1. ## Simplify this.. :D

Im unable to simplify this, ive tried but i just cant seem to get it to work.. it goes a little like this -

I understand that $tan(arcsinx) = \frac{x}{\sqrt{1-x^2}}$

But i need to simplify this -

$\frac{1}{256} [\frac{tan^3(arcsin\frac{x}{4)}}{3} + tan(arcsin\frac{x}{4})$

Which i think should eventually end up like this -

$\frac{x(x-4)(x+4)(x^2-24)}{384[16-x^2]^\frac{5}{2}}$

Can someone show me the steps please? Btw, i posted this in calculus, because it had origianlly started off as an integration question.

2. Originally Posted by brd_7
Im unable to simplify this, ive tried but i just cant seem to get it to work.. it goes a little like this -

I understand that $tan(arcsinx) = \frac{x}{\sqrt{1-x^2}}$

But i need to simplify this -

$\frac{1}{256} [\frac{tan^3(arcsin\frac{x}{4)}}{3} + tan(arcsin\frac{x}{4})$

Which i think should eventually end up like this -

$\frac{x(x-4)(x+4)(x^2-24)}{384[16-x^2]^\frac{5}{2}}$

Can someone show me the steps please?
well. $\tan \left( \arcsin \frac x4 \right) = \frac {\left( \frac x4 \right) }{\sqrt{1 - \frac {x^2}{16}}}$

and $\tan^3 \left( \arcsin \frac x4 \right) = \left( \tan \left( \arcsin \frac x4 \right) \right)^3 = \left( \frac {\left( \frac x4 \right) }{\sqrt{1 - \frac {x^2}{16}}} \right)^3$

3. Haha, i got the first bit, the first bit was as far as i could get up to. Its the further simplifying after that stage which i found difficult...

4. (1/256){(1/3)tan^3(arcsin(x/4)) +tan(arcsin(x/4))}

Angle arcsin(x/4):
opp side = x
hyp = 4
so,
tan(arcsin(x/4)) = opp/adj = x /sqrt(16 -x^2)

= (1/256){(1/3)[x /sqrt(16 -x^2)]^3 +x/sqrt(16 -x^4)}
= (1/256){[x^3 / 3(sqrt(16 -x^2)^3] +x/sqrt(16 -x^4)}

Combine the two fractions inside the curly brackets into one fraction only. The common denominator is 3(sqrt(16 -x^2))^3.

= (1/256){[x^3 +x*3(sqrt(16 -x^2))^2] / [3(sqrt(16 -x^2)^3]}
= (1/256){[x^3 +3x(16 -x^2)] / [3(sqrt(16 -x^2)^3]}
= (1/256){[x^3 +48x -3x^3] / [3(sqrt(16 -x^2)^3]}
= (1/256){[48x -2x^3] / [3(sqrt(16 -x^2)^3]}
= (1/256)(2/3){[24x -x^3] / [(sqrt(16 -x^2)^3]}
= (1/384){[x(24 -x^2)] / [(sqrt(16 -x^2)^3]}

or,
= x(24 -x^2) / 384(16 -x^2)^(3/2) -----------answer.

5. Hello, brd_7!

Basically the same as ticbol's great solution . . .

First I would factor it: . $\frac{1}{768}\tan\left(\arcsin\frac{x}{4}\right)\, \bigg[\tan^2\!\left(\arcsin\frac{x}{4}\right) + 3\bigg]$

$\text{Let }\:\theta \:=\:\arcsin\frac{x}{4}\quad\Rightarrow\quad\sin\t heta \:=\:\frac{x}{4} \:=\:\frac{opp}{hyp}$

. . $\text{Then: }\:adj \:=\:\sqrt{16-x^2} \quad\Rightarrow\quad\tan\theta \:=\:\frac{x}{\sqrt{16-x^2}} \quad\Rightarrow\quad \theta \:=\:\arctan\left(\frac{x}{\sqrt{16-x^2}}\right)$

$\text{Hence: }\:\tan\left(\arcsin\frac{x}{4}\right) \:=\:\tan\left(\arctan\frac{x}{\sqrt{16-x^2}}\right) \:=\:\frac{x}{\sqrt{16-x^2}}$

The problem becomes: . $\frac{1}{768}\cdot\frac{x}{\sqrt{16-x^2}}\bigg[\left(\frac{x}{\sqrt{16-x^2}}\right)^2 + 3\bigg] \;= \;\frac{1}{768}\cdot\frac{x}{\sqrt{16-x^2}}\bigg[\frac{x^2}{16-x^2} + 3 \bigg]$

. . $= \;\frac{1}{768}\cdot\frac{x}{\sqrt{16-x^2}}\bigg[\frac{x^2 + 48 - 3x^2}{16-x^2}\bigg] \;=\;\frac{1}{768}\cdot\frac{x}{\sqrt{16-x^2}}\bigg[\frac{48-2x^2}{16-x^2}\bigg] \;=$ . $\frac{1}{768}\cdot\frac{x}{\sqrt{16-x^2}}\bigg[\frac{2(24-x^2)}{16-x^2}\bigg]$

Therefore: . $\boxed{\frac{x(24-x^2)}{384(16-x^2)^{\frac{3}{2}}}}$