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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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- Thread starter garytse86
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- #1

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

- #2

Integral

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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?

- #3

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inverse of sine and cosine

ie.

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

- #4

Hurkyl

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

sin

cos

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.

- #5

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

- #6

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yeah I think I'd want it in radians

- #7

Integral

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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.

- #8

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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?

- #9

Hurkyl

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- #10

HallsofIvy

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" 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

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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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