+0

# Trigonometry question

+5
254
13
+6900

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

#6
+7159
+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.

?

asinus  Mar 12, 2017
Sort:

#1
+7159
+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
+7159
+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.

?

asinus  Mar 12, 2017
#2
+6900
+5

I did not mention $$\cos \beta = 1$$......

MaxWong  Mar 12, 2017
#3
+7159
+5

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

asinus  Mar 12, 2017
#4
+5565
+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
+5565
+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
+6900
+5

It does not work like that.

MaxWong  Mar 13, 2017
#8
+6900
+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
+5565
+2

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
#10
+6900
+5

Oh sorry it was a typo.

There is supposed to be a -1

MaxWong  Mar 14, 2017
#11
+5565
+7

Ohhh, in that case...

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

So

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

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

hectictar  Mar 14, 2017
#12
+6900
+6

lol

MaxWong  Mar 16, 2017
#13
+6900
+6

That is correct though :)

MaxWong  Mar 16, 2017

### 8 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details