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Given that \(\sin\beta -\cos^2 \beta\\=a\cos^8\beta +b\cos^6\beta+c\cos^4\beta-1\\=d\cos^{12}\beta + e\cos^{10}\beta + f\cos^8\beta + g\cos^6\beta\\ = 0\) 

 

Find \(\dfrac{ad+be+cf+g}{ag+bf+ce+d}\)

MaxWong  Mar 12, 2017

Best Answer 

 #6
avatar+7059 
+10

\(\sin\beta -\cos^2 \beta=0\)

 

\(\sqrt{1-cos^2\beta}=cos^2\beta\)

 

\(cos^2\beta=z\)

 

\(\sqrt{1-z}=z\)

 

\(z^2+z-1=0\)

 

\(z=cos^2\beta=-\frac{1}{2}+\sqrt{(\frac{1}{2})^2+1}=\frac{-1+\sqrt{1+4}}{2}\)

 

\(cos^2\beta=\frac{\sqrt{5}-1}{2}\)   (equal with hectictar)

 

MaxWong, please tell us how to continue.

 

smiley  ?

asinus  Mar 12, 2017
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13+0 Answers

 #1
avatar+7059 
+5

 

Given that

 

\(\sin\beta -\cos^2 \beta\\=a\cos^8\beta +b\cos^6\beta+c\cos^4\beta-1\\=d\cos^{12}\beta + e\cos^{10}\beta + f\cos^8\beta + g\cos^6\beta\\ = 0\)

 

Find

 

\(\dfrac{ad+be+cf+g}{ag+bf+ce+d}\)

 

\(cos\ \beta=x\)

 

\(ax^8+bx^6+cx^4-1=0\)

 

\(dx^{12}+ex^{10}+fx^8+gx^6=0\)

 

If x = 1 then

 

\(a+b+c-1=0\) 

 

\(d+e+f+g=0\)

 

Please tell me how it goes, dear smarter biscuit.

 

asinus :- )

asinus  Mar 12, 2017
 #6
avatar+7059 
+10
Best Answer

\(\sin\beta -\cos^2 \beta=0\)

 

\(\sqrt{1-cos^2\beta}=cos^2\beta\)

 

\(cos^2\beta=z\)

 

\(\sqrt{1-z}=z\)

 

\(z^2+z-1=0\)

 

\(z=cos^2\beta=-\frac{1}{2}+\sqrt{(\frac{1}{2})^2+1}=\frac{-1+\sqrt{1+4}}{2}\)

 

\(cos^2\beta=\frac{\sqrt{5}-1}{2}\)   (equal with hectictar)

 

MaxWong, please tell us how to continue.

 

smiley  ?

asinus  Mar 12, 2017
 #2
avatar+6810 
+5

I did not mention \(\cos \beta = 1\)......

MaxWong  Mar 12, 2017
 #3
avatar+7059 
+5

I've used cos β = x = 1, but I'm not progressing. Please give a tip.

asinus  Mar 12, 2017
 #4
avatar+4716 
+11

At first I was thinking like asinus that these equations were true for all values of beta, but then I realized that can't be because if beta were 90° that would make it 1 - 0 = 0, which is not true. So then I thought about this:

 

\(\sin \beta - \cos^2 \beta = 0 \\ - \cos^2 \beta = - \sin \beta \\ - \cos^2 \beta + 1 = - \sin \beta + 1 \\ 1 - \cos^2 \beta = - \sin \beta + 1 \\ \sin^2 \beta = - \sin \beta + 1 \\ 0 = - \sin^2 \beta - \sin \beta + 1 \\ 0 = \sin^2 \beta + \sin \beta - 1\)

Then use the quadratic formula or complete the square to find out that

\(\sin \beta = \frac{\sqrt{5}-1}{2}\)

and

\(\sin \beta = \frac{-\sqrt{5}-1}{2}\)

So

\( \beta =\arcsin( \frac{\sqrt{5}-1}{2} )\)

and

\(\beta = \arcsin (\frac{-\sqrt{5}-1}{2})\)

I don't really know if that helps figure out the rest of the problem or not.

hectictar  Mar 12, 2017
edited by hectictar  Mar 12, 2017
 #5
avatar+4716 
+11

Actually you can throw out the option of \(\sin \beta = \frac{-\sqrt{5}-1}{2}\)because sin can never be bigger than 1 or less than -1.

