how do i prove that

[rider]

∀x ∈]0, 1[, (1/2)arccos (2x − 1) + arcsin (√x) = π/2

[Lily]

I don't even think Google knows that. . .

[Radene]

I'd try to prove it graphically. As in, draw/plot a graph for the function.

[Masque]

Copy/paste it into the search bar on Wolfram Alpha, prefaced by "solve".

It even gives you a step-by-step solution, and plots a graph.

wolframalpha.com

[76561198278961361]

One thing that might help is rearranging the terms:
(1/2)arccos (2x − 1) = arcsin (√x) - π/2
From here, keep in mind that arccos and arcsin are inverse trigonometric functions, and that means the result of the function is an angle, so π/2 has something to do with it.
Presumably it has to do with how a cosine wave is just a sine wave shifted by π/2 radians.

Unfortunately this is the point where I no longer know what to do, because I don't know enough about the inverse trig functions. I could suggest looking into it from the perspective of the exp function, and e^(iθ).

[TruebornJester]

Ursprungligen skrivet av rider:

∀x ∈]0, 1[, (1/2)arccos (2x − 1) + arcsin (√x) = π/2 [/quote]
The question I would ask is. Why prove it?

[Schamyl006]

I think you should ask a mathemathican about that. I don't think anyone here knows that much about mathemathics. They wouldn't be here writing on forums If they were that good at mathematics, they would be preparing for a lecture at a university instead.

[iceman1980]

Ursprungligen skrivet av rider:

∀x ∈]0, 1[, (1/2)arccos (2x − 1) + arcsin (√x) = π/2 [/quote]


Plug in values see if it's equal... simple as that.

[76561198278961361]

Ursprungligen skrivet av Rolekte:

I think you should ask a mathemathican about that. I don't think anyone here knows that much about mathemathics. They wouldn't be here writing on forums If they were that good at mathematics, they would be preparing for a lecture at a university instead.

Not necessarily. Many people in sites like StackOverflow would probably be able to answer, and even I tried to give what I know despite not knowing enough to straight up solve the problem. So I think you could find some math enthusiasts or mathematicians who know more about it to give a better answer.

Ursprungligen skrivet av Product ∏:

Ursprungligen skrivet av rider:

∀x ∈]0, 1[, (1/2)arccos (2x − 1) + arcsin (√x) = π/2 [/quote]


Plug in values see if it's equal... simple as that.

Not really. That might tell you if it is, but it doesn't tell you why.

[iceman1980]

Ursprungligen skrivet av stupid(x)=e^x:

Ursprungligen skrivet av Rolekte:

I think you should ask a mathemathican about that. I don't think anyone here knows that much about mathemathics. They wouldn't be here writing on forums If they were that good at mathematics, they would be preparing for a lecture at a university instead.

Not necessarily. Many people in sites like StackOverflow would probably be able to answer, and even I tried to give what I know despite not knowing enough to straight up solve the problem. So I think you could find some math enthusiasts or mathematicians who know more about it to give a better answer.

It's actually really simple to answer. Plug in values from 0 - 1 see if the result is equal to pi / 2

The expression is also around the wrong way. ∀x ∈]0, 1[, (1/2)arccos (2x − 1) + arcsin (√x) = π/2

∀x ∈ (0,1) 0 to 1 being a set. this can include ALL numbers in this set unless there a limiting parameters such as ∀x ∈ is the set of rational numbers (numbers that can be expressed as a ratio of two numbers)

∀x ∈ (0,1) Prove that.
(1/2)arccos (2x − 1) + arcsin (√x) = pi / 2 (Prove the left is equal to the right) simple as that.

[76561198278961361]

Ursprungligen skrivet av Product ∏:

Ursprungligen skrivet av stupid(x)=e^x:

Not necessarily. Many people in sites like StackOverflow would probably be able to answer, and even I tried to give what I know despite not knowing enough to straight up solve the problem. So I think you could find some math enthusiasts or mathematicians who know more about it to give a better answer.

It's actually really simple to answer.

It doesn't really work like that. Discretely, you could give an answer, but if you want to make sure it's true for all values of x between 0 and 1, if you do it one by one you can always find a value in between that you haven't checked. To put it in perspective, there are more real numbers in between 0 and 1 than there are whole numbers at all.
And you might think it sounds a bit petty to make sure every single value is π/2, but that's just how it is. Plus, if instead of just solving it with a calculator you go through the process and reason why it's true, it's going to give you a lot more insight into the topic.

Math isn't always about the result. More often than not it's about figuring out how to get there.

Oh, and after seeing the edit, just a tip, (0,1) implies that it's only the values between the two, whereas [0,1] implies it's all the values from 0 to 1.

[iceman1980]

Ursprungligen skrivet av Rolekte:

Ursprungligen skrivet av Product ∏:

It's actually really simple to answer. Plug in values from 0 - 1 see if the result is equal to pi / 2

The expression is also around the wrong way. ∀x ∈]0, 1[, (1/2)arccos (2x − 1) + arcsin (√x) = π/2

∀x ∈ (0,1) 0 to 1 being a set. this can include ALL numbers in this set unless there a limiting parameters such as ∀x ∈ is the set of rational numbers (numbers that can be expressed as a ratio of two numbers)

∀x ∈ (0,1) Prove that.
(1/2)arccos (2x − 1) + arcsin (√x) = pi / 2 (Prove the left is equal to the right) simple as that.

I am not even capable of doing primary school math, so to me, this sounded like something only a renowned mathemathican could answer lol.

Not even close. I'm the process of teaching mysefl, high school to university mathematics off my own back. We've done basic mathematical induction at university but I wasn't satisfied with my level of understanding. What he is doing is called mathematical induction, he simple needs to do some manipulation using rules he should have learned at university / college if he hasn't he needs to go back to review that, it's that simple.

Good percentage of mathematics is just rules. When you get into Differential equations is when you can get mathematically creative.

From my vague understanding those arccos functions are simple cos^-1 = theta because thats what happens when you take the inverse of your cosine ratio you get an angle. rerpresented by the corner of a set of sides which is SOHCAHTOA - Sine(Opposite over hypotenuse), Cosine(Adjacent over hypotenuse), Tangent(Opposite over adjacent). Cosine(something) is going to return an angle)

In this case it's related to the unit circle not necessarily triangles.

If you are struggling with the basic mathematics you won't be able to try to even solve that equation. You must know the rules off by heart and know how to manipulate them. Looking at your previous posts, if you want to get into OpenGL you will require a lot of linear algebra.

Cross-products, Dot products, Scalar manipulations etc, without them 3D graphics will be but a dream.

[kbm demon.♡]

ahem wtf is that

[Rumpelcrutchskin]

42

[Q-T_3.14.exe]

My math teachers were right...

[rider]

thank you all i learned that arcsin(x) is (1/2)arccos(1-2x²) and by plugging √x into arcsin i got arcsin(√x)=(1/2)arccos(1-2x) which gave me (1/2)(arccos(2x-1)+arccos(1-2x)) which is (1/2)[-arcsin(2x-1)+arcsin(2x-1)+pi]=pi/2