This is probably the hardest concept we did so far in my algebra 2 class. I got a test in 3 days on this and everything else on the unit. I'm probably gonna watch this multiple times.
I'm a little confused with the 4 sin^2 x - 1 = 0 example. I understood every step to sin x equaling 1/2 and -1/2. But my understanding is that I would have to use arc sine in order to find the values of x. Doesn't arc sine only exist in Quadrants 1 and 4? So, wouldn't x equal 30 degrees (pi/6) and -30 degrees (-pi/6) since the range is [-pi/2, pi/2]? Please advise. As always, your videos are superb. Thank you for taking the time to make and upload them. Using them to study for my math certification exam (teaching).
If you look at the upper right corner at 5:35, there is a range where the angle can exist [0, 2pi]. I believe you can right that as long as there is a range given which exists on all quadrants.
How do we know when we are supposed to use the arc sin or any arc trig function in a problem? I mean if youre solving an equation and you have last that a trig function equals a value, i know that sine or cosine of the answer equals a value, but what about the restricted range of the trig function??? How do we know when we are supposed to not use the inverse function and when we do use the inverse. It just doesnt make sense to me since an equation requires you to apply the opposite function to solve it??? I really could use some answers or a video that highlights why you can include values outside of an inverse trig function's range
Hi, I’m not OP but here’s an explanation. You mainly use an arc trig function to find the angle that corresponds to that value. For example let’s just say you have sin x=1. You need to find what angle at sin x is equal to 1. You can’t find it using the regular sin function, so you need to use the arcsine function. So arcsinx of 1 is = pi/2. Now since we are using arcsine we are only able to evaluate inputs between [-1,1]. So, we are just able to evaluate what angle that is. with sinx=1. Let’s say we had to evaluate sinx=1.2. We need to use arcsine to find the angle that equals to 1.2. However our restricted domain for arcsine is [-1,1], so we are unable to find the angle giving us no solution. Btw keep in mind that the range of sine is the same as the domain of arcsine. And that the domain of sine is the same as the range of arcsine. So keeping that in mind, the domain of sine is (-00,00) which is the range of arcsine. So when we are looking for a trig value, the output is able to lie within the range because it can be anything. However, the domain of input is restricted to [-1,1]. Hope that makes sense
Oh also, you use the regular trig functions in order to find the function value. For example Tanx=45degrees. This question is asking us what is the value using the tangent function at 45 degrees. Tan is sin/cos. So rad2/2 divided by rad2/2 is equal to 1. So our function is equal to 1. In the video above, OP is specifically looking for the opposite. He is looking for what angle equals a solution which is where the arc functions come into play. It’s also the reason he said that you can’t use some solutions because they lie outside of the domain of the arc functions. Btw keep in the mind the domain of arc tangent is (-00,00) because the range of tangent is (-00,00), and the domain of tangent is (-pi/2,pi/2).
MAKE SOME VIDEO IN TRIGONOMETRIC EQUATIONS equations that connecting trigonometric functions including their reciplac in both three ways of solving trigonometric equations, either by; (a) Expressing the given equations into quadratic trigonometric equations or into polynomial equations. (b) By factorization method. (c) By t-formula, i.e. equations of the form
In 5:36, how do you get the (PI over 6 and 5PI over 6) and (7PI over 6 and 11PI over 6) without any mathematical solutions to prove it? PLEASE RESPOND!!!
firstly, sin associates with Y value. On the unit circle, Y value of sin only appears at PI/2, given 1, and 3PI/2, given -1. other than that sin value will go out of the range. lets say 7PI/2 gives a 630 degree, which lies on the Y -1, but it goes out of the range. you always have to according to the restriction between 0 and 2PI to justify the answers.
2sinx + 1 = 0. So subtract both sides by 1 which will then equal 2sinx = -1. To get rid of the 2 in front of "sinx", divide both sides by 2, which will equal sinx = -1/2. Remember that 2sinx is 2(times)sinx.
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This guy is legit like the only person I understand in math.
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Hallelujah ur a life savior, you covered some things that I struggled on 😭
This is probably the hardest concept we did so far in my algebra 2 class. I got a test in 3 days on this and everything else on the unit. I'm probably gonna watch this multiple times.
Hi there, freaky Fred
thank you so much!! My math class goes super fast/confusingly and you always help to simply explain - huge props to you
MR. Organic Chemistry Tutor thank you for solving Trigonometric Equations by Factoring and using Identities.
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literally have a test tomorrow and this saved me 😭THANK YOU
I forgotten some algebra there's no hope for me in this trig class
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Just watched the video and it was so helpful for my trigonometric equations. Thank you ✨
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u literally saved my life dude
A couple of those could have been solved much simpler.
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I'm a little confused with the 4 sin^2 x - 1 = 0 example. I understood every step to sin x equaling 1/2 and -1/2. But my understanding is that I would have to use arc sine in order to find the values of x. Doesn't arc sine only exist in Quadrants 1 and 4? So, wouldn't x equal 30 degrees (pi/6) and -30 degrees (-pi/6) since the range is [-pi/2, pi/2]? Please advise.
As always, your videos are superb. Thank you for taking the time to make and upload them. Using them to study for my math certification exam (teaching).
If you look at the upper right corner at 5:35, there is a range where the angle can exist [0, 2pi]. I believe you can right that as long as there is a range given which exists on all quadrants.
