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Mathematics 15 Online
OpenStudy (anonymous):

how do i find sin^-1(cos(pi)) and answer in radians in terms of pi

OpenStudy (ghazi):

cos pi= -1 and sin^-1(-1)= ??

OpenStudy (ghazi):

2n *pi+pi/2

OpenStudy (anonymous):

how did you get that?

OpenStudy (ghazi):

general solution of \[\sin^{-1}(-1)\] that means points at which sin is -1

OpenStudy (anonymous):

okay that makes sense but i think i might have written the problem wrong. its: sin^-1(cospi). then it ask you to write your answer in radians in terms of pi

OpenStudy (ghazi):

pi itself is in radian..and general solution is in radian too

OpenStudy (anonymous):

okay so the answer i got is 2n multiplied by pi+pi/2

OpenStudy (anonymous):

is that correct?

OpenStudy (anonymous):

actually 2 multiplied by (pi+pi)/2

OpenStudy (anonymous):

i think it must be (2k+1)pi + pi/2

OpenStudy (ghazi):

well just use n= any integer....

OpenStudy (anonymous):

does sin-1 =2pi

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