Question

If sin^-1x+sin^-1y+sin^-1z=pi,prove that

(i) x(1-x^2)^1/2 +y(1-y^2)^1/2 +z(1-z^2)^1/2=2xyz

(ii) x^4 +y^4 +z^4 +4x^2 y^2 z^2 =2(x^2 y^2 +y^2 z^2+z^2-- x^2

Answer 1

here it is................

Answer 2

The results (i) $x(1-x^2)^{\frac{1}{2}}+y(1-y^2)^{\frac{1}{2}}+z(1-z^2)^{\frac{1}{2}}=2xyz$

(ii) $x^4+y^4+z^4+4x^2y^2z^2=2(x^2y^2+y^2z ^2+z^2x^2)$ is proved

Step-by-step explanation:

Given that $sin^{-1}x+sin^{-1}y+sin^{-1}z=\pi$

To prove that

(i)  $x(1-x^2)^{\frac{1}{2}}+y(1-y^2)^{\frac{1}{2}}+z(1-z^2)^{\frac{1}{2}}=2xyz$

(ii) $x^4+y^4+z^4+4x^2y^2z^2=2(x^2y^2+y^2z^2+z^2x^2)$

Let $sin^{-1}x=A$

x=sinA

Similarly $sin^{-1}y=B$

y=sinB and $sin^{-1}z=C$

z=sinC

We have $sin^{2}\theta +cos^{2}\theta=1$

$cosA=\sqrt{1-sin^2A}$

$cosA=\sqrt{1-x^2}$ since x=sinA

and similarly we have $cosB=\sqrt{1-y^2}$ since y=sinB

$cosC=\sqrt{1-z^2}$ since z=sinC

From the given

$sin^{-1}x+sin^{-1}y+sin^{-1}z=\pi$ we have

(i) Enough to prove that sin2A+sin2B+sin2C=4sinAsinBsinC

Take LHS sin2A+sin2B+sin2C=2sin(A+B)cos(A-B)+2sinCcosC

$=2sin(\pi-C)cos(A-B)+2sinCcosC$

=2sinCcos(A-B)+2sinCcosC

=2sinC[cos(A-B)+cosC]

=2sinC(2sinBsinA)

=4sinAsinBsinC=RHS

sin2A+sin2B+sin2C=4sinAsinBsinC

LHS=RHS

2sinAcosA+2sinBcosB+2sinCcosC=4sinAsinBsinC

2(sinAcosA+sinBcosB+sinCcosC)=2(2sinAsinBsinC)

sinAcosA+sinBcosB+sinCcosC=2sinAsinBsinC

Therefore   $x(1-x^2)^{\frac{1}{2}}+y(1-y^2)^{\frac{1}{2}}+z(1-z^2)^{\frac{1}{2}}=2xyz$

(ii) $A+B+C=\pi$

$A+B=\pi-C$

Taking cos on both sides we get

$cos(A+B)=cos(\pi-C)$

$cosAcosB-sinAsinB=-cosC$  ( since cos(A+B)=cosAcosB-sinAsinB )

$\sqrt{1-x^2}\sqrt{1-y^2}-xy=-\sqrt{1-z^2}$

$\sqrt{1-x^2}\sqrt{1-y^2}=xy-\sqrt{1-z^2}$

Squaring on both sides

$(\sqrt{1-x^2}\sqrt{1-y^2})^2=(xy-\sqrt{1-z^2})^2$

$(\sqrt{1-x^2})^2(\sqrt{1-y^2})^2=(xy)^2+(\sqrt{1-z^2})^2-2(xy)(\sqrt{1-z^2}$

$(1-x^2)(1-y^2)=x^2y^2+1-z^2-2xy\sqrt{1-z^2}$

$1-y^2-x^2+x^2y^2=x^2y^2+1-z^2-2xy\sqrt{1-z^2}$

$x^2+y^2-z^2-2xy\sqrt{1-z^2}=0$

$x^2+y^2-z^2=2xy\sqrt{1-z^2}$

Squaring on both sides

$(x^2+y^2-z^2)^2=(2xy\sqrt{1-z^2})^2$

$x^4+y^4+z^4+2x^2y^2-2y^2z^2-2z^2x^2=4x^2y^2(1-z^2)$

$x^4+y^4+z^4+2x^2y^2-2y^2z^2-2z^2x^2=4x^2y^2-4x^2y^2z^2$

$x^4+y^4+z^4+2x^2y^2-2y^2z^2-2z^2x^2-4x^2y^2+4x^2y^2z^2=0$

$x^4+y^4+z^4+4x^2y^2z^2=2x^2y^2+2y^2z^2+2z^2x^2$

$x^4+y^4+z^4+4x^2y^2z^2=2(x^2y^2+y^2z^2+z^2x^2)$

Hence proved