How to Find the Sum of Sin-1(1/3) + Cos-1(1/2) in Terms of Pi?

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Discussion Overview

The discussion centers around finding the sum of the inverse sine of 1/3 and the inverse cosine of 1/2, specifically in terms of pi. Participants explore the interpretation of the functions and the possibility of expressing the result as a fraction of pi.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • Some participants seek clarification on the notation used for inverse sine and cosine functions, with one participant initially misinterpreting the notation.
  • It is noted that arcsin(1/3) and arccos(1/2) do not yield a standard fraction of pi, with one participant suggesting that arcsin(1/3) does not correspond to a recognizable fraction.
  • Another participant confirms that arccos(1/2) equals pi/3, based on memorized values of primary angles.
  • One participant questions the correctness of a book's answer stating that the sum equals (1/2)pi, arguing that numerical calculations yield a different value.
  • There is a suggestion that the problem might have been misquoted, proposing an alternative expression that could yield a simpler result involving pi.
  • Ultimately, one participant concludes that the original problem cannot be expressed in terms of pi, suggesting a potential error in the book.

Areas of Agreement / Disagreement

Participants generally disagree on whether the sum can be expressed in terms of pi, with some asserting it cannot while others reference the book's answer. The discussion remains unresolved regarding the validity of the book's claim.

Contextual Notes

Participants express uncertainty about the interpretation of the inverse functions and the specific values involved, leading to differing conclusions about the problem's solution.

garytse86
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does anyone know how to find sin-1(1/3) + cos-1(1/2) in terms of pi?
 
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I am not sure what the argument of the sin is supposed to be. Is that -1*(1/3)= -1/3? That is the only way I can read it, it that what you mean?

Even the meaning of the argument of the cos is not clear. Is that -1 degee? -1 radian?

Generaly the results of trig functions are not expressed in terms of Pi but the arguments are. Radians, the natural unit of angle measure is usually given in terms of Pi, is that what you want?
 
sorry the -1 was the inverse function.

inverse of sine and cosine

ie.

inversesin (1/3) + inversecos (1/2) in terms of pi
 

For ascii text, the traditional names of these functions are:

sin-1 = arcsin = asin
cos-1 = arccos = acos


Anyways, a quick calculation with my TI-89 hints that there is no way to express the answer as a fraction times π the fraction would have to equal .441506781303..., which is not one I recognize, and does not have a denominator less than 22.
[/size]
 
can you show me how to work it out by hand please?
 
yeah I think I'd want it in radians
 
OK,
So the problem is asin(1/3) + acos(1/2)

as Hurkyl pointed out, there is no standard fraction of pi for asin(1/3).

acos(1/2)= π/3

This is because I have memorized, as you should, the sin and cos of the primary angles.

Double check your first value prehaps it should be ([squ](1/3))or something of that nature.
 
the full question is:

find the value of the following in terms of pi

arcsin (1/3) + arccos (1/2)

and the answer in the book is (1/2)pi

how did the book get this answer?
 
The answer to what you wrote is certainly not π/2. Numerical calculation yields 1.387... while π/2 = 1.57...[/size]
 
  • #10
Is it at all possible that your problem said

" asin([sqrt](3)/2)+ acos(1/2)" ?

Since sin(pi/3)= [sqrt](3)/2 and cos(pi/3)= 1/2 that would is doable- although it does NOT give pi/2.

asin([sqrt](3)/2)+ asin(1/2)= pi/2.
 
  • #11
no, it said:

arcsin 1/3 + arccos 1/2

so it must be an error because it can't be expressed in terms of pi
 

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