Given that sin alpha 45 alpha in quadrant III and tan beta

Given that sin alpha = -4/5, alpha in quadrant III, and tan beta = 3, 0

step-1: Given

sin (alpha) = -4/5 (alpha is in III quadrant) - in III quadrant tan and cot are positive.

tan (beta) = 3, (beta lies in I quadrant with the given condition, [0<beta< (pi/2)]) - in I quadrant all trigonometric ratios are positive

Step-2: to find cos(alpha)

we have trigonometric identity

sin^2(alpha) + cos^2(alpha) = 1

so cos (alpha) = sqrt (1-sin^2(alpha))

cos(alpha) = sqrt(1-[-4/5]^2)

= sqrt(1-16/25)

= sqrt(9/25)

= +3/5 and -3/5

we take cos(alpha) to be -ve as it is alpha is in III quadrant

so cos (alpha) = -3/5

step-3: to find cos(beta)

First we find sec^(beta)

from trigonometric identity 1+tan^2(beta) = sec^2(beta)

sec^2(beta) = 1+tan^2(beta)

sec (beta) = sqrt[1+tan^2(beta) ]

= sqrt[1+3*3]

=sqrt(10)

=+3.162, -3.162

in I quadrant all ratios are positive so

sec (beta) = 3.162

hence cos (beta) = 1/3.162 = 0.3162

sin (beta+ = sqrt[1-cos^2(beta)]=sqrt[1-(1/9)]=sqrt(8/9) = 0.942

a) to find sin(alpha+beta)

sin (alpha+beta) = sin (alpha) cos (beta) + cos (alpha) sin (beta)

= (-4/5)*(0.3612)+(-3/5)*(0.942)

= -0.8541

b) cos(alpha/2) = sqrt[(1+cos(alpha)/2]

= sqrt [(1+(-3/5))/2]

= sqrt [1/5]

=0.4472