Trigonometry> if sin -1x +sin-1 y+ sin-1 z= pie ,prove ...

if sin -1x +sin-1 y+ sin-1 z= pie ,prove thata)x( root 1-x2)+y( root 1- y2)+ z(root 1-z2)= 2xyz

ashu , 11 Years ago

Grade 11

5 Answers

Rinkoo Gupta

Let sin^-1x =A then x= sinA

Let sin^-1y=B then y =sinB

Let sin^-1z=C then z=sinC

Then A+B+C =p

Xv(1-X²) +Yv(1-Y²) +Zv(1-Z²)=sinA v(1-sin²A)+sinBv(1-sin²B) +sinCv(1-sin²C)

=sinA.cosA+sinB.cosB+sinC.cosC

=1/2[sin2A+sin2B+sin2C]

=1/2 [4sinA.sinB.sinC]

=2sinAsinBsinC

=2xyz

Thanks & Regards

Rinkoo Gupta

AskIITians Faculty

Last Activity: 11 Years ago

Arun Kumar

Hi Ashu,

$\\for \,2kg \\T-f_1-10=2a \\for \,4kg \\20-f_2-T=4a \\since \,it \,slips \\f_1=5 \\f_2=24 \\a=-19/6 \\so \,it\,really\,slips$

Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty

Last Activity: 11 Years ago

Arun Kumar

Ashu

Above reply was a mistake.

Last Activity: 11 Years ago

Arun Kumar

Hi Ashu,

$\\sin^{-1}x+sin^{-1}y+sin^{-1}z=\pi \\sin^{-1}x=A \\sin^{-1}y=B \\sin^{-1}z=C \\seeing \,the\, expression \\sin(A)cos(A)+sin(B)cos(B)+sin(C)cos(C) \\=\sum_{r=A}^{C}sin(2r)/2 \\A+B+C=\pi$

$\\\sin(2A) + \sin(2B) + \sin(2C) & = 2 \sin(C) \cos(A-B) + 2 \sin(C) \cos(C)\\ & = 2 \sin(C) \left(\cos(A-B) + \cos(C) \right)\\ & = 2 \sin(C) \left(\cos(A-B) + \cos(\pi-(A+B)) \right)\\ & = 2 \sin(C) \left(\cos(A-B) - \cos(A+B) \right)\\ & = 2 \sin(C) \times 2 \sin(A) \sin(B)\\ & = 4 \sin(A) \sin(B) \sin(C)$