# solving for a variable inside a sin()

by Crusty
Tags: inside, solving, variable
 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.
 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)
 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?
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?
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,708 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.

 Related Discussions Precalculus Mathematics Homework 1 Precalculus Mathematics Homework 5 Calculus 7 Introductory Physics Homework 2 Introductory Physics Homework 2