How do you evaluate #sin^-1(cos((3pi)/4))# without a calculator?

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Answer 1 · 2018-08-03T19:32:25

#sin^-1(cos((3pi)/4))=-pi/4#

Let ,

#X=sin^-1(cos((3pi)/4))#

#X=sin^-1(cos(pi-pi/4))#

#X=sin^-1(-cos(pi/4))....to[because2^(nd)Quadrant]#

#X=sin^-1(-1/sqrt2)....to[becausesin^-1(-x)=-sin^-1x]#

#X=-sin^-1(1/sqrt2)#

#X=-pi/4#

Answer 2 · 2018-08-04T09:10:37

#- pi/4#

# cos theta = sin ( pi/2 - theta ), theta in ( -oo, oo ) radian #

#sin^(-1) cos ( 3/4pi ) = sin^(-1)sin (pi/2 -3/4pi)#

#= sin^(-1)sin(-pi/4)#

#= -pi/4#.

For related information:

Had it been

#sin^(-1) cos (- 5/4pi)#

#= sin^(-1)sin ( pi/2 + 5/4pi)#

#= sin^(-1)sin (9pi/4)#

#ne 9pi/4#, as it is #notin [ - pi/2, pi/2]#.

The calculators' answer is #- 45^o#

There is need for the piecewise bijective definition

#(sin)^(-1)x = kpi + (-1)^k sin^(-1)x, k = 0, +- 1, +- 2, +- 3, ...#.

Here, in this example,

#(sin)^(-1) sin ( 9/4pi) = 9/4pi#,

in accordance with #f^(-1)f(x) = x#