# Simplifying $\cos(2\arcsin(x))$ using only pythagorean trigonometric identity

I know that one can simplify $$\cos(2\arcsin(x))$$ using $$\cos(a+b)=\cos(a)\cdot\cos(b)-\sin(a)\cdot\sin(b)$$:

\begin{alignat}{1} \cos(2\arcsin(x))&=\cos^2(\arcsin(x))-\sin^2(\arcsin(x)) \\&=1-2\sin^2(\arcsin(x)) \\&=1-2x^2 \end{alignat}

I tried to make this simplification using only $$\sin^2(x)+\cos^2(x)=1$$: \begin{alignat}{1} \cos^2(2\arcsin(x))&=1-\sin^2(2\arcsin(x)) \\ \left|\cos(2\arcsin(x))\right|&=\sqrt{1-\sin^2(2\arcsin(x))} \\ &=\sqrt{1-\left(2x\sqrt{1-x^2}\right)^2} \\ &=\sqrt{1-4x^2 |1-x^2|} \\ &=\sqrt{1-4x^2(1-x^2)} \\ &=\sqrt{1-4x^2+4x^4} \\ &=\sqrt{\left(2x^2-1\right)^2} \\ &=|2x^2-1| \end{alignat} And then I could not figure out how to proceed. So, how to get rid of $$|\cdot|$$ in: $$\left|\cos(2\arcsin(x))\right|=|2x^2-1|$$ and get $$1-2x^2$$ ?

Note: this is a part of my attempt to solve the integral $$\int x^2\cdot\sqrt{1-x^2}\,\,\mathrm dx$$ by trigonometric substitution.

At this point, we know $$\cos(2\sin^{-1}x) = \pm(2x^2-1).$$ Since both these functions are continuous, we can just check one value in each region where $$2x^2-1 > 0$$ and $$2x^2-1 < 0$$ to verify the sign to use in that region. For $$|x|<1/\sqrt{2}$$, check $$x = 0$$, and for $$|x|>1/\sqrt{2}$$, check $$x = \pm 1$$. You'll find they all work out to $$\cos(2\sin^{-1}x) = 1-2x^2$$.

• I have a question: If i have an equation like, for example, $\left| {\frac{1}{2}z + 4} \right| = \left| {4z - 6} \right|$, I don't need to check regions. Why do I need to do it in this case and in my example I don't? – Vinicius ACP Apr 26 at 7:33
• In that equation, you're looking for a particular value of $z$. In this equation, you need it to be true for every value of $x$. – eyeballfrog Apr 26 at 12:58
• So is it really necessary to check more than one region? If we're talking about an identity, I think it is a contradiction to have an "piecewise identity", so checking only one region already gives me the identity, or am I wrong? – Vinicius ACP Apr 27 at 1:20
• You have to check it because when the function goes through zero, continuity no longer guarantees the signs match. As a simple example, it's certainly true that $|x| = \pm x$, but for $x < 0$, $|x| = -x$, while for $x > 0$, $|x| = +x$. – eyeballfrog Apr 27 at 3:20
• $|\cos(2\sin^{-1}x)| = |2x^2-1|$ is an identity, but it's not the one you're trying to prove. You're trying to prove $\cos(2\sin^{-1}x) = 1-2x^2$ using that identity, which requires you to check if the signs work out. In principle you'd have to check every $x$, but because the functions are continuous, you only need to check one value in each region between sign changes of one of the functions. – eyeballfrog Apr 27 at 4:40

$$\cos(2\arcsin x)$$ will be $$\ge0,$$ if

$$-\dfrac\pi2\le2\arcsin x\le\dfrac\pi2$$

$$\iff -\dfrac1{\sqrt2}\le x\le\dfrac1{\sqrt2}$$

In that case $$|2x^2-1|=-(2x^2-1)$$

Check if $$x>\dfrac1{\sqrt2}$$

or $$x<-\dfrac1{\sqrt2}$$

• But if $-1 \leq x \leq 1$, then $\frac{-\pi}{2}\leq \arcsin(x) \leq \frac{\pi}{2}$. And so $-\pi \leq 2 \arcsin(x) \leq \pi$? – Vinicius ACP Apr 26 at 5:25
• @ViniciusACP, Yes, what's the confusion? – lab bhattacharjee Apr 26 at 6:32
• I misinterpreted it. Now I think I get it: I have three "intervals of evaluation" (obtained by making $\cos(2\arcsin(x))=0$ and $2x^2-1=0$). $$x<-\dfrac1{\sqrt2}\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,-\dfrac1{\sqrt2}\le x\le\dfrac1{\sqrt2}\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,x>\dfrac1{\sqrt2}$$ In the first, I have: $$-\cos(2\arcsin(x))=+(2x^2-1)$$ In the second: $$+\cos(2\arcsin(x))=-(2x^2-1)$$ In the third: $$-\cos(2\arcsin(x))=+(2x^2-1)$$ – Vinicius ACP Apr 26 at 6:57