Math's problem killing me (1 Viewer)

[AmyLeeIsGod]

Can't make any sense at all out of this question:

Consider the geometric series 1 + sin(2x-1) + sin^2(2x-1) + sin^3(2x-1) ....... (x is in radians)

i) Write the constant ratio. (This is easy, it is sin(2x-1))

ii) Write out the first 3 values of x for which sin(2x-1)

These values form an arthimetic progression. Find:

(iii) the common difference

(iv) the 20th term

 

[garystewart51]

How can u be expected to get the common difference if its a Geometric Series... common difference is only for Arithmetic series.

Although I wouldn't worry too much about it, cause I'm pretty positive you will never get one that hard in the HSC.

 

[NJMORTIMER]

This looks like a very interesting question...you've got me hooked now
This piece of information
" ii) Write out the first 3 values of x for which sin(2x-1)" seems incomplete
Does it ask you to find the 3 values for which "sin(2x-1)=something"?

 

[AmyLeeIsGod]

Yeh sorry Part (ii) should read:


ii) Write out the first 3 values of x for which sin(2x-1) = 1

 

[ssglain]

Write out the first 3 values of x for which sin(2x - 1) = 1

i.e 2x - 1 = 2npi + pi/2 where n is an integer [use general solutions]

First 3 values: 2x - 1 = pi/2, 5pi/2, 9pi/2
Solve for x.

 

[webby234]

ii) Are you sure the question is complete now? By first three values of x do they mean the first three positive values of x?

2x - 1 = pi/2 + 2kpi, where k is an integer. (Sine repeats every 2pi)
2x = pi/2 + 1, 5pi/2 + 1, 9pi/2 + 1
x = pi/4 + 1/2, 5pi/4 + 1/2, 9pi/4 + 1/2

iii) The common difference is pi.
The first term (assuming i'm interpreting the question correctly) is pi/4 + 1/2
So work out the 20th term from that - I'm too lazy .

And there's no reason why that sort of question shouldn't appear in the later questions of the HSC - most question 10s are much more difficult than that.

 

[forevaunited]

I may have got this wrong because i dont like radian conversions but anyways here goes:

Sin(2x-1) = 1
2x-1 = 90 , 450 , 810
2x = pi/2 + 1, 5pi/2 + 1, 9pi/2 + 1
2x = 3pi/2, 7pi/2, 11pi/2
x = 3pi / 4, 7pi/4, 11pi/4

Therefore a = 3pi/4
and the common difference = pi

Tn= 3pi/4 + (n-1).pi

Therefore T20= 3pi/4 + 19pi

= 79pi / 4

Is that the answer? Only area of concern is the addition of 1 from (2x -1) because as i said, i get confused with radians and all that.

 

[AmyLeeIsGod]

ii) Are you sure the question is complete now? By first three values of x do they mean the first three positive values of x?

2x - 1 = pi/2 + 2kpi, where k is an integer. (Sine repeats every 2pi)
2x = pi/2 + 1, 5pi/2 + 1, 9pi/2 + 1
x = pi/4 + 1/2, 5pi/4 + 1/2, 9pi/4 + 1/2

iii) The common difference is pi.
The first term (assuming i'm interpreting the question correctly) is pi/4 + 1/2
So work out the 20th term from that - I'm too lazy .

And there's no reason why that sort of question shouldn't appear in the later questions of the HSC - most question 10s are much more difficult than that.

Yes it is now complete and thanks for the help. BTW was my answer to part (i) correct?

 

[filo_romz]

for the 20th term i got
(77pi + 2)/ 4

''

 

[SoulSearcher]

I may have got this wrong because i dont like radian conversions but anyways here goes:

Sin(2x-1) = 1
2x-1 = 90 , 450 , 810
2x = pi/2 + 1, 5pi/2 + 1, 9pi/2 + 1
2x = 3pi/2, 7pi/2, 11pi/2
x = 3pi / 4, 7pi/4, 11pi/4

Therefore a = 3pi/4
and the common difference = pi

Tn= 3pi/4 + (n-1).pi

Therefore T20= 3pi/4 + 19pi

= 79pi / 4

Is that the answer? Only area of concern is the addition of 1 from (2x -1) because as i said, i get confused with radians and all that.

(pi/2) + 1 = (pi + 2)/2, not 3pi/2.

That's pretty much the problem there.

 

[forevaunited]

(pi/2) + 1 = (pi + 2)/2, not 3pi/2.

That's pretty much the problem there.

yes youd be right there . . . . I is mega noob lol