# Thread: Simplifying the cosine equation

1. ## Simplifying the cosine equation

How do i simplify the following:

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

2. Hello, mx-!

Simplify: .$\displaystyle \sin\left[2\cos^{-1}(4x)\right]$
Recall that an inverse trig expression is just "some angle."

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

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

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

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

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

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

3. Originally Posted by mx-
How do i simplify the following:

$\displaystyle sin(2*cos^{-1}(4x))$
Realize that $\displaystyle \cos^{-1}(4x)$ is an angle, whose $\displaystyle \cos$ value is $\displaystyle 4x$ providing that $\displaystyle -1\le 4x\le 1$. Mathematically, $\displaystyle \cos^{-1}(4x)=y \Leftrightarrow 4x=\cos(y)$

Now use double angle formula:

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

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

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

Roy