• Support PF! Buy your school textbooks, materials and every day products Here!

Determining Arcsin(Sin[7Pi/5])

  • #1
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?

Comments suggestions would be appreciated.

Thanks.


PS: Sorry for lack of tex.
 

Answers and Replies

  • #2
21
0


I think you need to look at where the arcsin is defined.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,794
925


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?
 
  • #4


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.
 
  • #5
33,173
4,858


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).
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,794
925


Mark44, you mean "You are not being asked to find sin(7pi/5) don't you?
 
  • #7
33,173
4,858


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
 
  • #8


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)

So, just slap a minus sign infront of 2Pi/5?
 
  • #9
33,173
4,858


You're looking at the wrong identity. The sine function is symmetric about the line x = π/2. This means that sin(π/2 + a) = sin(π/2 - a). Find a value of x in [-π/2, π/2] such that sin(x) = sin(7π/5).
 

Related Threads for: Determining Arcsin(Sin[7Pi/5])

  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
9
Views
4K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
5K
Replies
4
Views
2K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
4
Views
31K
  • Last Post
Replies
12
Views
8K
Top