Ghettoguru: House gurus help with this questions pls :

1. Prove that sec (x) cos (3x) - cos (2x) = -2sin^2 x

2. (a) Solve the simultaneous equation for the value(s) of x and y.

x + y = 3
x√(y) - 3y +1 = 0

(b) Find the value(s) of (x - y)

I'll take question 1 first.
From the LHS:
We know that sec x = 1/cos x
Also, cos ( A + B ) = cos A cos B - sin A sin B
cos 2x = cos^2 x - sin^2 x
Also since, sin^2 x + cos^2 x = 1
cos^2 x = 1 - sin ^2 x
:. cos 2x = [1 - sin^2 x] - sin^2 x
cos 2x = 1 - 2sin^2 x

Same way,
cos 3x = cos 2x cos x - sin 2x sin x

= (1 - 2sin^2 x) cos x - sin 2x sin x
Using: sin (A + B ) = sin A cos B + cos A sin B,
sin 2x = 2sin x cos x

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

Now having known the expression for sec x = 1/cos x, cos 2x = (1 - 2sin^2 x) and cos 3x = cos x (1 - 4sin^2 x). Substituting on the LHS yields:

[1/ cos x ]. [ cos x (1 - 4sin^2 x)] - [1 - 2sin^2 x]
= 1 - 4sin^2 x - (1 - 2sin^2 x)
= 1 - 4sin^2 x -1 + 2sin^ x
= -2sin^ x.

Thus: sec (x) cos (3x) - cos (2x) = -2sin^2 x

Ghettoguru: House gurus help with this questions pls :

1. Prove that sec (x) cos (3x) - cos (2x) = -2sin^2 x

I meant "trig functionS"!

2nioshine: @doubleDX once again gr8 wuk....any who can should try this...2^x=4x.... question2
x+y+z=2, x^2+y^2+z^2=26,
x^3+y^3+z^3=38....plz no graphical soln..ans=4....Q2ans=4,-3 &1

Good one bro! I'll try 'em out later or tomorrow morning.

Ghettoguru: House gurus help with this questions pls :

1. Prove that sec (x) cos (3x) - cos (2x) = -2sin^2 x

2. (a) Solve the simultaneous equation for the value(s) of x and y.

x + y = 3
x√(y) - 3y +1 = 0

(b) Find the value(s) of (x - y)

Question 2

x + y = 3 ..... Equation 1
x√(y) - 3y +1 = 0 .... Equation 2

From Equation 1, x = 3 - y .... Equation 3.

Subtitute equation 3 in equation 2:
= ( 3 - y)√(y) - 3y +1 = 0
= ( 3 - y)√(y) = 3y - 1
squaring both sides to elimate the square root:

[( 3 - y)√(y)]^2 = ( 3y - 1 )^2
(3 -y)(3 - y).y = ( 3y - 1 ) ( 3y - 1 )
(9 - 6y + y^2)y = 9y^2 - 6y +1
9y - 6y^2 + y^3 = 9y^2 - 6y + 1

Rearranging and collecting like terms, yields:
y^3 - 15y^2 + 15y - 1 = 0
Factorising the polynomial gives:

(y - 1)(y^2 -14y - 1) =0
.: y - 1 = 0 or y^2 -14y - 1 =0

y1 = 1 . Lets solve y^2 -14y - 1 = 0 for y2 and y3 using the quadratic formula:
y^2 -14y - 1 = 0
a = 1, b = -14 and c = -1

y = - b ±√(b^2 - 4ac)/2a
y = - (-14)±√[ (-14)^2 -4(1)(-1)]/2(1)
y = 14±√(196 + 4)/2
y = 7±√(200)
y = 7±√(25).(4).(2)
y = 7 + 5(2)√2
:. y1 = 1, y2 = (7 + 10√2), y3 = (7 - 10√2).
We can now substitute each value of y in equation 3 to determine the corresponding value of x.

When y1 = 1
x = 3 - y
x = 3 - 1
x1 = 2

y2 = (7 + 10√2)
x2 = 3 - (7 + 10√2)
x2 = -4 - 10√2)

y3 = (7 - 10√2)
x3 = 3 - (7 - 10√2)
x3 = -4 +10√2
x3 = -4 +10√2

:. x1 = 2, x2 = (-4 - 10√2) and x3 = (-4 +10√2)

(b) Find the value(s) of (x - y)

x1 = 2, y1 =1
x1 - y1
= 2 - 1
= 1

x2 = (-4 - 10) , y2 = (7 + 10√2)
x2 - y2
= -4 - 10√2 - (7 + 10√2)
= -11 - 20√2

x3 = -4 +10√2, y3 = (7 - 10√2)
x3 - y3
= -4 +10√2 - (7 - 10√2)
= -4 + 10√2 - 7 + 10√2
= -11 + 20√2

:.The values of (x - y) are : 1, (-11 - 20√2) and (-11 + 20√2)

Use any of the equations to verify your answer by substuting the value of x and corresponding value y.

Eg.

From equation : x + y = 3
Check with x1 = 2 and y1 = 1

=2 + 1
= 3

Check with x2 = -4 - 10√2 and y2 = 7 + 10√2

-4 - 10√2 + 7 + 10√2
= 7 - 4 - 10√2 + 10√2
= 3

Check with x3 = (-4 + 10√2) and y3 = (7 - 10√3)
= -4 + 10√2 + 7 -10√2
= -4 + 10√2 + 7 - 10√2
= 3

ositadima1:

Boom!!! Memba mi tell yu.

Sum n=1 to 193,
(n+7)^3= 404009216

tnx bross buh can u pls explain further cos i didn't get u sir.. Thanks !

ositadima1:

Boom!!! Memba mi tell yu.

Sum n=1 to 193,
(n+7)^3= 404009216

tnx bros.. Buh can u explain further 2make it clear

lanre074: tnx bros.. Buh can u explain further 2make it clear

Unless u are extremely gifted, like newton, its goin to be pritty hard to fashion a formula for dat serie.

8^3+9^3+10^3+...+200^3

Assuming n starts at 1 then each term is like (n+7)^3

Now since u first term is 8^3 and last is 200^3 we got 200-7=193 term total.

So for n=1 to n=193, u gotta had all them (n+7)^3, now is where ur programmable calculator takes over... lets see what them math gurus will come up with, gurus over to ya...

^^^ actually u can expand (n+7)^3 and make a formula outta dat, I am sleepy, tomorrow...

BABE!:
I may be back with a proof to the trig. question.

You tried but there is an error when you got to the point of using quadratic equation on the variable z.

(z-1)(-z2-4z-1) = 0

z should have given you 1, -2+2√3 and -2-2√3

From these values of z, you will get y to be:
y=1, 8(2+√3), 8(2-√3).

Just check your quadratic equation of z to confirm. Thank you.

BABE!:


If you need more details or have more questions, ask.

You tried but there is an error when you got to the point of using quadratic equation on the variable z.

(z-1)(-z²-4z-1) = 0

z should have given you 1, -2+2√3 and -2-2√3

From these values of z, you will get y to be:
y=1, 8(2+√3), 8(2-√3).

Just check your quadratic equation of z to confirm. Thank you.

BABE!:


I meant "trig functionS"!

Good job, well done!

Mehn guys are tearing math here o!!!! @All, who have been helpful in solving questions here, much appreciated. God bless you all!

@doublddx baba your efforts are greatly appreciated. We are learning. Thanks.

BABE!:


I meant "trig functionS"!

Thanks BABE! God bless!

remmyz: solve the ones solvable from these:
http://exams4success.com/eknowledgebase/threads/12-Bearing-amp-Distance
http://exams4success.com/eknowledgebase/threads/27-Word-problem
http://exams4success.com/eknowledgebase/threads/25-Modular-arithmetic
http://exams4success.com/eknowledgebase/threads/26-simplify
http://exams4success.com/eknowledgebase/threads/24-Equation
http://exams4success.com/eknowledgebase/threads/23-Formulae
http://exams4success.com/eknowledgebase/threads/22-probability
http://exams4success.com/eknowledgebase/threads/21-error
http://exams4success.com/eknowledgebase/threads/20-series-amp-sequences
http://exams4success.com/eknowledgebase/threads/19-Number-Base
http://exams4success.com/eknowledgebase/threads/18-Sphere
http://exams4success.com/eknowledgebase/threads/13-Add-amp-subtract-of-areas-amp-volumes
http://exams4success.com/eknowledgebase/threads/11-Ratio-and

Next time, try and copy the questions here so that people will not need navigating away before they can answer your questions. Here is the first one:

1. A photographer is 350m away from a lion and want to get closer before he takes a photograph.There is a water-hole in the direct line between the lion and himself,so he moves at an angle of 8 degree to this line to a better position 200m further on. Calculate his distance from the lion.

Solution:
Let the required distance be x.
So, x² = 350²+200²-2 X 350 X 200 X cos 8^o
x² = 122500 + 40000 - 140000 X 0.9903
x² = 162500 - 138642
x² = 23858
x = 154.46
Answer: distance = 154.5 metres.

lanre074: Gurus pls solve dis na.. Find d sum of 8^3 + 9^3 + 10^3 + ... + 200^3

Back to this question

Each term => (n+7)^3
where n=1 to n=200-7=193

Meaning we have 193 terms from 8^3 to 200^3.

Summation of (n+7)^3 for n=1 to n=193

Sn= (1+7)^3+(2+7)^3+(3+7)^3+...+(193+7)^3

Let`s expand (n+7)^3
(n+7)^3=(n+7)(n+7)(n+7
=n^3+21n^2+147n+343

Let's see the summation of each term above

S= n^3= 1^3+2^3+3^3+...+nth= (1/4)(n^2)(n+1)^2


S=n^2= 1^2+2^2+3^2+...+nth= (1/6)(n)(n+1)(2n+1)

S=n =1+2+3+...+nth= (1/2)(n)(n+1)

S=1+1+1+...+nth= n

Now substituting back

Sn= (1/4)(n^2)[(n+1)^2]+21(1/6)(n)(n+1)(2n+1)+147(1/2)(n)(n+1)+343(n)

So, if u now carefully substitute n=193

Sn=404009216

Actually that equation can give u sum to any number of terms u want, hope this was helpful

2. http://exams4success.com/eknowledgebase/threads/27-Word-problem

D numerator of a fraction is 5 less than its denominator.If 6 is added to d numerator & 4 to denominator,d fraction is doubled.What is d fraction.

Solution:
Let the numerator be x and the denominator be y. So the fraction is x/y
Therefore, x = y-5
and (x+6)/(y+4) = 2(x/y)
Substituting y-5 for x in the second equation, you have
(y-5+6)/(y+4) = 2(y-5)/y
(y+1)/(y+4) = (2y-10)/y
On cross multiplying, you have:
y(y+1) = (y+4)(2y-10)
y²+y = 2y²-10y+8y-40
y²-2y²+y+2y+40 = 0
-y²+3y+40 = 0
y²-3y-40 = 0
(y+5)(y-eight) = 0
y=-5 or 8
So, x=y-5
x=-10 or 3
Since x/y is a fraction, therefore x cannot be -10 but 3 and y cannot -5 but 8.
So x/y = 3/8.

2nioshine: @doubleDX once again gr8 wuk....any who can should try this...2^x=4x.... question2
x+y+z=2, x^2+y^2+z^2=26,
x^3+y^3+z^3=38....plz no graphical soln..ans=4....Q2ans=4,-3 &1

Guy, u don't want graphing, ok
I am going to play with it small sha, shoot and miss style.

If u look at this one a^2+b^2+c^2=26

We can assume dat a, b, c are intigers based on this;

Nearest square to 26 is 25
5^2=25 off
4^2=16 ok
3^2=9 ok
2^2=4 ok
1^2=1 ok

Is clear dat if we add any three the best combo will be;
16+9+1=26

So, we have 4, 3 and 1
Like,
4, 3, 1
4,-3,-1
4,-3, 1
4, 3, -1
-4, 3, 1
-4, -3, -1
-4, 3, -1
-4,-3, 1

Now to a+b+c=2, 4-3+1=2
So, 4, -3, 1 is it.

To confirm whether we are right
a^3+b^3+c^3=38

This easy methold will only hold if a, b, c are INTEGERS!!!

arbitrage:

You tried but there is an error when you got to the point of using quadratic equation on the variable z.

