Q. Trig and Series (1 Viewer)

[ezzy85]

Whats the quickest method for the trig one?
Thanks

 

[Newbie]

x = 0, 90?

 

[ezzy85]

is there working or is it obvious?

 

[kass]

Try putting all your sinx's on one side and your cosx's on the other. Then expand the 2x's... Just an idea..

 

[ezzy85]

heres what im getting:

2cos²x - cosx - 1 = 2sinx cosx - sinx

= (cosx - 1)(2cosx + 1) = sinx(2cosx - 1)

 

[underthesun]

I got all the way to:

(cos2x - sin2x)(cosx + sinx) = cos(2x)

but, this seems to be a very tricky question. I'm not sure..

 

[ezzy85]

i used trig transformations to get x=0, 90. surely theres a quicker way?

 

[kass]

I can get to sin2x = 2sin^2(x)

 

[wogboy]

i) cos2x - sin2x = cosx - sinx

therefore,

sqrt(2) * [sin(pi/4)*cos2x - sin2x*cos(pi/4)] = sqrt(2) * [sin(pi/4)*cosx + cos(pi/4)*sinx]

so

sqrt(2) * sin(pi/4 - 2x) = sqrt(2) * sin(pi/4 - x)

we can get rid of the sqrt(2) on both sides, and get:

sin(pi/4 - 2x) = sin(pi/4 - x)

we know that if sinA = sinB, then:

1. A = 2*pi*n + B

OR

2. A = 2*pi*n + (pi - B)

(for any arbitrary integer n)

taking the first set of solutions (1), we get:

pi/4 - 2x = 2*pi*n + pi/4 - x

-2x = 2*pi*n - x
x = -2*pi*n

to get x in the range 0<=x<=2*pi, we choose n=0, n=-1 and get:

x = 0, x = 2*pi

Now for the 2nd set of solutions (2), we get:

pi/4 - 2x = 2*n*pi + pi - (pi/4 - x)
pi/4 - 2x = 2*n*pi + pi - pi/4 + x
p/4 - pi + pi/4 - 2*n*pi = 3x
3x = - pi/2 - 2*n*pi

x = -pi/6 - (2/3)*n*pi

to get x in the range 0<=x<=2*pi, we choose n=-1, n=-2, n=-3 and get:

x=pi/2, x= 7*pi/6, x=11*pi/6

so the complete set of solutions for 0<=x<=2*pi are:

x=0, x=pi/2, x=7*pi/6, x=11*pi/6, x=2*pi

ii)

1*10^-1 + 2*10^-2 + 3*10^-3 + 4*10^-4 + ...
= 1*(0.1)^1 + 2*(0.1)^2 + 3*(0.1)^3 + 4*(0.1)^4

(hopefully you can see a sort of pattern happening)

= [(0.1)^1] + [(0.1)^2 + (0.1)^2] + [(0.1)^3 + (0.1)^3 + (0.1)^3] + [(0.1)^4 + (0.1)^4 + (0.1)^4 + (0.1)^4] + ...

= {(0.1)^1 + (0.1)^2 + (0.1)^3 + (0.1)^4 + .... }
+ {(0.1)^2 + (0.1)^3 + (0.1)^4 + (0.1)^5 + ....}
+ {(0.1)^3 + (0.1)^4 + (0.1)^5 + (0.1)^6 + ....}
+ ....

( You can see that the above is a sum of infinite sums. You can work out each individual infinite sum by using that formula:
S = a/{r-1} )

= (0.1^1)/(1-0.1) + 0.1^2/(1-0.1) + 0.1^2/(1-0.1) + ....
= (0.1^1)/0.9 + (0.1^2)/0.9 + (0.1^3)/0.9 + (0.1^4)/0.9 + ....
= (1/0.9) * (0.1^1 + 0.1^2 + 0.1^3 + 0.1^4 + ....)

(again an infinite sum)

= (1/0.9) * (0.1)/(1-0.1)
= (1/0.9) * (0.1/0.9)
= 10/9 * 1/9
= 10/81

it shouldn't surprise you that if you convert this answer to a decimal, you'll get a recurring decimal which is:

0.123456789....

 

[ezzy85]

cool, thanks

 

[Richard Lee]

Originally posted by ezzy85
Whats the quickest method for the trig one?
Thanks

I think it's right. But I am not very sure.
1/10 + 2/10^2 + 3/10^4+...=8/99

 

[Richard Lee]

Originally posted by ezzy85
Whats the quickest method for the trig one?
Thanks

cos2x-cosx=sin2x-sinx
sin(x/2)=0 or tan(3x/2)=-1
therefore:
x=0, 2pie, pie/2, 7pie/6 and 11pie/6

 

[Rand]

huh????

Richard Lee, when you explain something, please explain it so it makes sense to us poorer students...

I mean, where the fuck did line 2 come from ?

 

[Jeo]

hahahahaha

 

[ND]

Originally posted by Rand

I mean, where the fuck did line 2 come from ?

Sum to product rule.

 

[Richard Lee]

Originally posted by Richard Lee
cos2x-cosx=sin2x-sinx
sin(x/2)=0 or tan(3x/2)=-1
therefore:
x=0, 2pie, pie/2, 7pie/6 and 11pie/6

Sorry about that!

You guys must know the formula:
cos (X+Y)=cosXcosY-sinXsinY (1)
cos (X-Y)=cosXcosY+sinXsinY (2)
(1)-(2):
cos(X+Y)-cos (X-Y)=-2sinXsinY (3)
Let A=X+Y and B=X-Y sub into (3):
cosA-cosB=-2sin[(A+B)/2]sin[(A-B)/2]

Same as:
sinA-sinB=2sin[(A+B)/2]sin[(A-B)/2]

Then, you will get sin(x/2)[tan(3x/2)+1]=0

Isn't clear, now!

Sorry about that!