Bruh i feel like my teachers missed a couple things when teaching this
Fucking love this guy, missed a lot of school and with this I’m catching up
Wow, this is very helpful!
Istg this factorization, been studying it for years now and i still dont get it smh
Ur a life saver
Sir plz solve this question... 5 sin square thita +14 cos thita=13
Watching this in math because this makes sense and my teacher doesn't
Thanks
How do we know when we are supposed to use the arc sin or any arc trig function in a problem? I mean if youre solving an equation and you have last that a trig function equals a value, i know that sine or cosine of the answer equals a value, but what about the restricted range of the trig function??? How do we know when we are supposed to not use the inverse function and when we do use the inverse. It just doesnt make sense to me since an equation requires you to apply the opposite function to solve it??? I really could use some answers or a video that highlights why you can include values outside of an inverse trig function's range
Hi, I’m not OP but here’s an explanation. You mainly use an arc trig function to find the angle that corresponds to that value. For example let’s just say you have sin x=1. You need to find what angle at sin x is equal to 1. You can’t find it using the regular sin function, so you need to use the arcsine function. So arcsinx of 1 is
= pi/2. Now since we are using arcsine we are only able to evaluate inputs between [-1,1]. So, we are just able to evaluate what angle that is. with sinx=1. Let’s say we had to evaluate sinx=1.2. We need to use arcsine to find the angle that equals to 1.2. However our restricted domain for arcsine is [-1,1], so we are unable to find the angle giving us no solution. Btw keep in mind that the range of sine is the same as the domain of arcsine. And that the domain of sine is the same as the range of arcsine. So keeping that in mind, the domain of sine is (-00,00) which is the range of arcsine. So when we are looking for a trig value, the output is able to lie within the range because it can be anything. However, the domain of input is restricted to [-1,1]. Hope that makes sense
Oh also, you use the regular trig functions in order to find the function value. For example Tanx=45degrees. This question is asking us what is the value using the tangent function at 45 degrees. Tan is sin/cos. So rad2/2 divided by rad2/2 is equal to 1. So our function is equal to 1. In the video above, OP is specifically looking for the opposite. He is looking for what angle equals a solution which is where the arc functions come into play. It’s also the reason he said that you can’t use some solutions because they lie outside of the domain of the arc functions. Btw keep in the mind the domain of arc tangent is (-00,00) because the range of tangent is (-00,00), and the domain of tangent is (-pi/2,pi/2).
a LITTERAL life saver
MAKE SOME VIDEO IN TRIGONOMETRIC EQUATIONS
equations that connecting trigonometric functions including their
reciplac
in both three ways of solving trigonometric equations, either by;
(a) Expressing the given equations into quadratic trigonometric equations or
into polynomial equations.
(b) By factorization method.
(c) By t-formula, i.e. equations of the form
But this guy makes us simple work in class . Things that can take for the 3 months we understand for single hour
The very first example why is sinx =π/2
Why do you lose a solution when you divide by cosx in the equation 2sinxcosx=cosx
Cuz it’s a piece of wood
7:32 tricky af!
In 5:36, how do you get the (PI over 6 and 5PI over 6) and (7PI over 6 and 11PI over 6) without any mathematical solutions to prove it?
PLEASE RESPOND!!!
Search "trigonometry unit circle" on youtube and watch some videos to find the explanation you prefer :)
why was ur first answer in radiants??? cuz usually we just do it normally so it would’ve been x=the inverse of sin 1
You can always convert the radians to degree
I’m failing this test lmao
Why does -1xosx change to 2?
How do you know the solutions at 8:45 - 9:01 ? Is that memorization?
probably not useful anymore, but yes. you just gotta know what radian values go along with which points on a graph to find the solutions
Clutch god
Ok
Hey come back
13:00
1:26 why the sine cannot be -3 how did you know that it's -1 to 1 anybody knows. sorry I'm just by my self learning this on my own
Go to the calculator and try sin any angle like 30 or 60 or 300 no way you can get -3 because the opp is smaller than the hyp
Microsoft whiteboard on a surface?
you're working in radians,
It’s easier to do that when doing trig.
How did he got the pi?
its 180
I need help! Why can't the answer be greater than 1 or less than negative 1???
That’s my question
Sin function only works in ranges of 1 to -1
If you look at a sign graph, the peaks in each interval only go up to 1 and down to -1
firstly, sin associates with Y value. On the unit circle, Y value of sin only appears at PI/2, given 1, and 3PI/2, given -1. other than that sin value will go out of the range. lets say 7PI/2 gives a 630 degree, which lies on the Y -1, but it goes out of the range. you always have to according to the restriction between 0 and 2PI to justify the answers.
I don't get
You need to know many math rules to understand this shht
4:56. Why do we subtract 1 then divide by 2?????
2sinx + 1 = 0. So subtract both sides by 1 which will then equal 2sinx = -1. To get rid of the 2 in front of "sinx", divide both sides by 2, which will equal sinx = -1/2.
Remember that 2sinx is 2(times)sinx.
at 9:43 how did you get tha the value of cosine^2
cos² +sin² = 1
3/2 is not greater than one.
It’s 1 and a half dude
???
🙏🙏🙏🙏
6:19 could u have done the same process for the previous problem? I tried to do that but I got the square root of 1/2
So you're supposed to just guess the value of x? Never explained how to figure out that cos(pi) is -1.
You can use the unit circle as your reference.
Does not a difference if you put 2 or -3 before in your cos problem.