Given as
\(sin({\frac{3π}{8}} -5) cos({\frac{π}{8}} + 5) + cos({\frac{3π}{8}} - 5) sin({\frac{π}{8}} + 5)\) = 1
Let us consider the LHS
\(sin({\frac{3π}{8}} -5) cos({\frac{π}{8}} + 5) + cos({\frac{3π}{8}} - 5) sin({\frac{π}{8}} + 5)\)
As we know that sin(A + B) = sin A cos B + cos A sin B
\(sin({\frac{3π}{8}} -5) cos({\frac{π}{8}} + 5) + cos({\frac{3π}{8}} - 5) sin({\frac{π}{8}} + 5)\) =
= sin 90°
= 1
= RHS
∴ LHS = RHS
Thus proved.
