 1 –2sin2x – sinx + 1 = 0 2sin2x + sinx -1 = 0 let x = sin (2x – 1)(x + 1) = 0 2x – 1 = 0 x + 1 = 0 2x = 1 x = -1 x = .5 sin x = .5 sin x = -1 x = 30° x = 270° x = 150°
How do you solve a quadratic trigonomic equation? Well it's pretty much the same as the regular quadratic equations, just pretend that the trig functions aren't there. Most of the time you will need to use your identities, then you set it equal to zero and factor.
Example 1: sin^2x-sinx=2
First you need to set everything equal to zero by subtracting 2---sin^2x-sinx-2=0 Then you need to factor---(sinx-2)(sinx+1) After you factor, set both equations equal to zero and solve for sinx---sinx=2 sinx=-1
Then you would take the inverses of these two equations to find x.
Example 2: 2sin^2x-3sinx+1=0
Since everthing is already equal to zero, you have to factor everything out---(2sinx-1)(sinx-1)=0 Then you need to set everything equal to zero and solve for sin x---sinx=1/2 sinx=1 After you solve for that, you have to take the inverses of both of these equations to get your answers for x---inverses: pi/6 and 5pi/6 and pi/2
One way of solving the quadratic trigonometry equations is by doing simple factoring that we learned in algebra 1. For Example: 4cos^2x - 3cosx -first you factor out a cosx, so.. cosx(4cosx - 3) -then you set both equal to 0, so.. cosx = 0 & 4cosx - 3 = 0 -now you solve for x in both equations.. x = cos^-1(0) & x = cos^-1(3/4) x = 90 degrees & x = 41.4 -next you use the unit circle to find the other degree values, so.. x = 90 degrees, 270 degrees, 41.4 degrees, and 318.6 degrees
Another way is to use identites that we learned last week and in chapter 8. For Example: cos^2(15) - sin^2(15) -to do this you just plug in the simple indenty of cos2alpha = cos^2alpha - sin^2alpha, so.. cos2(15) = cos30 -using the unit circle you find out the answer.. x = 30 degrees and 330 degrees
How to solve a quadratic trig equation? It just like a regular quadratic equation pretty much.
Example 1 : Solve 3x2 + x - 2 = 0 for x.
3x - 2)(x + 1) = 0 Use the principle of zero products, which says, if ab = 0, either a, b, or both must be equal to zero. 3x - 2 = 0, x + 1 = 0 3x = 2 , x = -1 x = (2/3) x = -1, (2/3)
Example 2: Solve 3x2 + 5x = 0 for x.
x(3x + 5) = 0 Use the principle of zero products. x = 0, 3x + 5 = 0 3x = -5 x = -(5/3) x = 0, -(5/3)
First start by referring back to algebra 2 and find that it is very similar to those quadratic equations. So here are some examples of quadratic equations.
Ex: Solve 3x2 + x - 2 = 0 for x. (3x - 2)(x + 1) = 0 Use the principle of zero products, which says, if ab = 0, either a, b, or both must be equal to zero. 3x - 2 = 0, x + 1 = 0 3x = 2, x = -1 x = (2/3) x = -1, (2/3)
Ex:Solve 3x2 + 5x = 0 for x. x(3x + 5) = 0 Use the principle of zero products. x = 0, 3x + 5 = 0 3x = -5 x = -(5/3) x = 0, -(5/3)
i realized like 3 people have the same trig equation as me, i just found it in my notebook and mayge that's what everyone else did too, i didn't copy because obviously you'd notice, buuutttt, i'll go and do another example, is that all i have to do? just a different example correctt?
This is Nathan with the week 2 Blog Prompt response.
ReplyDeleteTo solve a quadratic trig equation you basically go back to Algebra 2, but these equations involve stuff like sin, cos, and tan. You can factor out the original equation and group the two equations and set them equal to zero. Here is an example:
2sin^2x+sinx-1=0
Here, you just factor out and group.
(2sinx-1)(sinx+1)=0
Now you just, set each equation equal to zero and solve.
a.)2sinx-1=0
sinx=1/2
x=π/6 or 5π/6
Degrees value=30 and 150.
b.)sinx+1=0
sinx=-1
x=3π/2
x=270 degrees
If there is more than one trig function in the equation, identities are needed to reduce the equation to a single function for solving.