(z-1)(-z²-4z-1) = 0

z should have given you 1, -2+2√3 and -2-2√3

From these values of z, you will get y to be:
y=1, 8(2+√3), 8(2-√3).

Just check your quadratic equation of z to confirm. Thank you.

I had a hunch there was something off. I'll cross-check, again. Thanks.

Ghettoguru:

Thanks BABE! God bless!

You're welcome.

Pls All Maths Gurux Should Add Me On Whatsapp With 08050874552 ###Thanks you All

ositadima1:


Guy, u don't want graphing, ok
I am going to play with it small sha, shoot and miss style.

If u look at this one a^2+b^2+c^2=26

We can assume dat a, b, c are intigers based on this;

Nearest square to 26 is 25
5^2=25 off
4^2=16 ok
3^2=9 ok
2^2=4 ok
1^2=1 ok

Is clear dat if we add any three the best combo will be;
16+9+1=26

So, we have 4, 3 and 1
Like,
4, 3, 1
4,-3,-1
4,-3, 1
4, 3, -1
-4, 3, 1
-4, -3, -1
-4, 3, -1
-4,-3, 1

Now to a+b+c=2, 4-3+1=2
So, 4, -3, 1 is it.

To confirm whether we are right
a^3+b^3+c^3=38

This easy methold will only hold if a, b, c are INTEGERS!!!

I think without graphical solution the first question 2^x = 4x will be impossible to solve.

I have solved the other one!! Check below for the solution. I'm still coming with the values of x and y.

2nioshine: @doubleDX once again gr8 wuk....any who can should try this...2^x=4x.... question2
x+y+z=2, x^2+y^2+z^2=26,
x^3+y^3+z^3=38....plz no graphical soln..ans=4....Q2ans=4,-3 &1

I have suceeded in solving one of your questions. Follow the steps carefully and you will understand. Here it goes:

x + y + z = 2 .....equation1
x^2 + y^2 + z^2 = 26 ....equation 2
x^3 + y^3 + z^3 = 38 .... equation 3

Before we proceed. We have to create an expression for x^3 + y^3 in terms of (x + y)

Using add and substraction method to factorize the polynomial x^3 + y^3 yields:

x^3 + y^3 can be expressed as : (x + y)^3 - 3xy (x + y)

Also x^2 + y^2 can be expressed as : (x + y)^2 - 2xy

Now, having created an expressing for the above.

From equation 1, (x + y) = (2 - z) ....equation 4

Substituting the expression for x^2 + y^2 in equation 2 yields :
(x + y)^2 - 2xy + z^2 = 26 .... equation 5

Substituting the expression for x^3 + y^3 in equation 3 yields:
(x + y)^3 - 3xy (x + y) + z^3 = 38 .... equation 6

Since we already know the expression for x + y = (2-z) in terms of z. We can now find an expression for xy from equation 5 so we can substitute back in equation 6 to give us the values of z.

Computing xy in terms of z, subtitute x + y = (2 - z)

(2 - z)^2 - 2xy + z^2 = 26
2xy = (2 - z)( 2 - y) - 26 + z^2
2xy = 4 - 4z + z^2 - 26 +z^2
2xy = 2z^2 - 4z - 22
:. xy = z^2 - 2z - 11

Having known the expression for xy interms of z as written above. We can now substitute both expression (i.e the for (x + y) and xy) in terms of z in equation 6. x + y = (2 - z) and xy = z^2 - 2z - 11

(x + y)^3 - 3xy (x + y) + z^3 = 38
substituting yields :
(2 - z)^3 - 3 ( z^2 - 2z - 11) (2 - z) + z^3 = 38

Expanding and collecting like terms yields:

3z^3 - 6z^2 - 33z + 36 = 0
Simplifying further:

z^3 - 2z^2 - 11z + 12 =0
Factorizing the polynomial gives :
(z - 1)(z^2 -z -12) = 0
(z - 1)(z - 4)(z + 3) = 0

Hence the values of z are :
z1 = 1, z2 = 4 and z3 = -3

We can now subsitute the values of z in equations 1 and 2 to evalaute x and y.

I'll be right back with the remaining solution. I hope this helps.

