Biết \(\int\limits^{\frac{\pi}{2}}_0\left(e^{\sin x}+\cos x\right)\cos x\text{d}x=e-1+a\). Trong các khẳng định sau, khẳng định nào sai ?
\(\sin\left(\dfrac{3\pi}{4}+a-\alpha\right)=-\sin\alpha,\forall\alpha\). \(\cos\left(\dfrac{3\pi}{4}+a-\alpha\right)=-\cos\alpha,\forall\alpha\). \(\tan\left(\dfrac{3\pi}{4}+a-\alpha\right)=-\tan\alpha,\forall\alpha\). \(\cot\left(\dfrac{3\pi}{4}+a-\alpha\right)=-\cot\alpha,\forall\alpha\). Hướng dẫn giải:\(\int\limits^{\frac{\pi}{2}}_0\left(e^{\sin x}+\cos x\right)\cos x\text{d}x=\int\limits^{\frac{\pi}{2}}_0e^{\sin x}\cos x\text{d}x+\int\limits^{\frac{\pi}{2}}_0\cos^2x\text{d}x\)
\(=\int\limits^{\frac{\pi}{2}}_0e^{\sin x}\text{d}\left(\sin x\right)+\int\limits^{\frac{\pi}{2}}_0\dfrac{1+\cos2x}{2}\text{d}x\)
\(=e^{\sin x}|^{\frac{\pi}{2}}_0+\dfrac{1}{2}x|^{\frac{\pi}{2}}_0+\dfrac{1}{4}\int\limits^{\frac{\pi}{2}}_0\cos2x\text{d}\left(2x\right)\)
\(=\left(e^{\sin x}+\dfrac{1}{2}x+\dfrac{1}{4}\sin2x\right)|^{\frac{\pi}{2}}_0\)
\(=e+\dfrac{\pi}{4}-1\)
Suy ra \(e+\dfrac{\pi}{4}-1=e-1+a\)
\(\Rightarrow a=\dfrac{\pi}{4}\)
Vậy: \(\dfrac{3\pi}{4}+a=\dfrac{3\pi}{4}+\dfrac{\pi}{4}=\pi\)
Góc \(\dfrac{3\pi}{4}+a-\alpha\) và góc \(\alpha\) là hai góc bù nhau. Suy ra phát biểu \(\sin\left(\dfrac{3\pi}{4}+a-\alpha\right)=-\sin\alpha,\forall\alpha\) là sai.