# Trignometric functions

#### garytse86

does anyone know how to find sin-1(1/3) + cos-1(1/2) in terms of pi?

#### Integral

Staff Emeritus
Gold Member
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?

#### garytse86

sorry the -1 was the inverse function.

inverse of sine and cosine

ie.

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

#### Hurkyl

Staff Emeritus
Gold Member

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 &pi; the fraction would have to equal .441506781303..., which is not one I recognize, and does not have a denominator less than 22.

#### garytse86

can you show me how to work it out by hand please?

#### garytse86

yeah I think I'd want it in radians

#### Integral

Staff Emeritus
Gold Member
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)= &pi;/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.

#### garytse86

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?

#### Hurkyl

Staff Emeritus
Gold Member
The answer to what you wrote is certainly not &pi;/2. Numerical calculation yields 1.387... while &pi;/2 = 1.57...

#### HallsofIvy

Homework Helper
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.

#### garytse86

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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