What does sin(arccos(5))-3sec(arc sin(8)) equal?

2 Answers
Feb 14, 2016

Nothing meaningful.

Explanation:

The arguments for "arccos" and "arcsin" must be within the range [-1,+1]. The values given are not valid.

The argument of "arccos" must be a value which could be generated by the cos function and the "cos" function only generates values in the range {-1,+1].

Similarly for "arcsin".

Feb 14, 2016

Using definitions of Complex cos, sin, etc.:

sin(arccos(5)) - 3sec(arcsin(8))=(2sqrt(6)+sqrt(7)/7)i

Explanation:

While Alan's answer is correct for sin and cos considered as Real valued functions of Real arguments, it is possible to define them as Complex valued functions of Complex arguments:

cos z = (e^(iz)+e^(-iz))/2

sin z = (e^(iz)-e^(-iz))/(2i)

Note in passing that sin^2 z + cos^2 z = 1 for any Complex number z.

With these definitions, it is possible to define arccos and arcsin for values other than [-1, 1].

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If cos z_1 = 5 then sin^2 z_1 = 1 - cos^2 z_1 = 1 - 25 = -24

Hence sin(arccos(5)) = sqrt(-24) = 2sqrt(6)i

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If sin z_2 = 8 then cos^2 z_2 = 1 - sin^2 z_2 = 1 - 64 = -63

Hence cos(arcsin(8)) = sqrt(-63) = 3sqrt(7)i

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So:

sin(arccos(5)) - 3sec(arcsin(8))=2sqrt(6)i - 3/(3 sqrt(7)i)=(2sqrt(6)+sqrt(7)/7)i