How do you evaluate #sin^-1(1/sqrt2)# without a calculator?

2 Answers
Mar 7, 2018

#sin^-1(1/sqrt2)= 45^0 #

Explanation:

Let #sin^-1(1/sqrt2)=theta :. sin theta= 1/sqrt2#

In the right triangle #sin theta = P/H :. P=1; H= sqrt2; P# is

perpendiculur and #H# is hypotenuse. In right triangle

#H^2=P^2+B^2 :. B^2=H^2-P^2=2-1 :. B=1# . Since

#P=1 and B=1#, it is isosceles triangle #:. theta =45^0#

#:. sin^-1(1/sqrt2)= 45^0 # [Ans]

Mar 7, 2018

#pi/4 + 2kpi#
#(5pi)/4 + 2kpi#

Explanation:

#sin x = - 1/sqrt2 = - sqrt2/2#. Find arcsin x
Trig Table and unit circle give 2 solutions -->
arc #x = - pi/4# , and
#x = pi - (-pi/4) = pi + pi/4 = (5pi)/4#
Note. #x = (7pi)/4# is co-terminal to #x = (-pi/4)#
General answers:
#x = pi/4 + 2kpi#
#x = (5pi)/4 + 2kpi#