Richiez:

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

thanks once again! Sorry, are u a student of unibuja ? My e-mail:adams_itconsult@123mail.org

indoorlove: thanks richiez. Pls no 2.

Richiez, good job! All the gurus in the house respect!

@Indoorlove, richiez did a good job. I may post an alternative method for you later.

x-fire:

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

I used .gif.

3^x + 2^x = 5^x

To plot these graphs, you will need a table of values for 5^x and 3^x + 2^x

To create the table, you will give x a range of values, say from -2 to 2 and calculate the corresponding values of each of the equation by substituting each value of x in the equations.

For the graph 5^x,
Lets subtitute the value of x each from -2 to 2

When x = -2
5^x = 5^(-2) = 1/5^2 = 1/25
3^x + 2^x = 3^(-2) + 2^(-2) =1/9 +1/4 = 13/36

When x = -1
5^x = 5^(-1) = 1/5^1 = 1/5
3^x + 2^x = 3^(-1) + 2^(-1) =1/3 +1/2 = 5/6

When x = 0
5^x = 5^(0) = 1
3^x + 2^x = 3^0 +2^0 = 1+ 1 = 2

When x =1
5^x = 5^(1) = 5
3^x + 2^x = 3^1 + 2^1 = 5

When x = 2
5^x = 5^2 = 25
3^x + 2^x = 3^2 + 2^2 = 13

And so on.....

x = -2 , -1, 0, 1, 2
5^x = 1/25, 1/5, 1, 5, 25
3^x + 2^x = 13/36, 5/6, 2, 5, 13

If you plot these two graphs with the above table, they will intersect at x = 1, i.e when their values at x = 1 are the same. So the value of x at their point of intersection is your answer, which is 1. If you notice at x = 1, 3^x + 2^x = 5 and 5^x = 5 meaning that the two graphs will intersect at that point.

To the best of my knowledge, that's the only way that equation can be solved!

doubleDx:

Question 2

f(x) = sin^3 x cos 3x

First off, let us simplify cos 3x =>
cos(3x) = cos 2xcos x - sin 2x sin x
= cosx ( 1- 2sin^2 x) - 2sin^2 x cosx
= cos x (1- 4sin^2 x)

:. Inserting cos 3x into the question yields :

f(x) = sin^3x cos x (1- 4sin^2 x)
f(x) = cos x sin^3 x - 4sin^5 x cos x

Now, we can split the function into two (sin^3 x cos x) and (- 4 sin^5 x cos x)

Applying product and chain rule to -4sin^5 x cos x yields :

Put u = - 4sin^5 x, du/dx = - 20sin^4 x cos x
v = cos x, dv/dx = sin x

f'(x) of - 4sin^5 x cos x =>
= udv/dx + vdu/dx
= -4sin^5 x .(-sin x) + cos x (- 20sin^4 x cos x)
= 4sin^6 x - 20sin^4 x cos^2 x
since cos^2 x = (1 - sin^2 x) substituting yields
= 4sin^6 x - 20sin^4 x (1 - sin^2x)
= 4sin^6 x - 20sin^4 x + 20sin^6 x
= 24sin^6 x - 20sin^4 x .....(1)

For the derivative of the second part of f(x) which is cos x sin^3 x

Put u = cos x, du/dx = -sin x
v = sin^3 x, dv/dx = 3sin^2 x cos x

f'(x) for cos x sin^3 x =>
= udv/dx + vdu/dx
f'(x) = cos x . (3sin^2 x cos x) + sin^3 x (-sin x)
= 3sin^2 x cos^2 x - sin^4 x
since cos^2 x = (1 - sin^2 x) substituting yields =>
= 3sin^2 x (1 - sin^2 x) - sin^4 x
= 3sin^2 x - 3sin^4 x - sin^4 x
= 3sin^2 x - 4sin^4 x .... (2)

:.f'(x) for the expression cos x sin^3 x - 4sin^5 x cos x =>

= 3sin^2 x - 4sin^4 x + 24sin^6 x - 20sin^4 x

collecting like times =>

:.The first derivative of the function f(x) =>
f'(x) = 3sin^2 x - 24sin^4 x + 24sin^6 x

Second derivative of f(x) is the derivative of f'(x).

Since f'(x) = 3sin^2 x - 24sin^4 x + 24sin^6 x
f''(x) = 6sin x cos x - 96sin^3 x cos x + 144sin^5 x cos x.

