How do you find $cos ({sin}^{- 1} x - {cos}^{- 1} y)$ $cos(sin^-1x-cos^-1y)$ ?

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Answer 1 · 2016-09-30T19:52:41

$The Reqd. value= y \sqrt{1 - x^{2}} + x \sqrt{1 - y^{2}} .$ $"The Reqd. value="ysqrt(1-x^2)+xsqrt(1-y^2).$

Let, ${sin}^{- 1} x = α, and, {cos}^{- 1} y = β$ $sin^-1x=alpha, and,cos^-1y=beta$

We will consider only one case, namely, $0 \leq x, y \leq 1 .$ $0lex,yle1.$

Hence, $0 \leq α, β \leq \frac{π}{2} .$ $0 le alpha, beta le pi/2.$

Also, $sin α = x, cos β = y .$ $sinalpha=x, cosbeta=y.$

Now, reqd. value $= cos ({sin}^{- 1} x - {cos}^{- 1} y) = cos (α - β)$ $=cos(sin^-1x-cos^-1y)=cos(alpha-beta)$

$= cos α cos β + sin α sin β$ $=cosalphacosbeta+sinalphasinbeta$

$= y cos α + x sin β .$ $=ycosalpha+xsinbeta.$

Now, $sin α = x \Rightarrow cos α = \pm \sqrt{1 - {sin}^{2} α} = \pm \sqrt{1 - x^{2}}$ $sinalpha=x rArr cosalpha=+-sqrt(1-sin^2alpha)=+-sqrt(1-x^2)$

But, $0 \leq α \leq \frac{π}{2} \Rightarrow cos α = + \sqrt{1 - x^{2}}$ $0 le alpha le pi/2 rArr cosalpha=+sqrt(1-x^2)$

Similarly, from $cos β = y, we get, sin β = + \sqrt{1 - y^{2}} .$ $cosbeta=y", we get, "sinbeta=+sqrt(1-y^2).$ Hence,

$The Reqd. value= y \sqrt{1 - x^{2}} + x \sqrt{1 - y^{2}} .$ $"The Reqd. value="ysqrt(1-x^2)+xsqrt(1-y^2).$

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

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