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

1 answer

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.
Similar Questions
    1. answers icon 5 answers
  1. Prove the following:[1+sinx]/[1+cscx]=tanx/secx =[1+sinx]/[1+1/sinx] =[1+sinx]/[(sinx+1)/sinx] =[1+sinx]*[sinx/(sinx+1)]
    1. answers icon 3 answers
  2. Simplify #3:[cosx-sin(90-x)sinx]/[cosx-cos(180-x)tanx] = [cosx-(sin90cosx-cos90sinx)sinx]/[cosx-(cos180cosx+sinx180sinx)tanx] =
    1. answers icon 1 answer
  3. We are doing trig identities in school. I need help with these five:1.1+sinx/1-sinx=cscx+1/cscx-1 2.tanx+sinx/1+cosx=tanx
    1. answers icon 1 answer
more similar questions