Solve #cos(sin^-1(x))= 1/9# for #x#?

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Answer 1 · 2017-02-10T08:44:08

#x=+-sqrt80/9#

To solve #cos(sin^-1(x))= 1/9# for #x#, let us assume

#x=sintheta# and then #sin^-1x=theta#

and hence #cos(sin^-1(x))#

= #costheta=sqrt(1-sin^2theta)#

= #sqrt(1-x^2)#

As such #sqrt(1-x^2)=1/9#

#:.1-x^2=1/81# or #x^2-80/81=0#

or #(x-sqrt80/9)(x+sqrt80/9)=0#

Hence, #x=+-sqrt80/9#

Answer 2 · 2017-02-10T15:37:22

#+- 83^@62#

sin^-1 x --> arcsin x
cos (arcsin x) --> #cos x = 1/9#
Use calculator and unit circle:
#cos x = 1/9# --># x == +- 83^@62#