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

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Answer 1 · 2016-09-30T19:52:41

#"The Reqd. value="ysqrt(1-x^2)+xsqrt(1-y^2).#

Let, #sin^-1x=alpha, and,cos^-1y=beta#

We will consider only one case, namely, #0lex,yle1.#

Hence, #0 le alpha, beta le pi/2.#

Also, #sinalpha=x, cosbeta=y.#

Now, reqd. value#=cos(sin^-1x-cos^-1y)=cos(alpha-beta)#

#=cosalphacosbeta+sinalphasinbeta#

#=ycosalpha+xsinbeta.#

Now, #sinalpha=x rArr cosalpha=+-sqrt(1-sin^2alpha)=+-sqrt(1-x^2)#

But, #0 le alpha le pi/2 rArr cosalpha=+sqrt(1-x^2)#

Similarly, from #cosbeta=y", we get, "sinbeta=+sqrt(1-y^2).#Hence,

#"The Reqd. value="ysqrt(1-x^2)+xsqrt(1-y^2).#

We can deal with the other cases, like, #-1 le x le 0,# etc.,