How do you evaluate sin^-1(cos((3pi)/4))sin1(cos(3π4)) without a calculator?

2 Answers
Aug 3, 2018

sin^-1(cos((3pi)/4))=-pi/4sin1(cos(3π4))=π4

Explanation:

Let ,

X=sin^-1(cos((3pi)/4))X=sin1(cos(3π4))

X=sin^-1(cos(pi-pi/4))X=sin1(cos(ππ4))

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

Aug 4, 2018

- pi/4

Explanation:

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