only need answer 10 11 and 12 please show work In Exercises

only need answer 10, 11, and 12 please show work

In Exercises 8-12 you are asked to find the sine, cosine and tangent of the half angle 0/2, based on the given information on 0. Use exact values. 8. sin theta = -3/5,3pi/2

Solution

10 . tan = -7/24 , -3pi< < -2pi

is IVrth quadrant

cos = 24/25

Now use th double angle formula:

cos = 2cos^2(/2) -1

cos(/2) = sqrt{(1+cos)/2} = sqrt{(1+24/25)/2} =7/5sqrt2

cos = 1- 2sin^2(/2)

sin(/2) = sqrt{( 1- cos)/2} = sqrt{(1 -24/25/2} = 1/5sqrt2

tan(/2) = sin(/2)/cos(/2) = 1/7

11. cos = -1/2 ; 180 < <360

is in QIII

Now use th double angle formula:

cos = 2cos^2(/2) -1

cos(/2) = sqrt{(1+cos)/2} = sqrt( 1 -1/2)/2) =1/2

sin(/2) = sqrt{(1-cos)/2} = sqrt((1 +1/2)/2) = sqrt(3)/2

tan(/2) = sin/2/cos/2 = sqrt3

12) = arccos(-5/13)

Range of arccos [0, pi]

cos = -5/13    ; is IInd quadrant

sin = 12/13

cos = 2cos^2(/2) -1

cos(/2) = sqrt{(1+cos)/2} = sqrt( 1+ (-5/13))/2) = 2/sqrt13

sin(/2) = sqrt{(1-cos)/2} = sqrt( 1 -(-5/13))/2) = 3/sqrt13

tan(/2) = sin(/2)/cos(/2) = 3/2