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Solving for a variable inside a sin()

by Crusty
Tags: inside, solving, variable
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Crusty
#1
Apr4-06, 10:34 PM
P: 24
sin( degtorad( (180 - (180 - 360/x))/2 ) ) = y/z

degtorad(degrees) means the the degrees inside the parenthesis are converted to radians.

How do you solve for x?

Thank you.
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Euclid
#2
Apr4-06, 11:08 PM
P: 214
you can eliminate the 180's by using properties of sine
sin (180+ x) = -sin x = sin(-x)

also,
sin(90+x) = cos(x)
Crusty
#3
Apr4-06, 11:39 PM
P: 24
sin( degtorad( (180 - (180 - 360/x))/2 ) ) = y/z

Is this right?

sin( (180 - (180 - 360/x))/2 )
= sin( ( (180 - 360/x))/2 )
= sin( (180 - 360/x) /2 )

sin( (180 - 360/x) /2 )
= sin( ( 360/x) /2 )
= sin( ( 180/x) )
= sin( 180/x )

sin( 180/x ) = y/z

um, then what?

Euclid
#4
Apr4-06, 11:43 PM
P: 214
Solving for a variable inside a sin()

sin( (180 - 360/x) /2 ) = sin (90 - 180/x) = cos(-180/x) = cos (180/x)

Are you familiar with arccos (or cos^{-1})?

By the way, what exactly are y and z?
HallsofIvy
#5
Apr5-06, 06:03 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,310
Shorn of all the other things, arcsin( ) (also written sin-1( )) is defined as the inverse of sin( ) and arccos() (also written cos-1( )) is defined as the inverse of cos( ) (well, principal value). That is, arcsin(sin(x))= x and arccos(cos(x))= x.

You have to be a bit careful about that: since sin(x) and cos(x) are not "one-to-one" they don't have inverses, strictly speaking. Given an x between -1 and 1, there exist an infinite number of y such that sin(y)= x or cos(y)= x. Arcsin(x) always gives the value, y, between -pi/2 and pi/2 such that sin(y)= x and arccos(x) always gives the value, y, between 0 and pi such that cos(y)= x.


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