Simplifying the cosine equation

**mx-** · April 1st 2008, 08:11 AM

How do i simplify the following:

$sin(2*cos^{-1}(4x))$

**Soroban** · April 1st 2008, 09:00 AM

Hello, mx-!

Simplify: . $\sin\left[2\cos^{-1}(4x)\right]$

Recall that an inverse trig expression is just "some angle."

So we have: . $\sin(2\theta)$ , where: . $\theta \:=\:\cos^{-1}(4x)\quad\Rightarrow\quad \cos\theta \:=\:4x\;\;{\color{blue}[1]}$

. . This means: . $\cos\theta \:=\:\frac{4x}{1} \:=\:\frac{adj}{hyp}$

$\theta$ is in a right triangle with: $adj = 4x,\;hyp = 1$
Using Pythagorus: . $opp \:=\:\sqrt{1-16x^2}$
. . Hence: . $\sin\theta \:=\:\sqrt{1-16x^2}\;\;{\color{blue}[2]}$

We want: . $\sin2\theta \;=\;2\sin\theta\cos\theta$

Substitute [2] and [1]: . $\sin2\theta \;=\;2\left(\sqrt{1-16x^2}\right)(4x)$

Therefore: . $\boxed{\sin\left[2\cos^{-1}(4x)\right] \;=\;8x\sqrt{1-16x^2}}$

**roy_zhang** · April 1st 2008, 09:09 AM

Originally Posted by mx-

How do i simplify the following:

$sin(2*cos^{-1}(4x))$

Realize that $\cos^{-1}(4x)$ is an angle, whose $\cos$ value is $4x$ providing that $-1\le 4x\le 1$ . Mathematically, $\cos^{-1}(4x)=y \Leftrightarrow 4x=\cos(y)$

Now use double angle formula:

$\sin(2\cos^{-1}(4x))=2\sin(\cos^{-1}(4x))\cos(\cos^{-1}(4x))$

where $\cos(\cos^{-1}(4x))=4x$ and $\sin(\cos^{-1}(4x))=\sqrt{1-16x^2}$

hence $\sin(2\cos^{-1}(4x))=8x\sqrt{1-16x^2}$ for $\frac{-1}{4}\le x\le \frac{1}{4}$

Roy

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