Let sin A= -3/5 with 270°< A< 360°. Find the following.
sin A/2

1/sqrt10

Step-by-step explanation:

1) Find out cosA using formula (cosA)^2+(sinA)^2=1

The module of cosA= sqrt (1- (-3/5)^2)= sqrt 16/25=4/5

So cosA=-4/5 or cosA=4/5.

Due to the condition 270degrees< A<360 degrees, 0<cosA<1 that's why cosA=4/5.

2) Find sinA/2 using a formula cosA= 1-2sinA/2*sinA/2 where cosA=4/5.

(sinA/2)^2= 0.1

sinA= sqrt 0.1= 1/ sqrt10 or sinA= - sqrt 0.1= -1/sqrt10

But 270°< A< 360°, then 270/2°<A/2<360/2°

135°<A/2<180°, so sinA/2 must be positive and the only correct answer is

sin A/2= 1/sqrt10

b. 2 and 4

step-by-step explanation: