If θ is an angle in standard position that terminates in quadrant iii such that tanθ = 5/12, then sinθ/2 =

sin θ/2=5√26/26=0.196

Step-by-step explanation:

θ ∈(π,3π/2)

such that

θ/2 ∈(π/2,3π/4)

As a result,

0<sin θ/2<1, and

-1<cos θ/2<0

tan θ/2=sin  θ/2/cos θ/2

such that

tan θ/2<0

Let

t=tan θ/2

t<0

By the double angle identity for tangents

2 tan θ/2/1-(tanθ /2)^2  = tanθ

2t/1-t^2=5/12

24t=5 - 5t^2

Solve this quadratic equation for t :

t1=1/5 and

t2= -5

Discard t1 because t is not smaller than 0

Let s= sin θ/2

0<s<1.

By the definition of tangents.

tan θ/2= sin θ/2/ cos θ/2

Apply the Pythagorean Algorithm to express the cosine of θ/2 in terms of s. Note the cos θ/2 is expected to be smaller than zero.

cos θ/2 = -√1-(sin θ/2)^2 = - √1-s^2

Solve for s.

s/-√1-s^2 = -5

s^2=25(1-s^2)

s=√25/26 = 5√26/26

Therefore

sin θ/2=5√26/26=0.196

-362

step-by-step explanation:

(12)(4)+2(12)-34(12)+2(12)-3

  \/         \/         \/         \/

  48 +   24 -   408 + 24 - 3

      \   /               \   /

          72     -     432 - 3 = -363

any options?

step-by-step explanation: