Question #3504b

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Answer 1 · 2017-04-10T14:26:11

#cos(1/2 * sin^-1(sqrt(3) /2)) = sqrt(3) / 2#

I'm assuming the function is this;

#cos(1/2 * sin^-1(sqrt(3) /2))#

Then we can simplify this as,

#cos(1/2 * pi/3)# since, #sin(pi/3) = sqrt(3) /2#

#=> cos(pi/6)#

#=> sqrt(3)/2#

Answer 2 · 2017-04-10T14:32:53

#sqrt3/2#

#cos(1/2(color(blue)(arcsin(sqrt3/2))))#

#arcsin(color(red)(sqrt3/2))=color(blue)60# degrees, because #sincolor(blue)60=color(red)(sqrt3/2)#

#=cos(1/2(color(blue)60))#

#=cos(30)#

#=sqrt3/2#

Answer 3 · 2017-04-17T17:08:56

More generally:

#cos(theta)=2cos^2(theta/2)-1#

So:

#cos(theta/2)=sqrt((1+cos(theta))/2)#

Then:

#cos(1/2arcsin(sqrt3/2))=sqrt((1+cos(arcsin(sqrt3/2)))/2)#

Note that #cos(arcsin(sqrt3/2))=1/2#, since this refers to a #1-sqrt3-2# right triangle.

#cos(1/2arcsin(sqrt3/2))=sqrt((1+1/2)/2)=sqrt(3/4)=sqrt3/2#