^^^good , the (x+y) and xy trick was nice

doubleDx:

I have succeeded in solving one of those questions. Follow the steps carefully and you will understand. Here it goes:

x + y + z = 2 .....equation1
x^2 + y^2 + z^2 = 26 ....equation 2
x^3 + y^3 + z^3 = 38 .... equation 3

Before we proceed. We have to create an expression for x^3 + y^3 in terms of (x + y)

Using add and substraction method to factorize the polynomial x^3 + y^3 yields:

x^3 + y^3 can be expressed as : (x + y)^3 - 3xy (x + y)

Also x^2 + y^2 can be expressed as : (x + y)^2 - 2xy

Now, having created an expressing for the above.

From equation 1, (x + y) = (2 - z) ....equation 4

Substituting the expression for x^2 + y^2 in equation 2 yields :
(x + y)^2 - 2xy + z^2 = 26 .... equation 5

Substituting the expression for x^3 + y^3 in equation 3 yields:
(x + y)^3 - 3xy (x + y) + z^3 = 38 .... equation 6

Since we already know the expression for x + y = (2-z) in terms of z. We can now find an expression for xy from equation 5 so we can substitute back in equation 6 to give us the values of z.

Computing xy in terms of z, subtitute x + y = (2 - z)

(2 - z)^2 - 2xy + z^2 = 26
2xy = (2 - z)( 2 - y) - 26 + z^2
2xy = 4 - 4z + z^2 - 26 +z^2
2xy = 2z^2 - 4z - 22
:. xy = z^2 - 2z - 11

Having known the expression for xy interms of z as written above. We can now substitute both expression (i.e the for (x + y) and xy) in terms of z in equation 6. x + y = (2 - z) and xy = z^2 - 2z - 11

(x + y)^3 - 3xy (x + y) + z^3 = 38
substituting yields :
(2 - z)^3 - 3 ( z^2 - 2z - 11) (2 - z) + z^3 = 38

Expanding and collecting like terms yields:

3z^3 - 6z^2 - 33c + 36 = 0
Simplying:

z^3 - 2z^2 - 11z + 12 =0
Factorizing the polynomial gives :
(z - 1)(z^2 -z -12) = 0
(z - 1)(z - 4)(z + 3) = 0

Hence the values of z are :
z1 = 1, z2 = 4 and z3 = -3

We can now subsitute the values of z in equations 1 and 2 to evalaute x and y.

I'll be right back with the remaining solution. I hope this helps.

I am really impressed, are u still a student? Which level? Your answer is more sexy than mine...

Again u were able to factorize dat because answers were integers. There is a similar question on this thread dat solves to complex numbers, factorizing for dat will be a little complex.

Continuation....

With the values of z = 1, 4 and -3

x + y + z = 2
x^2 + y^2 + z^2 = 26

Substituting z = 1 in equation 1 yields:

x + y = 2 -1
x + y = 1
:. y = (1- x) ....equation 7

Substituting z = 1 in equation 2 yields:

x^2 + y^2 + 1^2 = 26
x^2 + y^2 = 25 ....equation 8

Now substitute equation 7 in 8 to evaluate x:

x^2 + (1 - x)^2 = 25
Expanding and collecting like terms yields :

2x^2 -2x - 24 = 0
Simplifying =>
x^2 - x -12 = 0
Factorizing yields:
(x - 4)(x + 3) = 0
:. x = 4 or -3

Substituting each value of x in equation 7 for corresponding values of y.

y = 1 - x
when x =4
y = 1 - 4
y = -3

When x = -3
y = 1 - (-3)
y = 4

Hence the corresponding values of x and y for z = 1 are [4, 3] and [-3, 4].

Subtitute z = 4 and find the corresponding values of x and y, then do it again, when z = -3 for corresponding values of x and y.

We have 3 values of z, so you can find the rest with z = 4 and z = -3.

I hope that helps.

