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If $\sin x+\sin^2x+\sin^3x=1$, then find $$\cos^6x-4\cos^4x+8\cos^2x$$

My Attempt \begin{align} \cos^2x&=\sin x+\sin^3x=\sin x\cdot\big(1+\sin^2x\big)\\ \text{ANS}&=\sin^3x\cdot\big(1+\sin^2x\big)^3-4\sin^2x\cdot\big(1+\sin^2x\big)^2+8\sin x\cdot\big(1+\sin^2x\big)\\ &=\sin x\cdot(1+\sin^2x)\bigg[\sin^2x\cdot(1+\sin^2x)^2-4\sin x\cdot(1+\sin^2x)+8\bigg]\\ &= \end{align} I don't think its getting anywhere with my attempt, so how do I solve it ?

Or is it possible to get the $x$ value that satisfies the given condition $\sin x+\sin^2x+\sin^3x=1$ ?

Note: The solution given in my reference is $4$.

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Given $$\sin x+\sin^2x+\sin^3x=1$$$$\sin x+\sin^3x=1-\sin^2x$$$$(\sin x+\sin^3x)^2=(1-\sin^2x)^2$$ $$\sin^2x+\sin^6x+2\sin^4x=\cos^4x$$ $$1-\cos^2x+(1-\cos^2x)^3+2(1-\cos^2x)^2=\cos^4x$$ $$1-\cos^2x+1-3\cos^2x+3\cos^4x-\cos^6x+1-4\cos^2x+2\cos^4x=\cos^4x$$$$\cos^6x-4\cos^4x+8\cos^2x=4$$

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    $\begingroup$ that's exactly the strategy @lab bhattacharjee proposed $\endgroup$ – weee Nov 9 '18 at 16:47
  • $\begingroup$ @weee That's OK $\endgroup$ – Key Flex Nov 9 '18 at 17:14
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Let us denote $y = \sin x$. The relation we have for $y$ is then $y + y^2 + y^3 = 1$, or also if we multiply by $y-1$, we get $y^4 = 2y - 1$. The idea is simply to write the expression in $\cos^2 x$ given in terms of $y$, and use the relation to simplify it. We have \begin{align*} \cos^6x-4\cos^4x+8\cos^2x &= (1 - y^2)^3 - 4 (1 - y^2)^2 + 8 (1 - y^2) \\ &= (1 - y^2) [(1 - 2y^2 + y^4) + (4y^2 - 4) + 8] \\ &= (1 - y^2) [5 + 2y^2 + y^4] \\ &= (1 - y^2) [5 + 2y^2 + (2y - 1)] \\ &= 2(1 - y^2) [2 + y + y^2] \\ &= 2 [2 + y + y^2 - 2y^2 - y^3 - y^4] \\ &= 2 [2 + y + (-y)(y + y^2 + y^3)] \\ &= 2 [2 + y - y] \\ &= 4. \end{align*}

Maybe there is some approach that is more straightforward using clever algebraic manipulation. However, this solution is quite clear from a theoretical point of view: you have a polinomial in $y$ that you want to simplify using the relation given. Then, you can divide it by the polynomial given in the relation and the remainder will be a polynomial with degree at most 2. In this specific case, it was the constant polynomial 4.

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$$\sin(x)(1+\sin^2(x))=1-\sin^2(x)$$

Square both sides & replace $\sin^2(x)$ with $1-\cos^2(x)$

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Write $s := \sin x, c := \cos x$.

Hint Using $s^2 + c^2 = 1$, we can write our expression as

$$c^6 - 4 c^4 + 8 c^2 = (1 - s^2)^3 - 4 (1 - s^2)^2 + 8 (1 - s^2) = -s^6 - s^4 - 3 s^2 + 5 .$$

Now, perform polynomial long division by $s^3 + s^2 + s - 1$.

Carrying out long division gives that our expression is $$(s^3 + s^2 + s - 1) p(s) + 4$$ for some cubic polynomial $p$. But we're given that $s^3 + s^2 + s = 1$, so the first term is zero, and thus $$\color{#df0000}{\boxed{c^6 - 4 c^4 + 8 c^2 = 4}} .$$

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Let $t=\sin x$ and solve the cubic $$t^3+t^2+t=1$$

Wolfram Alpha gives the real solution as $$t=(1/3) (-1 - 2/(17 + 3 \sqrt {33})^{1/3} + (17 + 3 \sqrt{33})^{1/3})$$ Plug the real solution of the above to get $$(1-t^2)^3 -4(1-t^2)^2+8(1-t^2) =4$$

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Start off with the identity $\sin^2(x)+\cos^2(x)=1$. Rearrange this to find $\sin(x)=\pm\sqrt{1-\cos^2(x)}$. Substitute this into your equation and then rearrange to find $\sqrt{1-\cos^2(x)}$. You should be able to take it from there.

Note that it's not an identity that $\cos^2(x)=\sin(x)+\sin^3(x)$.

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  • $\begingroup$ It is true based on the premise of the problem. It's not an identity. It's a result of the premise. $\endgroup$ – Mason Nov 9 '18 at 16:41
  • $\begingroup$ @Mason Even if that's true, you can still easily prove the original theorem as an identity. There's no need to complicate it by saying that $c^2=s+s^3$. $\endgroup$ – Jam Nov 9 '18 at 16:44
  • $\begingroup$ I don't think that your comment reads as "You are complicating in a way that is unneeded." And I don't think that's really helpful feedback. I think the OP did something very natural: Manipulating the premise to try and achieve the conclusion. Telling OP that her/his work is incorrect when it's not seems to be not encouraging or helpful. Now you comment reads that the statement is not an identity which the OP presumably already knows. So it just reads to me as condescending. $\endgroup$ – Mason Nov 9 '18 at 16:51
  • $\begingroup$ @Mason I'm not going to have an argument about this so this is my last comment. I also don't see why you're taking my comments so sensitively. The OP never stated that $c^2=s+s^3$ is part of the premise and it would be a mistake to use that in a proof because (1) it doesn't show the OP how to manipulate/solve wider trigonometric problems and (2) it limits the solution to a single $x$, when in fact the theorem is true for all $x$. If you can find a way to show that $c^2=s+s^3$ implies the desired theorem, then by all means provide us with an answer… $\endgroup$ – Jam Nov 9 '18 at 16:57
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    $\begingroup$ @Mason That's fair enough. Likewise, sorry if I was reactionary - I understand your intentions of encouraging the OP's education. But I'm sure you can also see my arguments $\endgroup$ – Jam Nov 9 '18 at 17:04

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