# Determining Arcsin(Sin[7Pi/5])

The problem is... What to put as my answer :P arcsin(Sin(7Pi/5))

http://mathworld.wolfram.com/TrigonometryAnglesPi5.html

The solution is -2Pi/5

The problem I'm having is this appears to me as a trick question. Would it be acceptable to say

sin(7pi/5) = -(sqrt(5/8+sqrt(5)/8)) make a note of an identity. Then simply say arcsin(-(sqrt(5/8+sqrt(5)/8))) =-2Pi/5?

Assuming this is correct, would not all the work I just cut out be acceptable because I'm not trying to prove the identity over again?

Thanks.

PS: Sorry for lack of tex.

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I think you need to look at where the arcsin is defined.

HallsofIvy
Homework Helper

This is much easier than you are making it. There is no reason to even think about the actual value of sin(7pi/5). As Hogger suggested, think about the definition of "arcsin". The only "tricky" point is getting the correct range for the answer. 7pi/5 itself is larger than pi. What's wrong with that?

I don't understand, am I missing an identity somewhere?

ArcSin[7Pi/5] is imaginary

Sin[7Pi/5] is not

And ArcSin[Sin[7Pi/5]] is not imaginary.

Mark44
Mentor

You are not asked to find arcsin(7pi/5).

If x is in the domain of the Sin function (which has a restricted domain, unlike the sin function), then ArcSin(Sin(x)) = x. 7pi/5 is not in the domain of the Sin function, so what you need to do is find a value of x in this restricted domain for which Sin(x) = sin(7pi/5).

HallsofIvy
Homework Helper

Mark44, you mean "You are not being asked to find sin(7pi/5) don't you?

Mark44
Mentor

No, I was responding to something Beer w/Straw wrote in post 4:
I don't understand, am I missing an identity somewhere?
ArcSin[7Pi/5] is imaginary

Well the definition ArcSin[Sinx]=x where -Pi/2=<x=<Pi/2 ...

But it equals -x ...

sin(7π/5), being sin (π + α) = - sinα and - sinα = sin(-α)