Simplifying =>

f''(x) = 6sin x cos x (1 - 16sin^2 x + 24sin^4 x)

@Ghettoguru, I made some mistakes before. Re-check now, I have corrected them.

Ok thanks sir, I get it now.

doubleDx:

I used .gif.

3^x + 2^x = 5^x

To plot these graphs, you will need a table of values for 5^x and 3^x + 2^x

To create the table, you will give x a range of values, say from -2 to 2 which and calculate the corresponding values of each of the equation by substituting each value of x in the equations.

For the graph 5^x,
Lets subtitute the value of x each from -2 to 2

When x = -2
5^x = 5^(-2) = 1/5^2 = 1/25
3^x + 2^x = 3^(-2) + 2^(-2) =1/9 +1/4 = 13/36

When x = -1
5^x = 5^(-1) = 1/5^1 = 1/5
3^x + 2^x = 3^(-1) + 2^(-1) =1/3 +1/2 = 5/6

When x = 0
5^x = 5^(0) = 1
3^x + 2^x = 3^0 +2^0 = 1+ 1 = 2

When x =1
5^x = 5^(1) = 5
3^x + 2^x = 3^1 + 2^1 = 5

When x = 2
5^x = 5^2 = 25
3^x + 2^x = 3^2 + 2^2 = 13

And so on.....

x = -2 , -1, 0, 1, 2
5^x = 1/25, 1/5, 1, 5, 25
3^x + 2^x = 13/36, 5/6, 2, 5, 13

If you plot these two graphs with the above table, they will intersect at x = 1, i.e when their values at x = 1 are the same. So the value of x at their point of intersection is your answer, which is 1. If you notice at x = 1, 3^x + 2^x = 5 and 5^x = 5 meaning that the two graphs will intersect at that point.

To the best of my knowledge, that's the only way that equation can be solved!

simple and straight forward, kudos

indoorlove: thanks once again! Sorry, are u a student of unibuja ? My e-mail:adams_itconsult@123mail.org

yep i'm a student of Uniabuja, i got your email addy and i'll reply soonest

Thanks for opening this thread @richiez.

1. Prove that cosec^2(x)cos(4x) - 15sin^2(x) + 6 = (cosec x + 3sin x)(cosec x - 3sin x)

2. Evaluate ∫ cos^2(x) sin^3(x) dx

Sorry, I missed some numbers in the first question. Corrected. Thanks

Richiez:

yep i'm a student of Uniabuja, i got your email addy and i'll reply soonest

thanks bruv, i'm a unibuja student too.....u and others are doing a great job here!

doubleDx:

Richiez, good job! All the gurus in the house respect!

@Indoorlove, richiez did a good job. I may post an alternative method for you later.

ok, thanks.

Ghettoguru: Thanks for opening this thread @richiez.

1. Prove that cosec^2(x)cos(4x) - cos^2(x) = (cosec x - 3cos x)(cosec x - cos x)

2. Evaluate ∫ cos^2(x) sin^3(x) dx

let me help with no 2 for now;

∫cos^2(x)sin^3(x)dx can be rewritten as;
∫cos^2(x)[sin(x)sin^2(x)dx........................(1)

from trig identities, sin^2(x) = 1 - cos^2(x)..............(2)
we now put eqn(2) in eqn(1)
∫cos^2(x)[1-cos^2(x)]sin(x)dx....................(3)

let us make u=cosx..............................(4)
this implies du/dx = -sinx
and dx= -du/sinx................................(5)

putting eqn(4) and (5) in eqn(3) we get

∫cos^2(x)[1-cos^2(x)]sin(x)dx = ∫u^2[1-u^2]sin(x)*-du/sinx

= -∫u^2[1-u^2]du
= ∫[u^4-u^2]du
= (u^5)/5 - (u^3)/3 + C

But recall that u=cosx
therefore the proper solution is;

1/5[cos^5(x)] - 1/3[cos^3(x)] + C

Ghettoguru: Thanks for opening this thread @richiez.

1. Prove that cosec^2(x)cos(4x) - 8sin^2(x) - 8 = (cosec x - 3cos x)(cosec x - cos x)

2. Evaluate ∫ cos^2(x) sin^3(x) dx

Sorry, I missed some numbers in the first question. Corrected. Thanks

QUESTION 2: Int cos^2(x)sin^3(x)dx. The above could be written as Int[(cos^2(x) sin^2(x)}sin(x)]dx. Now let u=cos(x), du=-sin(x)dx, and dx=-du/sin(x). Sin^2(x)=1-cos^2(x) -trig. Therefore: -Int[u^2{1-u^2}sin(x).du/sin(x)= -Int[u^2-u^4]du -[u^3/3-u^5/5]. Rem: u=cos(x), therefore, the final answer is -[cos^3(x)/3-cos^5(x)]+c ANS

Ghettoguru: Thanks for opening this thread @richiez.

1. Prove that cosec^2(x)cos(4x) - 8sin^2(x) - 8 = (cosec x - 3cos x)(cosec x - cos x)

2. Evaluate ∫ cos^2(x) sin^3(x) dx

Sorry, I missed some numbers in the first question. Corrected. Thanks

QUESTION 2: Int cos^2(x)sin^3(x)dx. The above could be written as Int[(cos^2(x) sin^2(x)}sin(x)]dx. Now let u=cos(x), du=-sin(x)dx, and dx=-du/sin(x). Sin^2(x)=1-cos^2(x) -trig. Therefore: -Int[u^2{1-u^2}sin(x).du/sin(x)= -Int[u^2-u^4]du -[u^3/3-u^5/5]. Rem: u=cos(x), therefore, the final answer is -[cos^3(x)/3-cos^5(x)]+c ANS. Note:Int means integral sign!

Thanks @richiez and indoorlove

1. Prove that cosec^2(x)cos(4x) - 15sin^2(x) + 6 = (cosec x + 3sin x)(cosec x - 3sin x)

There were some errors in question 1. I was coping two different questions at the same time math wan tear my head I've corrected them. Here is qestion 1 proper.

Richiez:

let me help with no 2 for now;

∫cos^2(x)sin^3(x)dx can be rewritten as;
∫cos^2(x)[sin(x)sin^2(x)dx........................(1)

from trig identities, sin^2(x) = 1 - cos^2(x)..............(2)
we now put eqn(2) in eqn(1)
∫cos^2(x)[1-cos^2(x)]sin(x)dx....................(3)

let us make u=cosx..............................(4)
this implies du/dx = -sinx
and dx= -du/sinx................................(5)

putting eqn(4) and (5) in eqn(3) we get

∫cos^2(x)[1-cos^2(x)]sin(x)dx = ∫u^2[1-u^2]sin(x)*-du/sinx

= -∫u^2[1-u^2]du
= ∫[u^4-u^2]du
= (u^5)/5 - (u^3)/3 + C

But recall that u=cosx
therefore the proper solution is;

1/5[cos^5(x)] - 1/3[cos^3(x)] + C

Kool, nice work bruv!

Ghettoguru: Thanks @richiez and indoorlove

1. Prove that cosec^2(x)cos(4x) - 15sin^2(x) + 6 = (cosec x + 3sin x)(cosec x - 3sin x)

There were some errors in question 1. I was coping two different questions at the same time math wan tear my head I've corrected them. Here is qestion 1 proper.

@ math wan tear my head lmao

Here is your solution=>

Cosec^2 x cos 4x - 15sin^2 x + 6 = (cosec x + 3sin x)(cosec x - 3sin x)

To solve this, you need the know of some basic trig identities like cos^2 x + sin^2 x = 1, Cos(A + B ) = cos A cos B - sin A sin B, cosec x = 1/sin x etc

From LHS Cosec^2 x cos 4x - 15sin^2 x + 6, lets look for an expression for cos 4x in terms of sin x. We are looking towards sinx and cosec x, so it's best we simplify in terms of sin x and also cosec x is the inverse of sin x.

cos 4x = cos (2x + 2x)
Applying Cos(A + B ) = cos A cos B - sin A sin B

Cos (2x+2x) = cos 2x cos 2x - sin 2x sin 2x ......(1)
Since cos 2x = cos (x + x)
= cos^2 x - sin^2 x
= 1 - sin^2 x - sin^2 x
cos 2x = 1 - 2sin^2 x

And sin 2x = sin x cos x + cos x sin x
= 2sin x cos x

Substituting cos 2x and sin 2x back in equation (1)
cos 4 x =>

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

substitute cos^2 x = 1 - sin^2 x

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

Thus Cosec^2 x cos 4x - 15sin^2 x + 6 =>
(1 - 6sin^2 x + 6sin^4 x)(1/sin^2 x) - 15sin^2 x + 6
= (1/sin^2 x) - 6 + 6sin^2 x - 15sin^2 x + 6
= cosec^2 x - 6 - 9sin^2 x + 6
= cosec^2x - (3^2)sin^2 x
Applying the difference of two squares (a^2 - b^2) = (a - b)(a + b)

Thus Cosec^2 x cos 4x - 15sin^2 x + 6 =>
= (cosec x + cos x)(cosec x - cos x)

LHS = RHS QED

Bravo!

bravo!

