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Nairaland Mathematics Clinic - Education (17) - Nairaland

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Re: Nairaland Mathematics Clinic by ositadima1(m): 1:57pm On Jan 11, 2013
digitalwale:

How can I PM you? I tried going to your profile and clicked to send email but the page that came up only showed a prepared and uneditable message.

Jeeze, that's 500 more..., thnx man. grin
Re: Nairaland Mathematics Clinic by digitalwale: 2:21pm On Jan 11, 2013
ositadima1:

07065104668

No MTN for 1000 naira so I gave you 1500 naira MTN. Please acknowledge receipt. Thanks
Re: Nairaland Mathematics Clinic by ositadima1(m): 2:25pm On Jan 11, 2013
digitalwale:

No MTN for 1000 naira so I gave you 1500 naira MTN. Please acknowledge receipt. Thanks

I owe you then..., again thnx man.
Re: Nairaland Mathematics Clinic by digitalwale: 2:32pm On Jan 11, 2013
ositadima1:

I owe you then..., again thnx man.

You owe me nothing man. Watch out for more questions
Re: Nairaland Mathematics Clinic by biolabee(m): 5:18pm On Jan 11, 2013
DoubleDx

I thot we could use product of a product to do the derivative as you have increased the degree from 3 to 4
Please review your workings again
I feel you have complicated things o although i think you have expressed a new
way of the same equation as trig functions have various ways of relating to each other
so that it will not be like our lecturers that only believed in one answer


I have only put the first derivative

1 Like

Re: Nairaland Mathematics Clinic by indoorlove(m): 5:42pm On Jan 11, 2013
indoorlove: Pls this integration has been given me prob. Question 1: integrate 1/cos(x-a)cos(x-b) with respect to x. Question2: Pls, I need explanation on complex analysis with some examples and their solutions. Thanks !
Re: Nairaland Mathematics Clinic by arbitrage: 6:00pm On Jan 11, 2013
doubleDx is good.

1 Like

Re: Nairaland Mathematics Clinic by Nobody: 6:39pm On Jan 11, 2013
biolabee: DoubleDx

I thot we could use product of a product to do the derivative as you have increased the degree from 3 to 4
Please review your workings again
I feel you have complicated things o although i think you have expressed a new
way of the same equation as trig functions have various ways of relating to each other
so that it will not be like our lecturers that only believed in one answer


I have only put the first derivative

Your first derivative: (3cos3x cos x sin^2x - 3sin^3x sin 3x) isn't wrong. The only difference is that, I actually simplified mine with the expression for cos 3x in terms of sin x and cos x before differentiating it. I think if your answer is expanded in terms if sin x only, it should be same as mine.

Thanks for the observation though. I'll use the method you used before expanding it in terms of sin x so you can see the result.

Why I actually simplified it in terms of sin x is because the options might not be in terms of cos 3x and sin 3x and that would leave the person confusion!

Here it goes=>

Expressing your answer : 3cos 3x cos x sin^2x - 3sin^3x sin 3 x in terms of sin x =>

cos 3x = cos 2x cos x - sin2x sin x
cos 3x = cos x(1 - 4sin^2 x)

sin 3x = sin x cos 2x + sin 2x cos x
sin 3x = sin x(1- 2sin^2x) + 2sin x (1 - sin^2 x)
sin 3x = 3sin x - 4 sin^3 x

If you substitute the expression for sin 3x and cos 3x in your answer you will have:

= 3 cos x(1 - 4sin^2 x) cos x sin^2x - 3sin^3 x ( 3sin x - 4 sin^3 x)

= 3cos^2 x sin^2 x (1 - 4 sin^2 x) - 9sin^4 x + 12sin^6 x
= 3cos^2 x sin^2 x - 12sin^4 x cos^2 x - 9sin^4 x + 12sin^6 x

But cos^2 x = 1 - sin^2 x, substituting yields:

= 3sin^2 x(1 - sin^2 x) - 12sin^4 x (1 - sin^2 x) - 9sin^4 x + 12sin^6 x

Expanding=>

3sin^2 x - 3sin^4 x - 12sin^4 x + 12sin^6 x - 9sin^4 x + 12sin^6 x

f'(x) = 3sin^2 x - 24sin^4 x + 24sin^6 x

So your answer and mine are the same!

