tanØ = -1
angle in standard position is 45° (tan^-1 (1) = 45° )
the tangent is negative in quads II and IV
so Ø = 180-45 = 135
or
Ø = 360-45 = 315°
check: tan135 = -1 and tan315 = -1
sin Ø/2 = 1/2
angle in standard postion is 30° and the sine is positive in I and II
Ø/2 = 30° or Ø/2 = 150°
Ø = 60° or Ø = 300°
check
sin (60/2) = 1/2 and sin (300/2) = 1/2
solve for theta
6. tan theta=-1
7. sin(theta/2)=1/2
thank you for your help!
2 answers
thank you so much! :)
2 answers