Answer:
Step-by-step explanation:
Given the expression 4sinB = 3sin(2A+B), we are to show that the expression 7cot(A+B) = cotA
Starting with the expression
4sinB= 3sin(2A+B)
Let us re write angle B = (A + B) - A
and 2A + B = (A + B) + A
Substituting the derived expression back into the original expression ww will have;
4Sin{(A + B) - A } = 3Sin{(A + B)+ A}
From trigonometry identity;
Sin(D+E) = SinDcosE + CosDSinE
Sin(D-E) = SinDcosE - CosDSinE
Applying this in the expression above;
4{Sin(A+B)CosA - Cos(A+B)SinA} = 3{Sin(A+B)CosA + Cos(A+B)sinA}
Open the bracket
4Sin(A+B)CosA - 4Cos(A+B)SinA = 3Sin(A+B)CosA + 3Cos(A+B)sinA
Collecting like terms
4Sin(A+B)CosA - 3Sin(A+B)cosA = 3Cos(A+B)sinA + 4Cos(A+B)sinA
Sin(A+B)CosA = 7Cos(A+B)sinA
Divide both sides by sinA
Sin(A+B)CosA/sinA= 7Cos(A+B)sinA/sinA
Since cosA/sinA = cotA, the expression becomes;
Sin(A+B)cotA = 7Cos(A+B)
Finally, divide both sides of the resulting equation by sin(A+B)
Sin(A+B)cotA/sin(A+B) = 7Cos(A+B)/sin(A+B)
CotA = 7cot(A+B) Proved!
