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Having a little trouble here. :/ I don't even know if you can prove this equivalence.

$$\frac{1 + \sin x + \cos x - \cos2x \sin x + \sin2x \cos x}{\cos x \sin x}= \frac{1 + sin^3x + cos^3x}{\cos x \sin x}$$

I've figured out that $-\cos^2x \sin x + \sin x = \sin^3x$ by this:

$$-\cos^2x \sin x + \sin x= \sin x (-\cos^2x + 1)$$ $$= \sin x (1 - \cos^2x)$$ $$= \sin x (\sin^2x)$$ $$= \sin^3x$$

But I can't figure out how how the remaining

$$\sin^2x \cos x + \cos x = \cos^3x$$

Does it even equal that? If it does, does it have to do anything with some kind of identity endemic to cosx? I was told that, up to my original equation (the long one, not the simplified one) that I was correct up to that point.

Please help! I've done as much as I can to get this far, but I can't quite finish it!

Thank you for any help!

-Jon

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  • $\begingroup$ Wait, since -cos(x) = cos(x)... Couldn't you just use sin^2x cosx - cosx? That DOES = cos^3x. Does that work? I'll be back to check after I get home tonight. $\endgroup$ – Werewoof Nov 16 '16 at 0:53
  • $\begingroup$ Please try format your question using Latex commands. It is difficult for users to understand what you are asking with questions like these... for example in line 1 do you mean $cos^2(x)$ or $cos^{2x}$ ? $\endgroup$ – Btzzzz Nov 16 '16 at 0:56
  • $\begingroup$ Sorry, I'm still not very good at formatting things. But indeed, I meant the former, cos^x(2) $\endgroup$ – Werewoof Nov 16 '16 at 4:24
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$\sin(x+y) = \sin(x)\cos(y) + \sin(y)\cos(x)$

$\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)$

Put $y=x$ in second equation to get

$\cos(2x) = \cos^{2}(x) - \sin^{2}(x)$

Put $y=-x$ in second equation to get

$\cos(0) = \cos^{2}(x) + \sin^{2}(x) \Rightarrow \cos^{2}(x) + \sin^{2}(x) = 1$

Using last equation, you can derive,

$\cos(2x) = 2\cos^{2}(x) - 1 = 1 - 2\sin^{2}(x)$

Note that $\sin^{2}(x)\cos(x) + \cos(x) \neq \cos^{3}(x)$. Instead, $-\sin^{2}(x)\cos(x) + \cos(x) = \cos^{3}(x)$.

Also, in general, $\sin(2x) \neq \sin(x^2) \neq (\sin(x))^{2} = \sin^{2}(x)$. Same holds for cosine.

You can follow from here, I guess.

I wasn't able to clearly understand the question so I've posted a bunch of formulas which you can use to solve

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  • $\begingroup$ Thanks for all of these!! Sorry about my formatting, I'm not very good at it. I will try to understand these as best I can, but you're right, sin^2(x)cos(x) + cos(x) doesn't = cos^3(x), but if cos(x) were negative it would. $\endgroup$ – Werewoof Nov 16 '16 at 4:40
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Both the identities stated in the title and in the question are incorrect.

It is false that $$ \sin x+\cos x-\cos^2x\sin x+\sin^2x\cos x=\sin^3x+\cos^3x $$ For instance, at $x=\pi/4$, the left-hand side is $$ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} -\frac{1}{2}\frac{1}{\sqrt{2}}+\frac{1}{2}\frac{1}{\sqrt{2}}=\sqrt{2} $$ while the right-hand side is $$ \frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}=\frac{1}{\sqrt{2}} $$ Similarly, evaluating the two sides of the alleged identity stated in the question at $x=\pi/4$, we get different values.

It is true, instead, that $$ \sin x+\cos x-\cos^2x\sin x-\sin^2x\cos x=\sin^3x+\cos^3x $$ because the left-hand side is $$ \sin x(1-\cos^2x)+\cos x(1-\sin^2x) $$

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