If tanx=5/12 then what is sinx/2?
Make a triangle.
By Pythagoras
r^2 = 5^2 + 12^2
r = 13
so sinx = 5/13 and cosx = 12/13
I will assume you meant sin(x/2)
using cosx = 1 - 2sin^2 (x/2)
12/13 = 1-2sin^2 (x/2)
2sin^2 (x/2) = 1/13
sin^2 (x/2) = 1/26
sin (x/2) = 1/√26 or √26/26
To find the value of sin(x/2), we need to use a trigonometric identity. Specifically, we can use the half-angle formula for sine, which states:
sin(x/2) = ±√[(1 - cosx) / 2]
To use this formula, we first need to find the value of cosx. Since we are given that tanx = 5/12, we can use another trigonometric identity:
tanx = sinx / cosx
Rearranging the equation, we have:
cosx = sinx / tanx
Plugging in the given value of tanx = 5/12, we can solve for cosx:
cosx = sinx / (5/12)
cosx = (12/5) * sinx
Now that we have the value of cosx, we can substitute it into the half-angle formula:
sin(x/2) = ±√[(1 - (12/5) * sinx) / 2]
Unfortunately, we do not have enough information to determine the exact value of sin(x/2) since the value of sinx is unknown. However, we can simplify the expression further using the given information if necessary.