\(sinx=\dfrac{2tan\dfrac{x}{2}}{tan^2\dfrac{x}{2}+1}\)
\(cosx=\dfrac{1-tan^2\dfrac{x}{2}}{1+tan^2\dfrac{x}{2}}\)
Đặt \(t=tan\dfrac{x}{2}\)
Khi đó pt: \(\Rightarrow a\cdot\dfrac{2t}{t^2+1}+b\cdot\dfrac{1-t^2}{1+t^2}=c\)
\(\Rightarrow2t\cdot a+\left(1-t^2\right)\cdot b=\left(1+t^2\right)\cdot c\)