2cos^2x+3sinx-3=0
By using the sin^2x+cos^2x=1 formula you get:
2(1-sin^2x)+3sinx-3=0
2-2sin^2x+3sinx-3=0
-2sin^2x+3sinx-1=0
Now, multiply the equation by -1 to get:
2sin^2x-3sinx+1=0
Now, you factor and solve.
(2sinx-1)(sinx-1)=0
a.)2sinx-1=0
sinx=1/2
x=π/6, 5π/6
b.)sinx-1=0
sinx=1
x=π/2
Blllllllaaaaaaaaaaaaaaaaaahhhhhhhhhhhhhhh!!!!!!!!!!!!!
this lawrence doing the blog prompt. i love these things. they help me pass this class somewhat.
ReplyDeleteit is very simple. you have to go back to algebra 2 and set everything equal to zero, but the ones in advanced math have sin, cos, tan, and whatever else they want to put in there. you also have to know your identities for some of them. yes that means going back to the part where most of us failed.
example 1)
1 - 2sin2x + sin x = 1
Set one side equal to 0
0 = 2sin2x - sin x
now you have to factor
0 = sin x(2sin x - 1)
sin x = 0 or 2sin x - 1 = 0
sin x = 0 or sin x = 1/2
x = sin-10 or x = sin-1(1/2)
x = 0, 180, 30, 150
example 2)
cos 2x - cos x = cos x - cos x
cos 2x - cos x = 0
Note: cos 2x = 2 cos ² x - 1
2 cos squared x - 1 - cos x = 0
2 cos squared x - cos x - 1 = 0
now you factor
(2 cos x + 1)(cos x - 1) = 0
Set each factor equal to zero and solve for x.
2 cos x + 1 = 0
2 cos x + 1 - 1 = 0 - 1
2 cos x = -1
(2 cos x)/2 = -1/2
cos x = -1/2
x = 120°, 240°
If a trig equation can be solved analytically, these steps will do it:
ReplyDelete1.Put the equation in terms of one function of one angle.
2.Write the equation as one trig function of an angle equals a constant.
3.Write down the possible value(s) for the angle.
4.If necessary, solve for the variable.
5.Apply any restrictions on the solution.
example 1
cos(4A) − sin(2A) = 0 ⇒
2sin²(2A) + sin(2A) − 1 = 0
You want to solve for sin(2A). You should recognize that the equation is really a quadratic,
2y² + y − 1 = 0, where y = sin(2A)
It can be factored in a straightforward way:
(sin(2A) + 1) (2sin(2A) − 1) = 0
From algebra you know that if a product is 0 then you solve by setting each factor to 0:
sin(2A) + 1 = 0 or 2sin(2A) − 1 = 0
sin(2A) = −1 or sin(2A) = 1/2
example 2
3tan²(B/2) − 1 = 0
To solve it, add 1 to both sides and divide by 3:
tan²(B/2) = 1/3
and then take square root of both sides:
tan(B/2) = ±√(1/3) or ±√(3)/3
It’s important to remember to use the plus-or-minus sign ± when taking the square root of both sides; otherwise you could overlook some solutions.
cite: http://oakroadsystems.com/math/trigsol.htm
How do you solve a quadratic formula??
ReplyDeletehmmmmm..let me see....
I have no idea, but wait let me look it up online. (lol)...
10 min later: ohhh... I remember this from Algebra 2.
Answer: You solve a quadratic formula by factoring.
Example#1
x2 + 5x + 6
(x + 2)(x + 3) = 0
x + 2 = 0 or x + 3 = 0
x = –2 or x = – 3
The solution to x2 + 5x + 6 = 0 is x = –3, –2
Example#2
(x – 3)(x – 4) = 0
x – 3 = 0 or x – 4 = 0
x = 3 or x = 4
Thanks to www.purplemath.com
I have a better understanding. :)
HEY IT IS TAYLOR HERE AND HERE IS MY COMMENT ON THIS WEEKS BLOG!!!
ReplyDeleteOk how to solve a quadratic trig equation. Well since the internet and my classmates gave me a clue as to what this was I will inform you.
First, you need to identify if the equation or any of its parts are trigonometric identities.
If so replace them in the formula, if not try to solve it . You need to look out for factoring and remembering that dividing can not be used for canceling. Also make sure to note if it is positive or negative and where it is in the quadrant plane if you want to get the correct answer. You basically see what you can do to find your answer.
EXAMPLES
4 sin^2 + 2 sin=0 Since there are no identities in this formula
2 sin( 2 sin + 1)=0 After you write the problem done you factor out a sin and a 2 and this is how your answer should look like.
