Trigonometry question

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

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

$cos^2\beta=z$

$\sqrt{1-z}=z$

$z^2+z-1=0$

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

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

MaxWong, please tell us how to continue.

  ?

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 :- )

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

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.

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 $(a+b)^3 = a^3 + 3a^2b+3ab^2+b^3$

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.