Solve Arsin(-1/2)

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Homework Statement



sin-1 (-1/2) = -pi/6

The Attempt at a Solution



I was wondering why.
Compare to cos-1(-sqrt(2)/2) = 3pi/4, this is done by pi/4 + pi/2

since cos is negative in Q2 AND Q3, this statement is valid, and the arcos is within -pi/2 and pi/2, so my answer is valid. well, i guess?

now, back to the problem with sin-1(-1/2)
how is it -pi/6? i know Q2 and Q2 are negative sin....

thanks.
 

Answers and Replies

  • #2
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The sine and cosine functions aren't 1-to-1, so don't have inverses. For example, sin(pi/6) = sin(5pi/6) = 1/2. So sine maps multiple inputs to 1/2. An inverse would map 1/2 to multiple values, which is enough to say that the inverse is not a function.

What is done to work around this problem is to restrict the domains of the trig functions to make them 1-to-1. The principal domain for sin is taken to be [-pi/2, pi.2]. The version of the sine function with restricted domain is sometimes represented as Sin(x), where Sin(x) = sin(x), -pi/2 <= x <= pi/2.

The same sort of thing is done for the cosine function, with Cos(x) = cos(x), 0 <= x <= pi.

As for your question about sin-1(-1/2), we have sin(-pi/6) = -1/2, so sin-1(-1/2) = -pi/6. The sine function is negative in Q3 and Q4 (not Q2 and Q2 as you stated). -pi/6 is in the restricted domain for Sin(x), hence in the range for sin-1(x).
 
  • #3
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Hi Mark, thank you very much for your help.
It is clear to me that I need to restrict the domain. What is not clear to me is -pi/6.

Before I worked on these problems, I created a table which has angles of sin, cos, and tan, from zero to pi/2. (0, pi/6, pi/4, pi/3, pi/2).

For sec, csc and cot I know how to produce them as well.

So when I was given this problem, cos-1(-sqrt(2)/2), I looked up at the table, I see that cos(pi/4) = sqrt(2)/2.
Since the domain of cos is restricted to [0, pi], so that I must added 90degree (pi/2) in order to get the next value of sqrt(2)/2. That is, moving to Q2. So I got cos-1(-sqrt(2)/2) = 3pi/4.


Back to the problem of sin-1(-1/2).
Since the domain of sin is [-pi/2, pi/2], and that sin(pi/6) = 1/2, and sin is negative only at Q3 and Q4, but in Q3, 7pi/6 is already out of the domain, and I realize I cannot find it using the method I did with cos...

I meant I did not know how -pi/6 was produced (although it may seem straightforward that we negate pi/6, and still within the domain...), but I want to know a way that can help me solve it....
 
  • #4
34,513
6,198


Hi Mark, thank you very much for your help.
It is clear to me that I need to restrict the domain. What is not clear to me is -pi/6.

Before I worked on these problems, I created a table which has angles of sin, cos, and tan, from zero to pi/2. (0, pi/6, pi/4, pi/3, pi/2).

For sec, csc and cot I know how to produce them as well.

So when I was given this problem, cos-1(-sqrt(2)/2), I looked up at the table, I see that cos(pi/4) = sqrt(2)/2.
Since the domain of cos is restricted to [0, pi], so that I must added 90degree (pi/2) in order to get the next value of sqrt(2)/2. That is, moving to Q2. So I got cos-1(-sqrt(2)/2) = 3pi/4.
That's not what happens. Adding pi/2 works in this case, but not in general. What's really going on is subtracting the angle from 180 degrees. Here the identity is cos(pi - x) = -cos(x). You can use this idea to help you understand why cos-1(-1/2) = 2pi/3.

cos(pi/3) = 1/2, cos(pi - pi/3) = cos(2pi/3) = -1/2

So cos-1(1/2) = pi/3, and cos-1(-1/2) = 2pi/3.
Back to the problem of sin-1(-1/2).
Since the domain of sin is [-pi/2, pi/2], and that sin(pi/6) = 1/2, and sin is negative only at Q3 and Q4, but in Q3, 7pi/6 is already out of the domain, and I realize I cannot find it using the method I did with cos...

I meant I did not know how -pi/6 was produced (although it may seem straightforward that we negate pi/6, and still within the domain...), but I want to know a way that can help me solve it....
 
  • #5
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Hi Mark. Thank you for your patient.
May I ask where can i get this identity?
I don't remember this identity at all.
I am interested in getting other identity for sin and tan (and maybe sec, cot, csc, which i think i can get it by using, eg, sec = 1/cos, use cos identity, and then take its reciprocal)
 
  • #7
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Hi, thank you.
After working through examples, here is a trend.
For sin-1, tan-1, and cot-1, if we restrict its domain to one section, i will just negate the positive.
eg. sin(-1/2) in the domain of -pi/2, pi/2, i will negate the original pi/6 to -pi/6

for cos, sec it follows the identity that pi-delta
 
  • #8
34,513
6,198


Hi, thank you.
After working through examples, here is a trend.
For sin-1, tan-1, and cot-1, if we restrict its domain to one section, i will just negate the positive.
eg. sin(-1/2) in the domain of -pi/2, pi/2, i will negate the original pi/6 to -pi/6
I know what you mean to say, but you're not writing what you mean. sin-1(-1/2) = -pi/2.

The sine function is odd, which means that sin(-x) = -sin(x). Restricting the domain to [-pi/2, pi/2] doesn't change this.

For Sin(x), the domain is [-pi/2, pi/2] and the range is [-1, 1].
For sin-1(x) (AKA arcsin(x)), the domain is [-1, 1], and the range is [-pi/2, pi/2]

For Cos(x), the domain is [0, pi] and the range is [-1, 1].
For cos-1(x) (AKA arccos(x)), the domain is [-1, 1], and the range is [0, pi]

For Tan(x), the domain is (-pi/2, pi/2) and the range is (-inf, inf).
For tan-1(x) (AKA arctan(x)), the domain is (-inf, inf), and the range is (-pi/2, pi/2)


for cos, sec it follows the identity that pi-delta
 
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