Given sinx= -1/8 and tanx<0. find sin2x

since sin<0 and tan<0, we are in QIV

if sinx = -1/8, then cosx = √63/8, so

sin2x = 2sinx*cosx = 2(-1/8)(√63/8) = -√63/32

makes sense, since if sinx is -1/8, x is close to 2pi, so 2x is close to 4pi, still in QIV.

To find sin2x, we can use the double-angle formula for sin:

sin2x = 2 * sinx * cosx

First, let's find cosx using the given information. We know that sinx = -1/8, and since sinx is negative and tanx is negative, we can determine that x lies in the third quadrant where both sinx and cosx are negative.

Using the Pythagorean identity, we can find cosx:

cosx = -√(1 - sin^2x)
= -√(1 - (-1/8)^2)
= -√(1 - 1/64)
= -√(63/64)
= -√63 / √64
= -√63 / 8

Now we can substitute sinx and cosx into the double-angle formula:

sin2x = 2 * sinx * cosx
= 2 * (-1/8) * (-√63 / 8)
= (-1/4) * (-√63)
= √63 / 4

Therefore, sin2x = √63 / 4.