Who can differentiate btw d kernel of homomorphism of a group and that of a ring

Thanks y'all @ Richiez, doubleD and indoorlove. Appreciated!

pls can someone help me with the following question ....find the periods of the folliwing question sinx/x , /sinx/,/sinxcosx/,cos(wx+alpha)

Prove that there are no simple groups of order 63.

Can sumone solve dis 4 me....IF Tan(x+y)=4/3 and tanx=1/2, evaluate tany....kindly show d workings.....another is...Simplify sinB +sin3B +sin5B/ cosB + cos3B + cos5B...note B=beta

Fetus: Can sumone solve dis 4 me....IF Tan(x+y)=4/3 and tanx=1/2, evaluate tany....kindly show d workings.....another is...Simplify sinB +sin3B +sin5B/ cosB + cos3B + cos5B...note B=beta

solution coming up shortly

Fetus: Can sumone solve dis 4 me....IF Tan(x+y)=4/3 and tanx=1/2, evaluate tany....kindly show d workings.....another is...Simplify sinB +sin3B +sin5B/ cosB + cos3B + cos5B...note B=beta

To solve this question, apply the trig identity for tan (x + y) which is => (tan x + tan y)/(1 - tan x tan y)

Since tan (x + y) = 4/3
(tan x + tan y)/(1 - tan x tan y) = 4/3
cross multiplying yields =>
4 (1 - tan x tan y) = 3 (tan x + tan y)
4 - 4tan x tan y = 3tan x + 3tan y

Collecting like terms =>

3tan y + 4tan x tan y = 4 - 3tan x
tan y (3 + 4tan x) = 4 - 3tan x
tan y = (4 - 3tan x)/(3 + 4tan x)
Now remember that tan x = 1/2, substituting yields =>

tan y = [4 - 3(1/2)]/[3 + 4(1/2)]
tan y = [4 - 3/2]/5
tan y = (5/2) .(1/5)
:. tan y = 1/2.

I'm on phone, I will post the solution to question 2 later.

Fetus: Can sumone solve dis 4 me....IF Tan(x+y)=4/3 and tanx=1/2, evaluate tany....kindly show d workings.....another is...Simplify sinB +sin3B +sin5B/ cosB + cos3B + cos5B...note B=beta

but for the 2nd part of the question, is it (sinB+sin3B+sin5B)/(cosB+cos3B+cos5B) or sinB+sin3B+(sin5B/cosB)+Cos3B+cos5B

pls reply quickly for prompt solution

doubleDx:

To solve this question, apply the trig identity for tan (x + y) which is => (tan x + tan y)/(1 - tan x tan y)

Since tan (x + y) = 4/3
(tan x + tan y)/(1 - tan x tan y) = 4/3
cross multiplying yields =>
4 (1 - tan x tan y) = 3 (tan x + tan y)
4 - 4tan x tan y = 3tan x + 3tan y

Collecting like terms =>

3tan y + 4tan x tan y = 4 - 3tan x
tan y (3 + 4tan x) = 4 - 3tan x
tan y = (4 - 3tan x)/(3 + 4tan x)
Now remember that tan x = 1/2, substituting yields =>

tan y = [4 - 3(1/2)]/[3 + 4(1/2)]
tan y = [4 - 3/2]/5
tan y = (5/2) .(1/5)
:. tan y = 1/2.