2 sin=0 2sin+1=0 Then you set both equations to equal zero.
Sin^-1 (½) Sin^-(-½) Then you divide by 2 for the first equation and set it to the inverse. Then you subtract 1 and divide by 2 for the 2nd equation and set it to inverse.
30 degrees Is the answer that you will get for both of them but since the first one is positive and the next one is negative you need to find 4 different answers. You just need to add 180 to 30, subtract 180 from 30, and subtract 30 from 360 to get your other answers.
Your final answers should be 30 degrees, 150 degrees, 210 degrees, and 330 degrees.
Cos^2- 1/2 cos = 0 Since there are no identities you will solve the problem normally.
Cos( cos -1/2 ) =0 After you write the problem done you factor out a cos and this is how your answer should look like.
Cos= 0 cos - 1/2 =0 Then you set both equations to equal zero.
Since one of our equations is zero we will use the unit circle. Cos is only zero at 90 degrees and 270 degrees. So we just need to work out the last equation.
Cos^1(½) This is on your trig chart and you should know that it is 60 degrees and since cos is positive we just need to find the 1st and second quadrant.
360 degrees - 60 degrees = 300 degrees You just need to subtract 60 degrees from 360 degrees to get your final answer.
Your answers should be 90 degrees, 270 degrees, 60 degrees, and 300 degrees.
That is how you solve quadratic trig equation. At least I think so lol.
http://www.regentsprep.org/Regents/math/algtrig/ATT10/trigequations2.htm
That site helped me a bit but I have no clue if I copied their equations manly because I was just looking at the equation formation.
WELL UNTIL NEXT TIME JA NE ( Japanese word for goodbye)
Mary'ss blogggg
ReplyDeletehow do you solve a quadratic trig equation?
good question... hahhaah.
You remember algebra 2 perhaps? well you are gonna need it! Its basically the same thing, you can sometimes pretend the trig function isn't there to solve, but most of the time, you will need identities to finish the equation.
Here's just a normal quadratic
example 1
6x^2+7x+2
i'm pretty sure you take the product of the coefficient and the last number, and find factors that add or subtract to give you the middle coefficient.
factors of 12 that add to give you 1 are -4 and 3. so you expand the equation like this
6x^2+3x+4x+2
then you group and factor out what you can
3x(2x+1)+2(2x+1)
(3x+2)=0 2x+1=0
set them both equal to zero and solve
x = -2/3 x=-1/2
Example with trig in it
2sin^2+sinx-1=0
factor and you get
(2sin^2-1)(sinx+1)
then you set them equal to 0
so you get
sinx=1/2 and sinx=-1
so you take the inverses, and i'm sure you can handle that on your own. its almost the same.
kbyyyeee!
to solve these equations, you have to apply the quadratic formula to find solutions to these equations. This is a multi-step process that starts with simplifying the equation. After the equation is simplified, you will be able to solve the quadratic formula and then enter that answer in to solve for X.
ReplyDeleteexample 1:
Solve. 0° ≤ x ≤ 360°
tan2x – 3tanx – 4 = 0 let x = tan x
x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0
x – 4 = 0 x + 1 = 0
x = 4 x = -1
tan x = 4 tan x = -1
x = 76° x = 135°
x = 256° x = 315
Example 2: Solve. 0° ≤ x ≤ 360°
2cos2x – sinx =1
Use the identity, cos2x = 1 – sin2x
2(1 – sin2x) - sinx = 1
2 – 2sin2x – sinx = 1
–2sin2x – sinx + 1 = 0
2sin2x + sinx -1 = 0 let x = sin
(2x – 1)(x + 1) = 0
2x – 1 = 0 x + 1 = 0
2x = 1 x = -1
x = .5
sin x = .5 sin x = -1
x = 30° x = 270°
x = 150°
How do you solve a quadratic trigonomic equation? Well it's pretty much the same as the regular quadratic equations, just pretend that the trig functions aren't there. Most of the time you will need to use your identities, then you set it equal to zero and factor.
ReplyDeleteExample 1: sin^2x-sinx=2
First you need to set everything equal to zero by subtracting 2---sin^2x-sinx-2=0
Then you need to factor---(sinx-2)(sinx+1)
After you factor, set both equations equal to zero and solve for sinx---sinx=2 sinx=-1
Then you would take the inverses of these two equations to find x.
