Question

Sin alpha+sin beta=-21/65 , cos alpha+cos beta=-27/65 then find cos(alpha-beta)/2

Answer 1

i guess the question is incorrect. because when we solve we get-
consider a triangle abc rt. angled at b
let angle C= ALPHA
let angle A= BETA
then, sin alpha= AB/AC and sin beta= BC/AC
= sin alpha+ sin beta = 21/65 (given)
therefore, AB+BC/AC = 21/65 
thus AB+BC=21
NOW, cos alpha= BC/AC and cos beta= AB/AC
also, cos alpha+ cos beta = 27/65 (given)
therefore, BC+AB= 27
this is also BC+AB= AB+BC
BUT when we solved earlier BC+AB CAME OUT TO BE 27 WHILE AB+BC CAME OUT TO BE 21

Answer 2

Answer:

firstly there must be condition in question that is

$\pi < \alpha - \beta < 3\pi$

Step-by-step explanation:

the follow the steps

hope it helps you