1 Like

Re: Nairaland Mathematics Clinic by Nobody: 6:41pm On Jan 11, 2013
arbitrage: doubleDx is good.

Thanks for the compliment bro!
Re: Nairaland Mathematics Clinic by Richiez(m): 7:43pm On Jan 11, 2013
digitalwale:

No MTN for 1000 naira so I gave you 1500 naira MTN. Please acknowledge receipt. Thanks

i must confess, i'm loving this grin grin grin grin grin
Re: Nairaland Mathematics Clinic by Nobody: 7:44pm On Jan 11, 2013
ositadima1:

Jeeze, that's 500 more..., thnx man. grin

Nice one bruv, you did a good job there!
Re: Nairaland Mathematics Clinic by biolabee(m): 8:32pm On Jan 11, 2013
nice one n i ir appreciate d cool way u took d feedbacck

I checked what u did and it looks tite so i am not condemning u
nice one once again
Re: Nairaland Mathematics Clinic by Nobody: 9:20pm On Jan 11, 2013
^Its ok bruv, your observation was helpful, it took me back and I realised I made some mistakes initially. I have corrected them though. I'm solving with the other method before expanding, I'm using a mobile now so I'll post that one later!
Re: Nairaland Mathematics Clinic by Nobody: 9:25pm On Jan 11, 2013
indoorlove: Pls this integration has been given me prob. Question 1: integrate 1/cos(x-a)cos(x-b) with respect to x. Question2: Pls, I need explanation on complex analysis with some examples and their solutions. Thanks !

Good questions. I'll check 'em out when I'm on PC!
Re: Nairaland Mathematics Clinic by Nobody: 9:45pm On Jan 11, 2013
Solve this and I will give the solution provider credit worth N3,000 (any network of his or her choice)
Note: Nobody I know has been able to solve this equation.
2^x + 3^x = 5^x.
Find x

Goodluck!!!
Re: Nairaland Mathematics Clinic by kasbeats(m): 1:26am On Jan 12, 2013
guys please help me with this.....prove that sin A+sin B=2[(sin(A+B)/2).cos(A-B)/2]....tanks broz,u guys have been really great......looking forward to a chemistry thread very soon
Re: Nairaland Mathematics Clinic by kasbeats(m): 1:41am On Jan 12, 2013
in how many ways can five letters be chosen from the letters of d word STATISTICS
Re: Nairaland Mathematics Clinic by biolabee(m): 7:07am On Jan 12, 2013
kasbeats: guys please help me with this.....prove that sin A+sin B=2[(sin(A+B)/2).cos(A-B)/2]....tanks broz,u guys have been really great......looking forward to a chemistry thread very soon

before attempting this question, one has to know the basic rules of sine addition, and the derivation of the factor formulae
NOW;

Sin(P+Q) = SinPCosQ + CosPSinQ......................(1)
Sin(P-Q) = SinPCosQ - CosPSinQ......................(2)

Subtracting eqn(2) from eqn(1),

Sin(P+Q) - Sin(P-Q) = CosPSinQ + CospSinQ

Sin(P+Q) - Sin(P-Q) = 2CosPSinQ ....................(3)

Now we have to do a little trick by letting;

P+Q = A ............................................(4)
P-Q = B ............................................(5)

adding eqns (4) and (5) we get;

2P = A + B
P = (A+B)/2 ........................................(6)

subtracting eqn(5) from eqn(4) we get;

2Q = A - B
Q = (A-B)/2 ........................................(7)

By appropriate substitution for A,B,P and Q in equation(3) we get ;

SinA - SinB = 2[Cos(A+B)/2*Sin(A-B)/2]

Q. E. D

kasbeat you raised the first q. confirm if u understood the first solution. If you do you will be able to solve this as that ivnvolved subtraction of trig functions which is much adder

However let me give you a hint Add equations 1 and 2 and go an sin no more

kasbeats: in how many ways can five letters be chosen from the letters of d word STATISTICS

