Asked by Jennifer
Rewrite cos(tan^-1(v)) as an algebraic expression in v.
Can someone help me with this problem? Thanks
Can someone help me with this problem? Thanks
Answers
Answered by
Mgraph
1)From definition -->tan(tan^-1(v))=v
2)cos^2(a)=1/(tan^2(a)+1)
cos(a)=+-1/sqrt(tan^2(a)+1)
cos(tan^-1(v))=+-1/sqrt(v^2+1)
2)cos^2(a)=1/(tan^2(a)+1)
cos(a)=+-1/sqrt(tan^2(a)+1)
cos(tan^-1(v))=+-1/sqrt(v^2+1)
Answered by
Jennifer
thanks for the help
Answered by
Mgraph
Because -pi/2<tan^-1(v)<pi/2
cos(tan^-1(v))=+1/sqrt(v^2+1) (only +)
cos(tan^-1(v))=+1/sqrt(v^2+1) (only +)
Answered by
Jennifer
Yeah I was looking at that, thanks for clarifing
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