Finding Formula without using any trig functions

[khuangg]

Find a formula for g(x)= sin(arccos(4x-1)) without using any trigonometric functions.

I have the answer key right in front of me, but i still get how to start it off or the steps in solving these kind of questions or how to do it at all :/

Thanks!

 

[soroban]

ello, khuangg!

Find a formula for $\,g(x)\,=\, \sin\big[\arccos(4x-1)\big]$
without using any trigonometric functions.
Not sure what that means.

Let $\theta \,=\,\arccos(4x-1)$

Then:$\:\cos\theta \:=\:\dfrac{4x-1}{1} \:=\:\dfrac{adj}{hyp}$

$\theta$ is in a right triangle with: $\,adj= 4x-1,\; hyp = 1.$

Pythagorus: $\:\text{(opp)}^2 + \text{(adj)}^2 \;=\; \text{(hyp)}^2$
$\qquad\qquad \text{(opp)}^2 + (4x-1)^2 \;=\;1^2$

And we have: $\: opp = \sqrt{8x-16x^2}$

Therefore: $\:\sin\theta \;=\;\dfrac{opp}{hyp} \;=\;\sqrt{8x-16x^2}$

 

[skeeter]

\(\displaystyle \theta = \arccos(4x-1) \implies \cos{\theta} = \frac{4x-1}{1}\)

\(\displaystyle g(x) = \sin{\theta} = \frac{y}{1} = \frac{\sqrt{1^2-(4x-1)^2}}{1} = \sqrt{8x-16x^2}\)

 

[Prove It]

As beautiful as it is to draw up a right-angle triangle and apply Pythagoras, in cases like these I prefer to use the Pythagorean Identity, simply because it is quite possible that the angle given is not in the first quadrant, and so the signs may be off...

$\displaystyle \begin{align*} \sin{ \left[ \arccos{ \left( 4x - 1 \right) } \right] } &= \pm \sqrt{ 1 - \left\{ \cos{ \left[ \arccos{ \left( 4x - 1 \right) } \right] } \right\} ^2 } \\ &= \pm \sqrt{ 1 - \left( 4x - 1 \right) ^2 } \\ &= \pm \sqrt{ 1 - \left( 16x^2 - 8x + 1 \right) } \\ &= \pm \sqrt{ 8x - 16x^2 } \end{align*}$

 

[skeeter]

As beautiful as it is to draw up a right-angle triangle and apply Pythagoras, in cases like these I prefer to use the Pythagorean Identity, simply because it is quite possible that the angle given is not in the first quadrant, and so the signs may be off...

$\displaystyle \begin{align*} \sin{ \left[ \arccos{ \left( 4x - 1 \right) } \right] } &= \pm \sqrt{ 1 - \left\{ \cos{ \left[ \arccos{ \left( 4x - 1 \right) } \right] } \right\} ^2 } \\ &= \pm \sqrt{ 1 - \left( 4x - 1 \right) ^2 } \\ &= \pm \sqrt{ 1 - \left( 16x^2 - 8x + 1 \right) } \\ &= \pm \sqrt{ 8x - 16x^2 } \end{align*}$

... and in that case, g(x) would not be a function.