ive been stuck need help please explain cleary

Question

If csc ( x ) = 8 , for 90 ∘ < x < 180 ∘ ,, then

sin(x/2)=

cos(x/2)=

tan(x/2)=

Answer 1

finding half angle values by formula has ambiguities

that can be resolved with a table that is a bit complicated.

If I know cosine than the easiest fomulas are

sin(x/2)=+/-√[1-cos(x))/2]

cos(x)=+/-√[1+cos(x))/2]

In your problem cos(x) can be determined so the +/- in front of square root

is the ambiguity to be resovled.

If some day it is more convenient,

and you just have sin(x) value there are two other formulas

sin(x/2)=(1/2)*[+/-√[1+sin(x)] +/-√[1-sin(x)]]

and

cos(x/2)=(1/2)*[+/-√[1+sin(x)] +/-√[1-sin(x)]]

let's try these and see the answer

sin(x/2)=1/2*[+√[1+1/8)] +√[1-1/8)]]=.998 (I used plus plus since the angle is in the second half of second quardrant).

let's check, our actual angle is 3.016 radians which is the bigger angle associated with sin^-1(1/8)

half =1.508 radians,

sin(x/2)=sin(1.508)=.998=1/2*[+√[1+1/8)] +√[1-1/8)]

and

cos(x/2)= 1/2*[+√[1+1/8)] -√[1-1/8)]]=.063 (I used plus minus since the angle is in the second half of second quardrant).

note.063²+.998²=1

tan(x/2)=.998/.063=15.84

here is table based on given value of sine and that it is know which quardrant the half-angle resides

A/2 is x/2 for this problem and

π/4 ≤ x/2 <3π/4

because π/2 ≤ x <3π/2,

if you notice for x/2 the period is x=4π (all results also valid for (A/2)+/- k*2π= (x/2)+/- k*2π)

which since the table ranges -π to +π for x/2

the range for x is -2π to +2π

R=+/-√(1+sinA)

T=+/-√(1-sinA)

_________________________________________________________

1.) 0 ≤ A/2 < π/4 for sin(A/2) use R=+, T= -, for cos (A/2) use R=+, T=+

2.) π/4 ≤ A/2 <3π/4 for sin(A/2) use R=+, T= +, for cos (A/2) use R=+, T= -

3.) 3π/4 ≤ A/2 <π for sin(A/2) use R= -, T=+, for cos (A/2) use R= -, T= -

4.) -π/4 ≤ A/2 <0 for sin(A/2) use R=+, T= -, for cos (A/2) use R=+, T=+

5.) -3π/4 ≤ A/2<-π/4 for sin(A/2) use R= -, T= -, for cos (A/2) use R=-, T=+

6.) -π ≤ A/2<-3π/4 for sin(A/2) use R= -, T=+, for cos (A/2) use R= -, T= -

if you need more clarification on the reaons fro the validity of the formulas, commment back to me.

Answer 2

make a right triangle in quadrant 2 ( which is where you would be if x is between 90-180 degrees) and csc =8, opposite side =1, hypotenuse =8 and adjacent =- 3sqr(7)

so sinx = 1/8 cosx= -3sqr(7)/8 and tan x= -1/3sqr(7)

Find "x";

arctan(tanx)=arctan(-1/3sqr(7)

x= -0.1253

Then:

sin(x/2)= sin ( -0.1253/2)=-0.0626 rads.

cos(x/2)=0.9980 rads

tan(x/2)=-0.0627 rads

Answer 3

sin(x) = 1/8, cos(x) = - √(1 - 1/64) = - 3√7/8 because x in Q II; x/2 in Q I

sin(x/2) = √(1 - cosx)/2 = √(8 + 3√7)/4

cos(x/2) = √(1 + cosx)/2 = √(8 - 3√7)/4

tan(x/2) = (1 - cosx)/sinx = (1 + 3√7/8)/(1/8) = 8 + 3√7