Question #caf8c

Question

739 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-10T15:13:52

Check

#tan x = -9/40#
#csc^2x = 1+cot^2x#

#csc^2x = 1+(-40/9)^2#

#csc^2x = 1681/81#

#csc x = \pm 41/9#

Because it should be in 4th quadrant we have

#csc x = -41/9#

#sinx = -9/41#

#2sin(x/2)cos(x/2) = -9/41#

#sin(x/2)sqrt(1-sin^2(x/2)) = -9/82#

Squaring both sides we get

#sin^2(x/2)(1-sin^2(x/2)) = (-9/82)^2#

Let #t = sin^2(x/2)#

#t(1-t)-(9/82)^2=0#

#t^2-t+(9/82)^2=0#

#t = (1\pm sqrt(1-4(9/82)^2))/2#

#t = (1\pm(40/41))/2#

#t = 81/82,1/82#

#sin(x/2) = -9/sqrt(82),-1/sqrt(82)#