Question

Cho hàm số \(f\left(x\right)=\dfrac{sinx}{sinx+cosx}\). Tính giá trị của biểu thức \(T=f\left(\dfrac{\pi}{2024}\right)+f\left(\dfrac{2\pi}{2024}\right)+f\left(\dfrac{3\pi}{2024}\right)+...+f\left(\dfrac{1011\pi}{2024}\right)\)

Answer 1

\(f\left(x\right)+f\left(\dfrac{\pi}{2}-x\right)=\dfrac{sinx}{sinx+cosx}+\dfrac{sin\left(\dfrac{\pi}{2}-x\right)}{sin\left(\dfrac{\pi}{2}-x\right)+cos\left(\dfrac{\pi}{2}-x\right)}\)

\(=\dfrac{sinx}{sinx+cosx}+\dfrac{cosx}{cosx+sinx}=1\)

\(\Rightarrow T=f\left(\dfrac{\pi}{2024}\right)+f\left(\dfrac{1011\pi}{2024}\right)+...+f\left(\dfrac{505\pi}{2024}\right)+f\left(\dfrac{507\pi}{2024}\right)+f\left(\dfrac{506\pi}{2024}\right)\)

\(=1+1+...+1+f\left(\dfrac{\pi}{4}\right)\) (505 số 1)

\(=505+\dfrac{sin\left(\dfrac{\pi}{4}\right)}{sin\left(\dfrac{\pi}{4}\right)+cos\left(\dfrac{\pi}{4}\right)}=505+\dfrac{1}{2}=...\)