Question

Given that Sin A = 4/5, 0 < A < π/2 and Cos B = -12/13, π< B < 3π/2, find Sin (A+B)

choices: a.) -56/65 b.) 56/65 c.) 63/16 d.) -63/16

Answer 1

use this formula Sin (A+B) = sin(A)cos(B) + cos(A)sin(B)

Answer 2

thank so much :)

Answer 3

now find cos(A) if given sin(A) = 4/5
use the identity :
cos^2 (A) = 1 - sin^2 (A)

cos^2 (A) = 1 - (4/5)^2
cos^2 (A) = 1 - 16/25
cos^2 (A) = 9/25
cos(A) = 3/5 (notice that A in the 1st quadrant, so cos be positive)

Answer 4

then find sin(B).
given Cos B = -12/13,

sin^2 (B) = 1 - cos^2 (B)
sin^2 (B) = 1- (-12/13)^2
sin^2 (B) = 1- 144/169
sin^2 (B) = 25/169
sin(B) = -5/13 (see that B in the 3rd quadrant, so sin be negative)

Answer 5

finally, plugin all values to formula sin(a+b)
sin (A+B) = sin(A)cos(B) + cos(A)sin(B)
sin (A+B) = 4/5 * (-12/13) + 3/5 * (-5/13) = ...
simplify it to get your answer

Answer 6

answer is -113/65 ?

Answer 7

how do you get -113 ?

Answer 8

-63 / 65 :) is the correct answer :) thank you my friend

Answer 9

case is closed :)

Answer 10

welcome

Answer 11

:D thank you ..help me again i have so many problems here ahahaha. Given that Sin A = 4/5, 0 < A < π/2 and Cos B = -12/13, π< B < 3π/2, find Tan (A+B) .. this time its tan(A+B) :)

Answer 12

well, it like before
we have sin(A) = 4/5, cos(A) = 3/5, sin(B) = -5/13, and cos(B) = -12/13
now find tan(A) and tan(B) respectively is ...

see tan = sin/cos, so

tan(A) = sin(A)/cos(A) = 4/5)/(3/5) = 4/3 (tan in the 1st quadrant is +)
tan(b) = sin(B)/cos(B) = (5/13)/(12/13) = 5/12 (tan in the 3rd quadrant is +)

Answer 13

the question is tan(A+B). use the identity :
tan(A+B) = (tan(A) + tan(B))/(1 - tan(A)tan(B))
so,
tan(A+B) = (4/3 + 5/12)/(1 - 4/3 * 5/12)
tan(A+B) = .... ?

Answer 14

63/16 is the answer :) thank you again

Answer 15

Given that Sin A = 4/5, 0 < A < π/2 and Cos B = -12/13, π< B < 3π/2, find Cos 2A