Given: sinA= 4/5 and cos B = -5/13; evaluate the following expression. (A: Quadrant I, B : Quadrant II.)?

Question

sin( A+ B )

16/65
33/65
56/65

cos( A- B)

-63/65
-33/65
33/65

tan( B+ A )

-46/63
-16/63
16/63

sin 2A

-40/65
6/25
24/25

cos 2B

-119/169
119/169
144/169

tan 2A

8/3
24/7
-24/7

tan½B

-3/2
2/3
3/2

sin 3A

44/125
12/5
64/125

Answer 1

Given that sinA = 4/5 and cosB = -5/13 (A: quadrant i, B:quadrant ii)

Apply pythagorean fomula c^2 = a^2+b^2

sinA = a/c = 4/5 where a=4 and c= 5

b^2 = c^2 - a^2

b^2 = 25 - 16 = 9

Therefore b = 3

cosA = b/c = 3/5

cosB = b/c = -5/13

Apply pythagorean fomula c^2 = a^2+b^2

a^2 = c^2 - b^2

a^2 = (13)^2 - (-5)^2

a^2 = 169- 25 = 144

Therefore a = 12

sinB = a/c = 12/13 where a = 12 and c= 13

tanA = 4/5 and tanB = -12/5

a) sin(A+B) = sinAcosB+cosAsinB

= (4/5) (-5/13) +(3/5)(12/13)

= (-4/13)+(36/65)

= -20+36/65 = 16/65

b) cos(A-B) = cosAcosB+sinAsinB

= (3/5)(-5/13) +(4/5)(12/13)

= -3/13+48/65

= (-15+48)/65 = 33/65

cos(A-B) = 33/65

c) tan(B+A) = (tanB+tanA)/1 - (tanA tanB)

= [(-12/5) +(4/3)]/ [1- (-12/5)(4/3)]

= [(-36+20)/15]/1- (-48/15)

= (-16/15)(15/63)= -16/63

tan(B+A) = -16/63

d)sin2A = 2sinAcosA

= 2 (4/5) (3/5)

= 2 (12/25) = 24/25

sin2A = 24/25

e) cos2B = 1- 2sin^2B

= 1 - 2(12/13)^2

= 1- (288/169)

= -119/169

cos2B = -119/169

f) tan2A = 2tanA/ 1- tan^2A

= 2(4/3)/1- (4/3)^2

= (8/3)/1- (16/9)

= (8/3)(-9/7) = -24/7

tan2A = -24/7

g) tan1/2B = (1- cosB)/sinB

= 1 - (-5/13)/ (12/13)

= (1+5/13)/12/13

=(18/13)(13/12)

= 18/12 = 3/2

tan1/2B = 3/2

h) sin3A = 3sinA - 4sin^3A

= 3(4/5) - 4 (4/5)^3

=(12/5) - 4(64/125)

= (12/5 - 256/125)

= 44/25

sin3A = 44/125.