Melody

921 Questions
34316 Answers

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Mathigon website

I just discovered a site called mathigon

https://mathigon.org/dashboard
और पढ़ें ..

Melody 2 Apr 2022

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Reporting problems.

There are a number of people having problems at the moment.
If you lodge an error can you please give all relevant information that you can think of.
Where you are in the world - the city or the country - is often very important information. और पढ़ें ..

Melody 16 Mar 2022

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How to make an empty rectangle in Latex

I just worked out how to make an empty box of various horizonal size easily in LaTex.
This is not a recommended method, but other methods that I have tried here have not worked.
This coding is really simple. Just put letters inside और पढ़ें ..

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Melody 15 Mar 2022

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HAPPY ANNIVERSARY CHRIS! (CPhill)

I just realized that you became a member here exactly 8 years ago today!

It has been great to have you here all that time

और पढ़ें ..

विषय से परेे

Melody 13 Mar 2022

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WELCOME BACK CPHILL !

Melody 11 Feb 2022

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Very interesting question. No high level maths required.

This question was posted a couple of days ago.
https://web2.0calc.com/questions/geometry_73561

I think it is really interesting. It looks much more difficult than it really needs to be. और पढ़ें ..

Melody 1 Feb 2022

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Members have a better experience. Their questions get more attention.

Almost everyone is posting as a guest.
If you post as a member your questions will have a far greater chance of a timely and helpful response.
If you are a genuine learner and interact with us, you will become known and answerers will do और पढ़ें ..

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Melody 9 Jan 2022

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HAPPY BIRTHDAY CHRIS!

और पढ़ें ..

Melody 20 Des 2021

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Some guidelines for question askers.

Please note:
1) Ask only one question per post.
2) When asking a question talk about what you have tried for yourself. Then people will take you more seriously and will be more able and willing to help.
3) If you get an answer और पढ़ें ..

विषय से परेे

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Melody 4 Nov 2021

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Remeniscing - a little historical humour.

I just found this old post that some of you, who have been around for a while, might find amusing.

https://web2.0calc.com/questions/look-at-the-new-leader-board

विषय से परेे

Melody 29 Okt 2021

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Sorry, this is not a question.

Melody6 Okt 2013

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This is a really good question.

a/bc = a/b*c
Multiply and divide are equally important. You just do them from left to right.
that is why it is a/b then multiply by c

Melody6 Okt 2013

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Sorry, your question doesn't make sense

Melody6 Okt 2013

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Hi scrutinizer,
It is no big deal but you missed a 2 in your final answer
(4 + 3)/2sqrt(3) = 7/ 2sqrt(3)
I also agree that 2+2/3 is an input error.

I also think that you have missed out half the solution.
I did it a bit differently
I said
7sin^2theta + 3cos^2 theta = 4
4sin^2theta + 3sin^2theta + 3cos^2 theta = 4
4sin^2theta + 3(sin^2theta + cos^2 theta) = 4
4sin^2theta + 3*1 = 4
4sin^2theta = 1
sin^2theta = 1/4
sin theta = +-1/2
theta = 30,150, 210, 330, ....degrees
cos theta = +-(root3) / 2
so sectheta+costheta = +-( 2/root3 + root3/2) = +- [7/(2root3) ] = +- ( 7root3/6 )

If you think I am wrong I would be happy to listen to your argument.

Melody5 Okt 2013

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I can't answer this without more information. sorry

Melody5 Okt 2013

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firstly
They are not 'reverse' functions. They are INVERSE FUNCTIONS.

say you want inverse cos (0.5)
sometimes written as cosa(0.5) where a stands for arc which is the same as inverse
sometimes written as cos ^-1(0.5) ----- NOTE: this is totally different from [cos(0.5)] ^-1

Anyway, to put this in the web 2.0 calculator input
2nd, acos, 0.5 =

Other cals may be inv, cos ^-1, 0.5

REMEMBER
Your calculator must be set to degrees or to radians - whichever you are working in. This is easy to overlook.
The number must be between -1 and 1 inclusive. Otherwise there can be no answer and you will get an error message.

I hope this helps.

Melody5 Okt 2013

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Melody5 Okt 2013

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Your question does not make any sense.

Melody5 Okt 2013

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Your question: show the sin^-1 (cosx)= x + pi/2

answer, you can't because it is not true, at least it is not true for any positive value of x because arcsin (anything) must be between -pi/2 and p/2 and this would give an answer above this range.
It may be true for x values between -pi and 0, I'll have to think about that some more.

I've only really thought this through for acute values of x (0 is less than x is less than pi/2 radians)
but
sin^-1 (cosx)= pi/2 - x (in radians)

Method 1 (0 is less than x is less than pi/2 radians)
You can show this by drawing a right angled triangle. One acute angle is x so the other one has to be pi/2-x
Let the side adjacent to x be p and the one opposite be q and the hypotenuse be h.
cos x = p/h
sine of what angle equals p/h? Sine of the angle at the other vertex equals p/h and the angle at the other vertex is pi/2-x
Try drawing the triangle - it should make it clearer.

Method 2
You can also consider it algebraically (0 is less than x is less than pi/2 radians)
arcsin[cos(x)]
= arcsin[sin(π/2 -x)]
= π/2 - x

It you want me to think about it more let me know.

Melody5 Okt 2013

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Our guest is correct.
If you have 2 b's and you ADD 1 b you do get 2b+1b = 3b
BUT
your question is multiply NOT add

2b*b = 2*b*b= 2*b^2 = 2b^2 that is, 2bsquared

Melody5 Okt 2013