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

Answer 1

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

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#
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Answer 2

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

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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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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