 

Then you can use the pythagorean theorem to figure out that

\((\cos \beta)^2 + (\sin \beta)^2 = 1^2 \\ (\cos \beta)^2 = 1 - (\sin \beta)^2 \\ \cos \beta = \sqrt{1 - (\sin \beta)^2} \\~\\ \cos \beta = \sqrt{1 - (\frac{-1+\sqrt{5}}{2})^2} \\~\\ \cos \beta = \sqrt{\frac{\sqrt{5}-1}{2}}\)

hectictar  Mar 12, 2017
 #7
avatar+6810 
+5

It does not work like that.

MaxWong  Mar 13, 2017
 #8
avatar+6810 
+5

You need not solve for beta and you can also get the answer.

\(\sin \beta - \cos^2 \beta = 0\\ \sin \beta = \cos^2 \beta --- (1) \\\sin^2 \beta = \cos^4 \beta ---(2) \\\sin \beta - (1 - \sin^2 \beta) = 0\\ \therefore \sin \beta + \sin^2\beta = 1 ---(3) \\\text{Substitute (1),(2) into (3)} \\\cos^4 \beta +\cos^2 \beta = 1\\\)

Then what does \((\cos^4\beta+\cos^2\beta)^3 =\)?

Use the identity \((a+b)^3 = a^3 + 3a^2b+3ab^2+b^3\)

MaxWong  Mar 13, 2017
edited by MaxWong  Mar 13, 2017
 #9
avatar+4716 
+1

HMMMM ok...

So continuing from where you left off at:

\(\cos^4 \beta + \cos^2 \beta = 1 \\ (\cos^4 \beta + \cos^2 \beta)^2 = 1^2 \\ \cos^8 \beta + 2\cos^6 \beta + \cos^4 \beta = 1 \\ \cos^8 \beta + 2\cos^6 \beta + \cos^4 \beta - 1 = 0\)

 

And that matches this form: \(a\cos^8 \beta + b\cos^6 \beta + c \cos^4 \beta -1 = 0\)

Which means a = 1, b = 2, and c = 1

 

Next:

\(\cos^4 \beta + \cos^2 \beta = 1 \\ (\cos^4 \beta + \cos^2 \beta)^3 = 1^3 \\ \cos^{12} \beta + 3cos^{10} \beta +3\cos^8 \beta + \cos^6 \beta= 1 \\ \cos^{12} \beta + 3cos^{10} \beta +3\cos^8 \beta + \cos^6 \beta - 1 =0\)

 

That looks almost exactly like: \(d\cos^{12} \beta + e\cos^{10} \beta + f\cos^8 \beta + g\cos^6 \beta = 0\)

But its not exactly the same because there's a - 1 in one and not in the other.

 

So now I'm stuck again. frown

hectictar  Mar 13, 2017
 #10
avatar+6810 
+5

Oh sorry it was a typo.

 

There is supposed to be a -1 

MaxWong  Mar 14, 2017
 #11
avatar+4716 
+7

Ohhh, in that case...

a = 1, b = 2, c = 1, d = 1, e = 3, f = 3, g = 1

 

So

\(\frac{ad+be+cf+g}{ag+bf+ce+d}=\frac{(1)(1)+(2)(3)+(1)(3)+1}{(1)(1)+(2)(3)+(1)(3)+1} \\~\\ = \frac{1+6+3+1}{1+6+3+1} = \frac{11}{11} = 1\)

 

laughlaughlaugh

(All that work and the answer is just 1!)

hectictar  Mar 14, 2017
 #12
avatar+6810 
+6

lol

MaxWong  Mar 16, 2017
 #13
avatar+6810 
+6

That is correct though :)

MaxWong  Mar 16, 2017

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