please explain clearly thank you this is precalculus by the way

Question

If csc ( x ) = 7 , for 90 ∘ < x < 180 ∘ , then

sin ( x/2 ) =

cos ( x/2 ) =

tan ( x/2 ) =

Answer 1

sinx = 1/7; cosx = - √(1 - 1/49) = - 4√3/7 because x in Q II, x/2 in Q I

sinx/2 = √(1 - cosx)/2 = √(1 + 4√3/7)/2 = √(7 + 4√3)/14

cosx/2 = √(1 + cox)/2 = √(1 - 4√3/7)/2 = √(7 - 4√3)/14

tanx/2 = (1 - cosx)/sinx = (1 + 4√3/7)/(1/7) = 7 + 4√3

Answer 2

cscx = 7 = h/o = hypotenuse/opposite side. adjacent side squared = hypotenuse squared minus opposite side squared. a^2 = h^2 - o^2 = 49 -1 = 48. a = -sqr48 (negative for quadrant II)

or

find cosx using the trig identity cos^2x + sin^2x = 1.

cos2x = 1-sin^2x = 1-(1/7)^2 = 1-1/49 = 48/49

cosx = sqr(48/49) - sqr(48)/7

sinx= 1/cscx = o/h = 1/7

cosx = a/h = -sqr48/7

tanx = sinx/cosx = o/a = -1/sqr48

then use half angle formulas

sin(x/2) = sqr(1-cosx)/2) = sqr(1/2 +sqr48/14)

cos(x/2) = sqr(1+cosx)/2) = sqr(1/2 -sqr48/14)

tan(x/2) = sin(x/2)/cos(x/2)

x is in quadrant II, so x/2 is in quadrant I and all trig functions for x/2 >0, so use only the positive square roots in the half angle formulas