Continuation:

x + y + z = 2 .... equation 1
x^2 + y^2 + z^2 = 26 ....equation 2

Substituting z = -3 in equation 1 yields:

x + y = 2 + 3
x + y = 5
:. y = (5 - x) ....equation 9

Substituting z = -3 in equation 2 yields:

x^2 + y^2 + (-3)^2 = 26
x^2 + y^2 = 17 ....equation 10

Now substitute equation 9 in 10 to evaluate x:

x^2 + (5- x)^2 = 10
Expanding and collecting like terms yields :

x^2 + 25 - 10x + x^2 = 17
2x^2 - 10x + 8 = 0
Simplifying =>
x^2 - 5x + 4 = 0
Factorizing yields:
(x - 4)(x -1) = 0
:. x = 4 or 1

Substituting each value of x in equation 9 for corresponding values of y.

y = 1 - x
when x = 4
y = 1 - 4
y = -3

When x = 1
y = 1 - (1)
y = 0

Hence the corresponding values of x and y for z = -3 are [4, -3] and [1, 0].

Subtitute the last one for z = 4 and find the corresponding values of x and y as done above.

ositadima1:


I am really impressed, are u still a student? Which level? Your answer is more sexy than mine...

Again u were able to factorize dat because answers were integers. There is a similar question on this thread dat solves to complex numbers, factorizing for dat will be a little complex.

Nah, I graduated a few years back.

Thanks for the compliment!

Truman155: If a+b+c=8, a^2+b^2+c^2=16 and a^3+ b^3+c^3=64 find a,b, and c.

Try this baby, lets c

doubleDx:

Nah, I graduated a few years back.

Thanks for the compliment!

Same here, I luv math, for me any math course nah straight A, back then.

^Tnks

doubleDx:

I have succeeded in solving one of those questions. Follow the steps carefully and you will understand. Here it goes:

x + y + z = 2 .....equation1
x^2 + y^2 + z^2 = 26 ....equation 2
x^3 + y^3 + z^3 = 38 .... equation 3

Before we proceed. We have to create an expression for x^3 + y^3 in terms of (x + y)

Using add and substraction method to factorize the polynomial x^3 + y^3 yields:

x^3 + y^3 can be expressed as : (x + y)^3 - 3xy (x + y)

Also x^2 + y^2 can be expressed as : (x + y)^2 - 2xy

Now, having created an expressing for the above.

From equation 1, (x + y) = (2 - z) ....equation 4

Substituting the expression for x^2 + y^2 in equation 2 yields :
(x + y)^2 - 2xy + z^2 = 26 .... equation 5

Substituting the expression for x^3 + y^3 in equation 3 yields:
(x + y)^3 - 3xy (x + y) + z^3 = 38 .... equation 6

Since we already know the expression for x + y = (2-z) in terms of z. We can now find an expression for xy from equation 5 so we can substitute back in equation 6 to give us the values of z.

Computing xy in terms of z, subtitute x + y = (2 - z)

(2 - z)^2 - 2xy + z^2 = 26
2xy = (2 - z)( 2 - y) - 26 + z^2
2xy = 4 - 4z + z^2 - 26 +z^2
2xy = 2z^2 - 4z - 22
:. xy = z^2 - 2z - 11

Having known the expression for xy interms of z as written above. We can now substitute both expression (i.e the for (x + y) and xy) in terms of z in equation 6. x + y = (2 - z) and xy = z^2 - 2z - 11

(x + y)^3 - 3xy (x + y) + z^3 = 38
substituting yields :
(2 - z)^3 - 3 ( z^2 - 2z - 11) (2 - z) + z^3 = 38

Expanding and collecting like terms yields:

3z^3 - 6z^2 - 33c + 36 = 0
Simplying:

z^3 - 2z^2 - 11z + 12 =0
Factorizing the polynomial gives :
(z - 1)(z^2 -z -12) = 0
(z - 1)(z - 4)(z + 3) = 0

Hence the values of z are :
z1 = 1, z2 = 4 and z3 = -3

We can now subsitute the values of z in equations 1 and 2 to evalaute x and y.

I'll be right back with the remaining solution. I hope this helps.

d truth is that i actually tried same method as u did bt i got stock somewere after those identity stuff(factorizatn)...must confes av been seen as d bst n ma sku...bt i off my cap 4 u tnks.@ositadime i apreciate u also

hello my math gurus, sorry for my dissapearing act. i'm back so burst my brains with your questions

c u, Doubledx don take ur position... U got to work harder