How do you use the half angle formula to solve for sin(x/2) when given sin(x)=(24/25)?

2 Answers
jos · Ratnaker Mehta
Jun 17, 2018

#sin(x/2)=3/5 or sqrt(17)/5#

Explanation:

angle x can be in first quadrant or second quadrant
1. first quadrant (#0 < x < pi/2 #)
#cos x = 1 - sin^2(x) = sqrt(25^2-24^2)/25=7/25#
2. second quadrant (#pi/2 < x < pi #)
#cos x = -7/25#

#cos x = 1-2*sin^2(x/2)#
#sin(x/2)=sqrt((1-cos (x))/2)#
(angle #x/2# is in first quadrant. #sin(x/2)>0#)
#sin (x/2)=sqrt(1/2((25+-7)/25)#
#sin(x/2)=3/5 or 4/5#

Ratnaker Mehta
Jun 17, 2018

# sin(x/2)=3/5, or, 4/5#.

Explanation:

We know that, #cosx=1-2sin^2(x/2)..................(star)#.

But, #sinx=24/25#.

# :. cos^2x=1-sin^2x=1-(24/25)^2=(25^2-24^2)/25^2#.

#:. cos^2x={(25+24)(25-24)}/25^2=49/25^2#.

#:. cosx=+-7/25#.

Utilising this value of #cosx# in #(star)#, we get,

#2sin^2(x/2)=1-(+-7/25), i.e., #

#2sin^2(x/2)=18/25, or, 2sin^2(x/2)=32/25#.

#:. sin^2(x/2)=9/25, or, 16/25#.

Let us note that, #sinx gt 0 :. 0 lt x lt pi :. 0 lt x/2 lt pi/2#.

#:. sin(x/2)" must be "+ve#.

#:. sin(x/2)=+3/5, or, +4/5#.

#color(red)("Enjoy Maths.!")#

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