Statistics has 10 letters
5 letters words to be spelt will involves a permutation rather than combi as ab is different from ba

10 P 5 = 10!/5! = 10 * 9 * 8 *7 * 6 = 30240

1 Like

Re: Nairaland Mathematics Clinic by Nobody: 7:39am On Jan 12, 2013
@biolabee, good hint. Lemme give him the alternative solution.

kasbeats: guys please help me with this.....prove that sin A+sin B=2[(sin(A+B)/2).cos(A-B)/2]....tanks broz,u guys have been really great......looking forward to a chemistry thread very soon


sin A+sin B = 2[(sin( A
+B )/2)cos( A - B )/2)]

This is easy: from the RHS =>

= 2[(sin(A
+B )/2).cos(A-B )/2]
= 2[(sin(A
/2+B/2)cos(A/2-B/2)]

Put P = A/2 and Q = B/2

= 2sin(P +Q)cos(P - Q)
Since sin(P + Q) = sinPcosQ + cosPsinQ
And cos(P - Q) = CosPcosQ + sinPsinQ

substituting =>
= 2[(sinPcosQ + cosPsinQ)(cosPcosQ + sinPsinQ)]

Expanding =>

= 2[sinPcosPcos^2 Q + sin^2P sinQcosQ + sinQcosQcos^2P + sinPcosPsin^2 Q]

Collecting like terms =>

2[sinPcosPcos^2 Q + sinPcosPsin^2 Q + sin^2PsinQcosQ + sinQcosQcos^2P]

Factorizing =>

2{ sinPcosP (cos^2Q + Sin^2Q) + sinQcosQ (sin^2P + cos^2P) }

Since cos^2x + sin^2x = 1

Simplifying yields =>
= 2[ sinPcosP (1) +sinQcosQ(1)]
= 2sinPcosP + 2sinQcosQ

Remember that : Put P = A/2 and Q = B/2. Replacing back gives =>

= 2sin(A/2)cos(A/2)+ 2sin(B/2)cos(B/2)

The above expression is a sum of two half angle formula for sinA and sinB respectively since sinA = sin(A/2 + A/2) = 2sin(A/2)cos(A/2) and sinB = sin(B/2 +B/2) = 2sin(B/2)cos(B/2)

:. RHS = sinA + sinB QED.
Re: Nairaland Mathematics Clinic by Nobody: 7:49am On Jan 12, 2013
x-fire:
Solve this and I will give the solution provider credit worth N3,000 (any network of his or her choice)
Note: Nobody I know has been able to solve this equation.
2^x + 3^x = 5^x.
Find x

Goodluck!!!

This can only be solved graphically. The answer is 1

Re: Nairaland Mathematics Clinic by Nobody: 8:53am On Jan 12, 2013
doubleDx:

This can only be solved graphically. The answer is 1
I downloaded the graphical solution but I couldn't view it. What format (programme) did you save it as?
Re: Nairaland Mathematics Clinic by indoorlove(m): 9:44am On Jan 12, 2013
doubleDx:

Good questions. I'll check 'em out when I'm on PC!
thanks in anticipation......
Re: Nairaland Mathematics Clinic by Nobody: 10:04am On Jan 12, 2013
x-fire:

I downloaded the graphical solution but I couldn't view it. What format (programme) did you save it as?

Plot two graphs : 3^x + 2^x and that of 5^x. Their point of intersection which is 1 is the answer to the equation!

I'll explain how a table of values for 5^x and 3^x + 2^x can be created and their point of intersection as the solution to the question.
Re: Nairaland Mathematics Clinic by Nobody: 10:05am On Jan 12, 2013
indoorlove: thanks in anticipation......

