Find the function value

Question

Find sin θ/2, given that sin θ = 3/5 and θ terminates in quadrant I.

Answer 1

Recall sin(θ/2) = ±√[(1-cosθ)/2], so we'll need to determine cosθ:

sinθ = opposite/hypotenuse = 3/5

cosθ = adjacent/hypotenuse = 4/5

Now compute sin(θ/2):

sin(θ/2)

= ±√[(1-cosθ)/2]

= ±√[(1-4/5)/2]

= ±√[(1/5)/2]

= ±√(1/10)

Because θ terminates in Quadrant I, then the y-coordinate is positive, so √(1/10) is correct here.

Hope this helped!

Answer 2

for sin(x/2) when given sin(x)

formula not so hard to derive

sin(x/2)=(1/2)*[±√(1+sin(x))±√(1-sin(x))],,,,don't neglect/forget the (1/2) coefficient.

have sin(x/2)=(1/2)[±√(8/5)±√(2/5)]

the first ± is + or the result is less than zero, the second ± is - or else result is almost 1, too large

sin(x/2)=(1/2)[+√(8/5)-√(2/5)]=(1/2)[√(16/10)-√(4/10)]=1/√10=(√10)/10