I'm on phone, I will post the solution to question 2 later.

oh there you are...good job doubledx

Richiez:

but for the 2nd part of the question, is it (sinB+sin3B+sin5B)/(cosB+cos3B+cos5B) or sinB+sin3B+(sin5B/cosB)+Cos3B+cos5B

pls reply quickly for prompt solution

(sinB+sin3B+sin5B)/(cosB+cos3B+cos5B)...yea dis one

doubleDx:

To solve this question, apply the trig identity for tan (x + y) which is => (tan x + tan y)/(1 - tan x tan y)

Since tan (x + y) = 4/3
(tan x + tan y)/(1 - tan x tan y) = 4/3
cross multiplying yields =>
4 (1 - tan x tan y) = 3 (tan x + tan y)
4 - 4tan x tan y = 3tan x + 3tan y

Collecting like terms =>

3tan y + 4tan x tan y = 4 - 3tan x
tan y (3 + 4tan x) = 4 - 3tan x
tan y = (4 - 3tan x)/(3 + 4tan x)
Now remember that tan x = 1/2, substituting yields =>

tan y = [4 - 3(1/2)]/[3 + 4(1/2)]
tan y = [4 - 3/2]/5
tan y = (5/2) .(1/5)
:. tan y = 1/2.

I'm on phone, I will post the solution to question 2 later.

....Tank u so much...God bless u...

tanks guys .......i really do appreciate ur efforts...buh i hope someone can explain to me d Laplace transformation........wats it about.....and is dere any awoite on dis thread,buzz me on 07030195065,add me on 2go-kareem806

Show that the ring of polynomials F[x ] is an Euclidean domain , if Ø(f) = deg F

Fetus: (sinB+sin3B+sin5B)/(cosB+cos3B+cos5B)...yea dis one

(sinB+sin3B+sin5B )/( cosB+cos3B+cos5B )

To simplify this, you need a know some basic trig identities and expressions like sin2B = 2sinBcosB, cos2B = cos²B - sin²B or 1 - 2sin²B or 2cos²B - 1, cos²B + sin²B = 1

Lets simplify the numerator =>
sinB + sin3B + sin5B

sin 3B = sin (2B + B )
= sin 2B cos B + cos 2B sin B
Subsituting the expressions for sin2B and cos2B

Sin3B = cosB ( 2sinBcosB ) + sin B (cos²B - sin²B )
= sin B (1- 2sin² B ) + 2sin B (1 - sin² B )
sin 3B = 3sin B - 4 sin³ B

cos3B
= cos (2B + B ) = cos 2B cos B - sin2B sin B
= cos B(1 - 4sin² B )

Now lets simplify sin 5B =>

Sin (2B + 3B ) = sin2B cos3B + cos 2Bsin3B
Subsituting the expressions for sin2B, sin3B, cos2B and cos3B

Sin5B = 2sinBcosB.cosB(1 - 4sin² B ) + (1 - 2sin²B )( 3sinB - 4sin³B )
Expanding and collecting like terms =>

Sin 5B = 16sin⁵B - 20sin³B + 5sinB

Now, the numerator sinB + sin3B + sin5B =>
= sinB + 3sin B - 4 sin³B + 16sin⁵B - 20sin³B + 5sinB
Numerator = 16sin⁵B - 24sin³B + 9sinB

Lets take the denominator cosB+ cos3B + cos5B =>

We already know cos3B = cos B ( 1 - 4sin³B ) if we simplify it in terms of cos B,

cos 3B = 4cos³B - 3cos B

Cos5B = cos (2B + 3B )
= cos2B cos3B - sin2B sin 3B
Subsituting the expressions for sin2B, sin3B, cos2B and cos3B
cos5B = (2cos²B -1 )(4cos³B - 3cosB ) - 2sinBcosB (3sinB - 4sin³B )

Expanding and collecting like terms,

Cos5B = 16cos⁵B - 20cos³B + 5cosB

:. The denominator cosB+ cos3B + cos5B =>

= cosB + 4cos³B - 3cosB + 16cos⁵B - 20cos³B + 5cosB

Collecting like terms=>

Denominator = 16cos⁵B -16cos³B + 3cosB

Therefore the fraction (sinB+sin3B+sin5B )/( cosB+cos3B+cos5B ) =>

(16sin⁵B - 24sin³B + 9sinB )/ (16cos⁵B -16cos³B + 3cosB )

= sinB/cosB [ (16sin⁴B - 24sin²B +9)/(6cos⁴B -16cos²B + 3) ]

= tanB [ (16sin⁴B - 24sin²B +9)/(6cos⁴B -16cos²B + 3) ]

The above expression could still be simplified further either in terms of sinB only or cos B only. But I'll stop there.

I hope that helps.

Please kindly solve this jamb question for me: b=a +cp and r=ab + 1/2cp^2, express b^2 in terms of a, c and r