y = sin(arctan(x/sqrt3))

My real problem is that I have no idea how to deal with arctangent. All help is appreciated.

Printable View

- Sep 3rd 2012, 05:28 PMJsteelSimplifying a trig function.
y = sin(arctan(x/sqrt3))

My real problem is that I have no idea how to deal with arctangent. All help is appreciated. - Sep 3rd 2012, 05:41 PMSworDRe: Simplifying a trig function.
Arctan is just the inverse of tan. All functions of the form

trig(arctrig(x))

Can be simplified into an algebraic equation that involves division and square roots.To do that, draw a right triangle, and use the triangle SOH-CAO-TOA definitions. In this case, define an angle θ, and define the opposite side to be x, and the adjacent side to be sqrt(3), so that tan(θ) = x/sqrt(3), and arctan(x/sqrt(3)) = θ. Then find the hypotenuse using the Pythagorean theorem, and find sin(θ). This will be the expression you want.

My explanation might have sucked so tell me if this made sense xD - Sep 3rd 2012, 06:00 PMJsteelRe: Simplifying a trig function.
This really isn't doing much for me. Can you write out steps or something?

- Sep 3rd 2012, 08:26 PMSorobanRe: Simplifying a trig function.
Hello, Jsteel!

Quote:

$\displaystyle \text{Simplify: }\:y \:=\:\sin\left(\arctan\frac{x}{\sqrt{3}}\right)$

Do you knowabout arc-functions?*anything*

If not, you need more help than we can provide.

$\displaystyle \text{Let }\theta \,=\, \arctan\frac{x}{\sqrt{3}}$

$\displaystyle \text{Then: }\:\tan\theta \:=\:\frac{x}{\sqrt{3}} \:=\:\frac{opp}{adj}$

$\displaystyle \theta\text{ is in a right triangle with: }\,opp = x,\;adj = \sqrt{3}$

$\displaystyle \text{Pythagorus tells us that: }\,hyp = \sqrt{x^2+3}$

. . $\displaystyle \text{Hence: }\:\sin\theta \:=\:\frac{opp}{hyp} \:=\:\frac{x}{\sqrt{x^2+3}}$

$\displaystyle \text{Answer: }\:y \:=\:\frac{x}{\sqrt{x^2+3}}$