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

• garytse86
In summary: The conversation is discussing the value of arcsin(1/3) + arccos(1/2) and how it can be expressed in terms of pi. The experts conclude that there is no way to express the answer as a fraction times pi and that the answer given in the book (1/2)pi is incorrect. The correct answer is not possible to express in terms of pi.
garytse86
does anyone know how to find sin-1(1/3) + cos-1(1/2) in terms of pi?

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

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)= &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.

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 &pi;/2. Numerical calculation yields 1.387... while &pi;/2 = 1.57...

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.

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

## 1. What are the basic trigonometric functions?

The basic trigonometric functions are sine, cosine, and tangent. These functions are used to calculate the relationship between the sides and angles of a right triangle.

## 2. What is the unit circle and how is it related to trigonometric functions?

The unit circle is a circle with a radius of 1 centered at the origin of a coordinate plane. It is related to trigonometric functions because the coordinates of any point on the unit circle can be used to calculate the values of sine, cosine, and tangent.

## 3. How are trigonometric functions used in real life?

Trigonometric functions are used in many real-life applications, such as navigation, engineering, and physics. They are used to calculate distances, angles, and heights of objects, as well as in the design of structures and machines.

## 4. What is the inverse trigonometric function?

The inverse trigonometric function is the opposite of the basic trigonometric functions. It is used to find the angle measure when given the ratio of sides of a right triangle. The inverse trigonometric functions are arcsine, arccosine, and arctangent.

## 5. How do you graph trigonometric functions?

To graph trigonometric functions, you first need to identify the period, amplitude, and vertical shift. Then, you can plot points using the unit circle or a table of values. Finally, you can connect the points to create a wave-like graph.

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