Asked by Anonymous
I am trying to solve 0=1-asin(a) for a, but I have reached a dead end at sin(a) = 1/a. From here I'm not sure what to do.
Answers
Answered by
MathMate
asin(a)-1=0
asin(a)=1
a=(2k+1/2)π where k=integer
Use the graph of sin(x) to solve for a.
Note:
asin(a) does NOT equal 1/sin(a).
asin(a)=x means sin(x)=a
asin(a)=1
a=(2k+1/2)π where k=integer
Use the graph of sin(x) to solve for a.
Note:
asin(a) does NOT equal 1/sin(a).
asin(a)=x means sin(x)=a
Answered by
Anonymous
I think I should restate this, I am trying to solve 0=1-a(sin(a). I didn't mean asin to mean arcsin, it's supposed to be a(sin(a)).
Answered by
MathMate
Are you doing Calculus or pre-calculus?
I believe that to solve for
sin(x)=1/x
requires numerical methods, such as Newton's method, or other approximation methods.
You can also plot sin(x) and 1/x and see where they intersect or as approximations.
See:
http://img543.imageshack.us/img543/7304/1338947466.png
I believe that to solve for
sin(x)=1/x
requires numerical methods, such as Newton's method, or other approximation methods.
You can also plot sin(x) and 1/x and see where they intersect or as approximations.
See:
http://img543.imageshack.us/img543/7304/1338947466.png
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