Example 2: 2sin^2x-3sinx+1=0
Since everthing is already equal to zero, you have to factor everything out---(2sinx-1)(sinx-1)=0
Then you need to set everything equal to zero and solve for sin x---sinx=1/2 sinx=1
After you solve for that, you have to take the inverses of both of these equations to get your answers for x---inverses: pi/6 and 5pi/6 and pi/2
Charlie's Blog Thingy:
ReplyDeleteOne way of solving the quadratic trigonometry equations is by doing simple factoring that we learned in algebra 1.
For Example:
4cos^2x - 3cosx
-first you factor out a cosx, so..
cosx(4cosx - 3)
-then you set both equal to 0, so..
cosx = 0 & 4cosx - 3 = 0
-now you solve for x in both equations..
x = cos^-1(0) & x = cos^-1(3/4)
x = 90 degrees & x = 41.4
-next you use the unit circle to find the other degree values, so..
x = 90 degrees, 270 degrees, 41.4 degrees, and 318.6 degrees
Another way is to use identites that we learned last week and in chapter 8.
For Example:
cos^2(15) - sin^2(15)
-to do this you just plug in the simple indenty of cos2alpha = cos^2alpha - sin^2alpha, so..
cos2(15) = cos30
-using the unit circle you find out the answer..
x = 30 degrees and 330 degrees
http://www.purplemath.com/modules/solvtrig.htm
This comment has been removed by the author.
ReplyDeleteHow to solve a quadratic trig equation? It just like a regular quadratic equation pretty much.
ReplyDeleteExample 1 :
Solve 3x2 + x - 2 = 0 for x.
3x - 2)(x + 1) = 0
Use the principle of zero products, which says, if ab = 0, either a, b, or both must be equal to zero.
3x - 2 = 0, x + 1 = 0
3x = 2 , x = -1
x = (2/3)
x = -1, (2/3)
Example 2:
Solve 3x2 + 5x = 0 for x.
x(3x + 5) = 0
Use the principle of zero products.
x = 0, 3x + 5 = 0
3x = -5
x = -(5/3)
x = 0, -(5/3)
http://library.thinkquest.org/20991/alg2/quad.html
First start by referring back to algebra 2 and find that it is very similar to those quadratic equations. So here are some examples of quadratic equations.
ReplyDeleteEx: Solve 3x2 + x - 2 = 0 for x.
(3x - 2)(x + 1) = 0
Use the principle of zero products,
which says, if ab = 0, either
a, b, or both must be
equal to zero.
3x - 2 = 0, x + 1 = 0
3x = 2, x = -1
x = (2/3)
x = -1, (2/3)
Ex:Solve 3x2 + 5x = 0 for x.
x(3x + 5) = 0
Use the principle of zero products.
x = 0, 3x + 5 = 0
3x = -5
x = -(5/3)
x = 0, -(5/3)
This is Feroz.
ReplyDeleteHow do you solve a quadratic trig equation?
I just learned how to do this so bear with me.
ex 1. 2sin^2x+sinx-1=0
This is where factoring comes into play.
= (2sinx-1)(sinx-1)
Set up two equations, and set them equal to zero.
2 sinx - 1 = 0 and sinx +1 = 0
Divide by 2 subtract 1
sinx = 1/2 sinx = -1
If you can use a calc, find the inverse, if not, use the trig chart.
x = 30, 150 x = 270, 3pi/2
= pi/6 or 5pi/6
Now I'll show you how to do it with two trig functions, using identities.
ex 2. cos 2x - cosx = cosx - cosx
Like terms, blah blah.
cos 2x - cos x = 0
So..
2cos^2x - 2 - cosx = 0
2cos^2x - cosx - 1
Take care of that ^2 (factor)
(2cosx + 1)(cosx - 1) = 0
Set equations, like before.
2 cosx + 1 = 0
2cosx = -1
cosx = -1/2
x = 120, 240
1. Some of these are copied from each other. You will not get credit if you have copied any part of some else's blog.
ReplyDelete2. Many of you forgot there are two angles for each inverse with exceptions at 1 and -1. sin^-1 (1/2) has two quadrants where sin is positive.
3. If you gave examples without trig functions redo them with trig functions in a new comment to get credit.
i realized like 3 people have the same trig equation as me, i just found it in my notebook and mayge that's what everyone else did too, i didn't copy because obviously you'd notice, buuutttt, i'll go and do another example, is that all i have to do? just a different example correctt?
ReplyDelete