How do you simplify #cos(sin^-1x)#?

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Answer 1 · 2017-08-11T11:51:14

After using #u=Arcsin(x)# or #x=sinu# transforms,

#cos(arcsinx)#

=#cosu#

=#sqrt[1-(sinu)^2]#

= #sqrt(1-x^2)#

Answer 2 · 2017-08-11T12:37:29

#costheta=sqrt(1-x^2)#

#cos(sin^-1x)#

Let # # #sin^-1x=theta#

Then, #x=sintheta#

Now put the value for #x# in #cos(sin^-1x)#

#=>cos(sin^-1(sintheta))#

So the equation becomes,

#=>costheta#

We know that #sin^2theta+cos^2theta=1#

#=>cos^2theta=1-sin^2theta#

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

We have already found that #x=sintheta,# then #x^2=sin^2theta.# Now put #x^2# in the place for #sin^2theta.#

#costheta=sqrt(1-x^2)#