If cos−1(a)=0 and sin−1(6)=b, then the exact value of sin(a+b) is

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- Oct 9th 2008, 04:53 PM #1

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- Oct 9th 2008, 05:19 PM #2

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arccos(a) = 0

So, a = cos(0) = 1

And pi(a) = pi(1) = pi

So,

sin[pi(a)] = sin(pi) = 0

cos[pi(a)] = cos(pi) = -1

arcsin(pi/6) = b

So, pi/6 = sin(b)

And, cos(b) = sqrt[1 -sin^2(b)] = sqrt[1 -(pi/6)^2]

sin(pi(a) +b)

= sin(a*pi)cos(b) +cos(a*pi)sin(b)

= (0)(sqrt[1 -(pi/6)^2] +(-1)(pi/6)

= -pi/6 ---------------------answer.

- Dec 2nd 2009, 05:39 AM #3