It's alright bruv, I'll soon post it. I'm still on mobile, so typing is kinda slow!
Re: Nairaland Mathematics Clinic by beey2(m): 10:08am On Jan 12, 2013
I'm not a Math Guru but a fan if permitted. U guys are really Gurus, so I'll just take a seat n learn more 4rm @ Double Dx, Ositadima1, Biolabee et al. How I wish there'll b thread for other aspect of learning like statistics n Chemistry etc. Once again I confirm this as the BEST on NL.
Re: Nairaland Mathematics Clinic by biolabee(m): 10:11am On Jan 12, 2013
bee-y:
I'm not a Math Guru but a fan if permitted. U guys are really Gurus, so I'll just take a seat n learn more 4rm @ Double Dx, Ositadima1, Biolabee et al. How I wish there'll b thread for other aspect of learning like statistics n Chemistry etc. Once again I confirm this as the BEST on NL.

THANKS but osita and doubledx are the gurus
if u need a thread for those subjects, just create one for each of them and before u know everyone is rolling
Re: Nairaland Mathematics Clinic by ositadima1(m): 10:49am On Jan 12, 2013
biolabee:

THANKS but osita and doubledx are the gurus
if u need a thread for those subjects, just create one for each of them and before u know everyone is rolling

Nah, Richiez, Biolabee and Doubledx are d gurus here.
Re: Nairaland Mathematics Clinic by Richiez(m): 10:56am On Jan 12, 2013
indoorlove: Pls this integration has been given me prob. Question 1: integrate 1/cos(x-a)cos(x-b) with respect to x. Question2: Pls, I need explanation on complex analysis with some examples and their solutions. Thanks !

To solve the first question, we need a little trick

let b-a = (x-a)-(x-b)..................(1)

∫1/cos(x-a)cos(x-b)dx can be rewritten as;

1/sin(b-a)∫sin(b-a)/[cos(x-a)cos(x-b)]dx............(2)

from eqn(1) and by the sine subtraction formula we have;
sin(b-a) = sin(x-a)cos(x-b)-sin(x-b)cos(x-a)..............(3)

putting eqn(3) in eqn(2) we get;

1/sin(b-a)∫tan(x-a)-tan(x-b)dx..............(4)

using the priciple of ∫f(x)-g(x) = ∫f(x) - ∫g(x) eqn(4) becomes;

1/sin(b-a)[∫tan(x-a)dx - ∫tan(x-b)dx]

= 1/sin(b-a)[-lncos(x-a) + lncos(x-b)] + C

2 Likes

Re: Nairaland Mathematics Clinic by Richiez(m): 11:02am On Jan 12, 2013
ositadima1:

Nah, Richiez, Biolabee and Doubledx are d gurus here.

No one person has been able to keep this thread as lively as it is, it's our collective efforts, thumb ups to ositadima1, Doubledx, Biolabee and some other humble gurus here, let's keep the thread on fire till we kill the spirit of maths terrorism
Re: Nairaland Mathematics Clinic by biolabee(m): 11:18am On Jan 12, 2013
good job guys!!
Re: Nairaland Mathematics Clinic by indoorlove(m): 11:58am On Jan 12, 2013
Richiez:

To solve the first question, we need a little trick

let b-a = (x-a)-(x-b)..................(1)

∫1/cos(x-a)cos(x-b)dx can be rewritten as;

1/sin(b-a)∫sin(b-a)/[cos(x-a)cos(x-b)]dx............(2)

from eqn(1) and by the sine subtraction formula we have;
sin(b-a) = sin(x-a)cos(x-b)-sin(x-b)cos(x-a)..............(3)

putting eqn(3) in eqn(2) we get;

1/sin(b-a)∫tan(x-a)-tan(x-b)dx..............(4)

using the priciple of ∫f(x)-g(x) = ∫f(x) - ∫g(x) eqn(4) becomes;

1/sin(b-a)[∫tan(x-a)dx - ∫tan(x-b)dx]

= 1/sin(b-a)[-lncos(x-a) + lncos(x-b)] + C

thanks richiez. Pls no 2.
Re: Nairaland Mathematics Clinic by Richiez(m): 3:09pm On Jan 12, 2013
indoorlove: thanks richiez. Pls no 2.

graphical methods would be employed when explaining complex analysis, especially the aspect of complex plane ( in complex mapping from x-y plane to u-v plane ) so the best i can do is to recommend good textbooks such as any advanced engineering maths textbook or better still use the internet. but you can give me ur email incase i come across a good material for you.
thanks

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