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How can one prove the validity of this trigonometric identity? \begin{equation} 2\arccos\sqrt{x} = \frac{\pi }{2}-\arcsin(2x-1) \end{equation}

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Let $\theta=\arccos(\sqrt{x})$. Now use a familiar trigonometric identity for $\cos(2\theta)$. –  André Nicolas Oct 4 '11 at 5:51
    
So it does hold. Thanks! –  liberias Oct 4 '11 at 6:08
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5 Answers 5

up vote 5 down vote accepted

EDITED in response to valdo's answer.

Your identity $$ 2\arccos \sqrt{x}=\frac{\pi }{2}-\arcsin (2x-1),\qquad 0\le x\le 1\tag{0}, $$ may be rewritten as $$ \arcsin (2x-1)=\frac{\pi }{2}-2\arccos \sqrt{x},\qquad 0\le x\le 1\tag{1}. $$

For identity $(1)$ to be valid$^1$ it is enough that

$$ \sin \left( \arcsin (2x-1)\right) =\sin \left( \frac{\pi }{2}-2\arccos \sqrt{ x}\right).\tag{2} $$ The LHS of $(2)$ is $$\sin \left( \arcsin (2x-1)\right) =2x-1,\tag{3}$$ and the RHS, $$ \sin \left( \frac{\pi }{2}-2\arccos \sqrt{x}\right) =\cos \left( 2\arccos \sqrt{x}\right) =2\cos ^{2}\left( \arccos \sqrt{x}\right) -1\tag{4}. $$ And so, it is enough that we have

$$2x-1 =2\cos ^{2}\left( \arccos \sqrt{x}\right) -1 =2\left( \sqrt{x}\right) ^{2}-1 =2x-1,\qquad x\ge 0,\tag{5} $$

which is indeed an identity. Consequently, all the previous identities are valid e so, also the given identity $(0)$.

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$^1$ See valdo's detailed explanation in his answer.

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Correct, but incomplete. Please see my post –  valdo Oct 4 '11 at 9:56
    
@valdo: Thanks! I added a note pointing to your explanation. –  Américo Tavares Oct 4 '11 at 10:52
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Hints:

  1. $\arcsin(x)+\arccos(x)=\frac{\pi}{2}$

  2. $\cos\,2x=2\cos^2x-1$

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I fully agree with Américo Tavares's solution, except one little moment.

If you prove that sin(a) = sin(b) this does not automatically mean that a = b. Strictly speaking the consequence of sin(a) = sin(b) is the following:

a = b + 2πn (n - any integer)

or

a = π - b + 2πn (n - any integer)

The proof would be complete if we prove that only (1) is possible, whereas n=0.

Let's start with LHS. We have arcsin(2x−1). The arcsin function's image is [-π/2, π/2]. It's defined only for x in [0, 1]. For x=0 we have -π/2, for x=1 we have π/2. And it's also easy to see that the function is ascending across all the defined range of x.

Now let's look at RHS. arccos(x^[1/2]) is defined for x>=0. Intersecting this with the domain of LHS we restrict the analysis for x in [0, 1]. The image is [0, π]. For x=0 we have π/2, and for x=1 we have 0. And it's also clear that the function is descending.

Taking into account the whole RHS we get -π/2 for x=0, and π/2 for x=1 (and it's ascending). Which equals to the LHS.

Worth to add that both sides are continuous and smooth functions inside the domain (excluding the endpoints).

From all that we can deduce that indeed LHS and RHS are equal on the defined domain

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OK, just for fun let's try another method. If you know calculus, show that both sides have the same derivative with respect to $x$ (it's $-1/\sqrt{x-x^2}$) and also show that the two expressions are equal when $x=1$ (or when $x=0$ or whatever).

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$\cos(2\arccos(\sqrt{x})) =\cos(\frac{\pi}{2}-\arcsin(2x-1))$

Since $ \cos(\frac{\pi}{2}-\alpha)=\sin(\alpha)$ we may write:

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

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

$2x-1=2x-1$

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Your first line seems to assume the identity is true. –  Michael Hardy Oct 4 '11 at 6:07
    
$\cos u =\cos v$ does not in general imply that $u=v$. Your "proof" would work if $\pi/2$ was replaced by $5\pi/2$, but the result would be false. But your idea can be made to work with an additional line or so. –  André Nicolas Oct 4 '11 at 6:14
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