This discussion is closed.
islandguy
Badges: 0
#1
Report Thread starter 16 years ago
#1
i got stck on thin trigonometric sum, can anybody help me please?
sum: 5cosxsin2x+4sinsquaredx=4
0
MalaysianDude
Badges: 4
Rep:
?
#2
Report 16 years ago
#2
(Original post by islandguy)
i got stck on thin trigonometric sum, can anybody help me please?
sum: 5cosxsin2x+4sinsquaredx=4
ekeke dunno if im right but
since you didnt give me the range...ill simplify the equation till its solvable by you

5CosXSin2X + 4Sin²X = 4
5CosX(2SinXCosX) + 4(1-Cos²X) = 4
10SinXCos²X + 4 - 4Cos²X = 4
10SinXCos²X - 4Cos²X = 0

(divide the equattion by Cos²X)

10SinX - 4 = 0
SinX = 4/10

X = 23.6, 156.4, 203.6, 336.4
i guess thats the way of doing it at condition that 0<X<360
0
islandguy
Badges: 0
#3
Report Thread starter 16 years ago
#3
10s very much, i owe u one!
0
It'sPhil...
Badges: 2
Rep:
?
#4
Report 16 years ago
#4
(Original post by MalaysianDude)
ekeke dunno if im right but
since you didnt give me the range...ill simplify the equation till its solvable by you

5CosXSin2X + 4Sin²X = 4
5CosX(2SinXCosX) + 4(1-Cos²X) = 4
10SinXCos²X + 4 - 4Cos²X = 4
10SinXCos²X - 4Cos²X = 0

(divide the equattion by Cos²X)

10SinX - 4 = 0
SinX = 4/10

X = 23.6, 156.4, 203.6, 336.4
i guess thats the way of doing it at condition that 0<X<360
You should have factorised - as is you have ignored the possibilty that cosx=0 and hence only have some of the solutions
0
islandguy
Badges: 0
#5
Report Thread starter 15 years ago
#5
Determine the values of x and z if:

3lnz - 4lnx = ln2 and 9^(z-2x) = 3^x




I thank u in advance
0
Nylex
Badges: 10
Rep:
?
#6
Report 15 years ago
#6
(Original post by islandguy)
Determine the values of x and z if:

3lnz - 4lnx = ln2 and 9^(z-2x) = 3^x




I thank u in advance
ln z^4 - ln x^4 = ln 2

(z^4)/x^4 = 2 (using ln A - ln B = ln A/B and nln A = ln A^n, then exponentiating both sides)

9^(z - 2x) = 3^x

(3^2)^(z - 2x) = 3^x

Bases the same, so powers must be the same

2(z - 2x) = x

Solve those simultaneously.
0
Evolin
Badges: 1
Rep:
?
#7
Report 15 years ago
#7
(Original post by islandguy)
Determine the values of x and z if:

3lnz - 4lnx = ln2 and 9^(z-2x) = 3^x




I thank u in advance
wot level is this? i dont know what the first one is

The second one is (i think)
9^(z-2x) = 3^x
3^2(z-2x)=3^x

so x is
2(z-2z)=x

z is
2z-4z=x
-2z=x
z=x/-2
0
Evolin
Badges: 1
Rep:
?
#8
Report 15 years ago
#8
(Original post by Tazzie)
wot level is this? i dont know what the first one is

The second one is (i think)
9^(z-2x) = 3^x
3^2(z-2x)=3^x

so x is
2(z-2z)=x

z is
2z-4z=x
-2z=x
z=x/-2
i dint realise it was part of one question...sorry , well idunno wot it is then
0
Nylex
Badges: 10
Rep:
?
#9
Report 15 years ago
#9
(Original post by Tazzie)
wot level is this? i dont know what the first one is

The second one is (i think)
9^(z-2x) = 3^x
3^2(z-2x)=3^x

so x is
2(z-2z)=x
You've made a mistake there, it should be 2(z - 2x) = x
0
islandguy
Badges: 0
#10
Report Thread starter 15 years ago
#10
the function f(x) is defined by
f(x) = (2 + sin2x)/(2 + cosx)
Ignoring terms in x^3 and higher powers of x, obtain a quadratic approximation to the function f(x) for small values of x.


i tried to work out and got to an answer but i dont think its good becuase it was very easy, can somebody work it out to check pls?




i thank u in advance Kris
0
Fermat
Badges: 8
Rep:
?
#11
Report 15 years ago
#11
I get f(x) = 2/3 + (2/3).x - x²/9

Looking at the original expression, you can see that f(x) -> 2/3 as x-> 0, which tends to confirm the expression I got above.
0
kindersurprise
Badges: 0
Rep:
?
#12
Report 15 years ago
#12
OMG i hate e's and logs lol

xox
0
shiny
Badges: 12
Rep:
?
#13
Report 15 years ago
#13
(Original post by Fermat)
I get f(x) = 2/3 + (2/3).x - x²/9

Looking at the original expression, you can see that f(x) -> 2/3 as x-> 0, which tends to confirm the expression I got above.
Agree with Fermat, except isn't the coefficient of x^2 positive?
0
Fermat
Badges: 8
Rep:
?
#14
Report 15 years ago
#14
(Original post by shiny)
Agree with Fermat, except isn't the coefficient of x^2 positive?
Yes! I missed out a sign in the binomial expansion thing.

f(x) = (2 + sin2x)/(2 + cosx)

sinx = x - x³/3! + x^5/5! - ...
cosx = 1 - x²/2! + x^4/4! - ...

excluding powers of x³ and higher, then substitute for these series for sinx and cosx into the expression for f(x).

f(x) = (2+2x)/(2+1-x²/2)
f(x) = (4+4x)/(6-x²)
f(x) = (1/6).(4+4x)/(1-(x/root(6))²)

Now, using the binomial series we can expand the expression (1 + x)^n to get,

(1-(x/root(6))²)^-1 = 1 + (-1).(-x/root(6))² + ... ) (this is where I had the sign wrong!
(1-(x/root(6))²)^-1 = 1 + (-1).(-x/root(6))²
(1-(x/root(6))²)^-1 = 1 + (-1).(-x²/6)
(1-(x/root(6))²)^-1 = 1 + (x²/6)

Now,

f(x) = (1/6).(4 + 4x).(1 + x²/6)
f(x) = (1/6).(4 + (4/6)x² + 4x + (4/6)x³), now ignore the term in x³
f(x) = (1/6).(4 + 4x + (2/3)x²)
f(x) = 2/3 + (2/3).x + (1/9).x²
===================
0
X
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • Sheffield Hallam University
    Get into Teaching in South Yorkshire Undergraduate
    Wed, 26 Feb '20
  • The University of Law
    Solicitor Series: Assessing Trainee Skills – LPC, GDL and MA Law - London Moorgate campus Postgraduate
    Wed, 26 Feb '20
  • University of East Anglia
    PGCE Open day Postgraduate
    Sat, 29 Feb '20

Do you get study leave?

Yes- I like it (5)
50%
Yes- I don't like it (1)
10%
No- I want it (3)
30%
No- I don't want it (1)
10%

Watched Threads

View All