again on the inverse function stuff

[she18]

if y= 2 times arcsin (square root 2 divided by 2) find y

 

[soroban]

Hello, she18!

This one isn't bad . . . if you know your basics.
I'll assume they want the smallest positive value.
. . Otherwise, you can generalize the answer, okay?

$\displaystyle \text{If }y\:=\:2\cdot\arcsin\left(\frac{\sqrt{2}}{2}\right),\;\text{ find }y.$

$\displaystyle \arcsin\left(\frac{\sqrt{2}}{2}\right)$ is just some angle, $\displaystyle \theta$, such that: $\displaystyle \sin\theta\,=\,\frac{\sqrt{2}}{2}$

You are expected to know that: $\displaystyle \sin\left(\frac{\pi}{4}\right)\,=\,\frac{\sqrt{2}}{2}$ . . . Hence: .$\displaystyle \arcsin\left(\frac{\sqrt{2}}{2}\right)\,=\,\frac{\pi}{4}$

Therefore: .$\displaystyle y\:=\:2\cdot\frac{\pi}{4}\:=\:\frac{\pi}{2}$

 

[she18]

quick question though

how come in the general solution when dealing with arcsin and all, i get left with whatever number times 2pi K???
im just not understanding how you work out the entire solution

 

[stapel]

Re: quick question though

how come in the general solution when dealing with arcsin and all, i get left with whatever number times 2pi K?

Trigonometric functions, such as sine, are periodic, which means they repeat in regularly-spaced intervals.

Eliz.

 

[pka]

“how come in the general solution when dealing with arcsin and all, i get left with whatever number times 2pi K???”

As Ms Stapel pointed out, sin(x)=sin(x+2k[pi]) for any integer k.
Some older textbooks use that convention for solutions sets in problems such as these. However, in almost all current practice authors use the domain of the arcsine and arccosine functions to be −1≤x≤1.
The range of the arccosine is 0 to [pi].
The range of the arcsine is −[pi]/2 to [pi]/2.