# If $\sin A+\sin^2 A=1$ and $a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0$

If $\sin A+\sin^2 A=1$ and $a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0$.

Find the value of $2b + \dfrac {c}{a}$.

My Attempt: $$\sin A+\sin^2 A=1$$ $$\sin A + 1 - \cos^2 A=1$$ $$\sin A=\cos^2 A$$

Now,

$$a(\cos^2 A)^{6}+b(\cos^2 A)^{4}+c(\cos^2 A)^{3}=1$$ $$a\sin^6 A+ b\sin^4 A+c\sin^3 A=1$$

How do I proceed further?

• sobstitute $\sin^2 A = 1-\sin A$ Mar 1, 2017 at 13:52
• $\sin A=\cos^2A$ Aug 4, 2017 at 4:43

I write it step by step:

With $\sin A+\sin^2 A=1$ we have $\sin A=1-\sin^2 A=\cos^2A$ so \begin{eqnarray} && a\cos^{12} A+b\cos^{8} A+c\cos^{6} A-1=0\\ && a(\cos^2A)^6+b(\cos^2A)^4+c(\cos^2A)^3-1=0\\ && a\sin^6A+b\sin^4A+c\sin^3A-1=0\\ && a(1-\cos^2A)^3+b(1-\cos^2A)^2+c\sin A(1-\cos^2A)-1=0\\ && a(1-\sin A)^3+b(1-\sin A)^2+c\sin A(1-\sin A)-1=0\\ && a(1-3\sin A+3\sin^2A-\sin^3A+b(1-2\sin A+\sin^2A)+c(\sin A-\sin^2 A)-1=0\\ && a-3a\sin A+3a\sin^2A-a\sin^3A+b-2b\sin A+b\sin^2A+c\sin A-c\sin^2 A-1=0\\ && a-3a\sin A+3a\sin^2A-a\sin^3A+b-2b\sin A+b\sin^2A+c\sin A-c\sin^2 A-1=0\\ && a-3a\sin A+3a(1-\cos^2A)-a\sin A(1-\cos^2A)+b-2b\sin A+b(1-\cos^2A)+c\sin A-c(1-\cos^2A)-1=0\\ && a-3a\sin A+3a(1-\sin A)-a\sin A(1-\sin A)+b-2b\sin A+b(1-\sin A)+c\sin A-c(1-\sin A)-1=0\\ && a-3a\sin A+3a-3a\sin A-a\sin A+a\sin^2A+b-2b\sin A+b-b\sin A+c\sin A-c+c\sin A-1=0\\ && a-3a\sin A+3a-3a\sin A-a\sin A+a-a\sin A+b-2b\sin A+b-b\sin A+c\sin A-c+c\sin A-1=0\\ && (a+3a+a+b+b-c-1)+(-3a-3a-a-a-2b-b+c+c)\sin A=0\\ && (5a+2b-c-1)+(-8a-3b+2c)\sin A=0 \end{eqnarray}

• @ My Glasses, what is $2b+\dfrac {c}{a}$?
– pi-π
Mar 2, 2017 at 13:31
• @LeonhardEuler This last equation has infinite solutions, but if we restrict it with some conditions, like $a$, $b$ and $c$ be integer, so we can answer that!. Meanwhile, I tried find this expression $2b+\dfrac{c}{a}$ independent from equation, but it is very complicated in general form. Mar 2, 2017 at 15:32
• @ My Glasses, then what would be the answer?. Could you please elaborate?
– pi-π
Mar 2, 2017 at 16:23

$$\sin A= \cos ^2A→\sin^2A=\cos^4A→1-\cos^2A=\cos^4A\\ (\cos^2A)^2+(\cos^2A)-1=0→\cos^2A=\frac{-1+ \sqrt{5}}{2}$$

So,

$$a\left(\frac{-1+ \sqrt{5}}{2}\right)^6+b\left(\frac{-1+ \sqrt{5}}{2}\right)^4+c\left(\frac{-1+ \sqrt{5}}{2}\right)^3=1\\ a(9-4\sqrt{5})+b\left(\frac{7-3\sqrt{5}}{2}\right)+c(\sqrt{5}-2)=1\\ (9a+7b/2-2c)+(-4a-3b/2+c)\sqrt{5}=1$$

Maybe the question want:

$$9a+\frac{7b}{2}-2c=1→18a+7b-4c=2\\ -4a-\frac{3b}{2}+c=0→8a+3b=2c$$

But on that case we get:

$$b=2-2a\\ c=a+3$$

And then

$$2b+\frac{c}{a}=5-4a+\frac{3}{a}$$

and the result depends on $a$. Otherwise I don't have another guess.

• @ Arnaldo, How is $$\textrm a(\dfrac {-1+\sqrt {5}}{2})^{6}=a(9-4\sqrt {5})$$, $....$?
– pi-π
Mar 2, 2017 at 3:50
• You just have to do the calculation! Mar 2, 2017 at 12:39
• [+1] very good answer if you assume that the looked for coefficients are rationals. May 17, 2017 at 1:30

@MyGlasses .. In the third line from below you forgot that there is $(-a)$ twice in your attempt to prove the required identity, resulting in a mistake in your answer.

• $a\sin^{2}A=a(1-\cos^{2}A)=a(1-\sin A)$ Mar 2, 2017 at 11:10
• Yes, you are right, thanks. Mar 2, 2017 at 11:22

HINT:

$$\cos^4A=(\cos^2A)^2=\cdots=1-\cos^2A\iff\cos^4A+\cos^2A-1=0$$

Divide $a\cos^{12}A+b\cos^8A+c\cos^8A-1$ by $\cos^4A+\cos^2A-1$

• @ lab bhattacharjee, could you please elaborate the very beginning step?
– pi-π
Mar 1, 2017 at 13:56
• @LeonhardEuler, Was planning to use en.wikipedia.org/wiki/Polynomial_remainder_theorem Mar 1, 2017 at 14:06
• @ lab bhattacharjee, where did you use that? Could you please show?
– pi-π
Mar 1, 2017 at 14:13
• @LeonhardEuler, In the division suggested, the remainder has to be $0$ Mar 1, 2017 at 14:14
• @ lab bhattacharjee, then how do you get $cos^4 A+cos^2A-1=0$? and is the division to be done by actual division method?
– pi-π
Mar 1, 2017 at 14:17