Question

if cos(A+B)=0 then sin(A-B) is reduced to ?​

Answer 1

Answer:

sin(a-b)=cos{2b

Step-by-step explanation:

Given that cos(a+b)=0

then we have to find the reduced form of sin(a-b)

As, cos(a+b)=0cos(a+b)=0

⇒ a+b=cos^{-1}(0)=90^{\circ}a+b=cos

−1

(0)=90

∘

⇒ a=(90-b)^{\circ}a=(90−b)

∘

sin(a-b)=sin(90-b-b)=sin(90-2b)=cos{2b}sin(a−b)=sin(90−b−b)=sin(90−2b)=cos2b

Hence, sin(a-b)=cos{2b}

hope it helps you