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Trigonometry question

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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

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$$\sin\beta -\cos^2 \beta=0$$

$$\sqrt{1-cos^2\beta}=cos^2\beta$$

$$cos^2\beta=z$$

$$\sqrt{1-z}=z$$

$$z^2+z-1=0$$

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

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

MaxWong, please tell us how to continue.

?

asinus  Mar 12, 2017
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#1
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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

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

$$cos\ \beta=x$$

$$ax^8+bx^6+cx^4-1=0$$

$$dx^{12}+ex^{10}+fx^8+gx^6=0$$

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
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$$\sin\beta -\cos^2 \beta=0$$

$$\sqrt{1-cos^2\beta}=cos^2\beta$$

$$cos^2\beta=z$$

$$\sqrt{1-z}=z$$

$$z^2+z-1=0$$

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

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

MaxWong, please tell us how to continue.

?

asinus  Mar 12, 2017
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I did not mention $$\cos \beta = 1$$......

MaxWong  Mar 12, 2017
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I've used cos β = x = 1, but I'm not progressing. Please give a tip.

asinus  Mar 12, 2017
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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
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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
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It does not work like that.

MaxWong  Mar 13, 2017
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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
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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.

hectictar  Mar 13, 2017
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Oh sorry it was a typo.

There is supposed to be a -1

MaxWong  Mar 14, 2017
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Ohhh, in that case...

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

So

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

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

hectictar  Mar 14, 2017
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lol

MaxWong  Mar 16, 2017
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That is correct though :)

MaxWong  Mar 16, 2017

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