Question

if sin A = 4/5 , sin B = 12/13 , and A and B are acute angles find sin(A-B)cos(A-B)?​

Answer 1

Given,

Given,sinA =3/5 and sinB=5/13 where A and B are acute angles.

Given,sinA =3/5 and sinB=5/13 where A and B are acute angles.SinA= 3/5 then, Cos A= 4/5. ( By Pythagoras theorem)

Given,sinA =3/5 and sinB=5/13 where A and B are acute angles.SinA= 3/5 then, Cos A= 4/5. ( By Pythagoras theorem)SinB = 5/13. Then ,Cos B= 12/13.(By Pythagoras theorem)

Given,sinA =3/5 and sinB=5/13 where A and B are acute angles.SinA= 3/5 then, Cos A= 4/5. ( By Pythagoras theorem)SinB = 5/13. Then ,Cos B= 12/13.(By Pythagoras theorem)Cos(A+B)= CosA.CosB – SinA.SinB

Given,sinA =3/5 and sinB=5/13 where A and B are acute angles.SinA= 3/5 then, Cos A= 4/5. ( By Pythagoras theorem)SinB = 5/13. Then ,Cos B= 12/13.(By Pythagoras theorem)Cos(A+B)= CosA.CosB – SinA.SinBThen put the value

Given,sinA =3/5 and sinB=5/13 where A and B are acute angles.SinA= 3/5 then, Cos A= 4/5. ( By Pythagoras theorem)SinB = 5/13. Then ,Cos B= 12/13.(By Pythagoras theorem)Cos(A+B)= CosA.CosB – SinA.SinBThen put the valueCos(A+B)= 4/5×12/13 – 3/5×5/13

Given,sinA =3/5 and sinB=5/13 where A and B are acute angles.SinA= 3/5 then, Cos A= 4/5. ( By Pythagoras theorem)SinB = 5/13. Then ,Cos B= 12/13.(By Pythagoras theorem)Cos(A+B)= CosA.CosB – SinA.SinBThen put the valueCos(A+B)= 4/5×12/13 – 3/5×5/13Cos(A+B) = 48/65 – 15/65

Given,sinA =3/5 and sinB=5/13 where A and B are acute angles.SinA= 3/5 then, Cos A= 4/5. ( By Pythagoras theorem)SinB = 5/13. Then ,Cos B= 12/13.(By Pythagoras theorem)Cos(A+B)= CosA.CosB – SinA.SinBThen put the valueCos(A+B)= 4/5×12/13 – 3/5×5/13Cos(A+B) = 48/65 – 15/65Cos(A+B)=33/65. ( Answer).

Answer 2

Step-by-step explanation:

by using sum and difference formula of trigonometry .