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cos ( 2cos^-1 x + sin^-1 x) at x= 1/7 = ?
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cos ( cos^-1 1/7+ cos^-1 1/7 + sin^-1 1/7)
=cos ( cos^-1 1/7 + pie/2)
It seems wrong (IMHO of course, may be not...)
cos (cos^ -1 1/7) cos pie/2 - sin (cos^ -1 1/7) sin pie/2
cos pi/2 = 0
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sin x = sqrt(1-cos^2 x) ..maybe this helps..
Is this correct
0 - sin ( pie/2 - sin^-1 1/7)
use this, u will get it sin^-1 t + cos^-1 t =pi/2
-sin (cos^ -1 1/7) = - sqrt ( 1- cos(cos-1 (1/7) ^2 ) ) = -sqrt ( 1- 1/49) was that right ?
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how does sqrt come into picture ??
i didnt Understand hw u reached here
just to eliminate the inverse term,i used sinx = sqrt(1- cos^2 x) ..
The answer is correct
lol....Yes....)...thxxxx
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-4sqrt3/7
yep..the very same..
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