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!


  • $\begingroup$ 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

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

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

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

  • $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.