Suppose that sin(a)=(3/5) and 0 < a < pi/2 Find the exact values of... A) sin(a/2) B) tan(a/2)?

sIn^2+cos^x=1,,,,(3/5)^2+cos^2=1
cos^2=1-9/25=16/25,, cosx =4/5
cosx=2cos^2(x/2)-1
4/5=2cos^2(x/2)-1
1+4/5=2cos^2(x/2)
9/4=2cos^2(x/2),,,cos^2(x/2)=(9/4)/2=9/8
cos(x/2)=3/sqr(8)
sinx=2sin(x/2)(cos(x/2)
(3/5)=2(sin(x/2)3/sqr(8)
(3/5)/(2(3/sqr8=sqr8/5(2)=2sqr2/2(5)=sqr2/5
tan(x/2)=sqr2/5)/(3sqr/8=sqr2/5/(3(2sqr2)=

1 st quadrant angle
sin a = 3/5
cos a = 4/5
tan a = 3/4
cos a = 1 - 2 sin ² a/2
2 sin² a/2 = 1 - cos a
2 sin² a/2 = 1 - 3/5 = 2/5
sin a/2 = 1/5^(1/2)
tan a/2 = 1/2

sin(a) = 3/5 = opp/hyp
adj = sqrt(hyp^2-opp^2) = sqrt(5^2-3^2) = sqrt(25-9)=sqrt(16)=4
cos(a) = adj/hyp = 4/5

sin^2(a/2) = (1-cos(a)) /2 = (1- 4/5) / 2 = 1/10
sin(a/2) = 1/sqrt(10)
cos^2(a/2) = (1+cos(a)) /2 = (1+4/5) /2 = 9/10
cos(a/2) = 3/sqrt(10)

tan(a/2) = sin(a/2) /cos(a/2) = [1/sqrt(10)] /[3/sqrt(10)] = 1/3

sin(a) = 3/5
cos(a) = 4/5

sin²(a/2) = (1-cos(a))/2
= (1-4/5)/2
= (1/5)/2
= 1/10
sin(a/2) = 1/√10
:::::
tan(a) = sin(a)/cos(a) = 3/4
tan²(a/2) = (1-cos(a))/(1+cos(a))
= (1-4/5)/(1+4/5)
= (1/5)/(9/5)
= 1/9
tan(a/2) = ⅓

　
Since 0 < a < π/2, then 0 < a/2 < π/4
Therefore, cos(a), sin(a/2), and tan(a/2) are all > 0

sin(a) = 3/5
cos(a) = 4/5

A)

sin²(a/2) = 1/2 (1 − cos(a)) = 1/2 (1 − 4/5) = 1/10
sin(a/2) = 1/√10

B)

tan(a/2) = (1−cos(a))/sin(a) = (1−4/5)/(3/5)
tan(a/2) = 1/3