What is the value of sin inverse of cos 53pi/5??

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Answer 1 · 2018-03-29T08:47:23

#sin^-1(cos((53pi)/5))=-pi/10#

We know that,

#color(red)((1)cos(2kpi+theta)=cos theta#

#color(red)((2)cos(pi/2+theta)=-sintheta#

#color(red)((3)sin^-1(-x)=-sin^-1x#

#color(red)((4)sin^-1(sin(x))=x, where,x in [-pi/2,pi/2]#

Here,

#(53pi)/5=10pi+(3pi)/5......1^(st)#.quadrant

#:.sin^-1(cos((53pi)/5))=sin^-1(cos(10pi+(3pi)/5))#

#=sin^-1(cos((3pi)/5)).....toApply(1)#

#=sin^-1(cos((6pi)/10))#

#=sin^-1(cos(pi/2+pi/10))#

#=sin^-1(-sin(pi/10))........toApply(2)#

#=-sin^-1(sin(pi/10)).......toApply(3)#

#=-pi/10..........toApply(4)#

#=>sin^-1(cos((53pi)/5))= -pi/10in[-pi/2,pi/2]#