Question

Chứng minh rằng với mọi \(\alpha\), ta luôn có :

a) \(\sin\left(\alpha+\dfrac{\pi}{2}\right)=\cos\alpha\)

b) \(\cos\left(\alpha+\dfrac{\pi}{2}\right)=-\sin\alpha\)

c) \(\tan\left(\alpha+\dfrac{\pi}{2}\right)=-\cot\alpha\)

d) \(\cot\left(\alpha+\dfrac{\pi}{2}\right)=-\tan\alpha\)

Answer 1

a)\(sin\left(\alpha+\dfrac{\pi}{2}\right)=cos\left[\dfrac{\pi}{2}-\left(\alpha+\dfrac{\pi}{2}\right)\right]=cos\left(-\alpha\right)=cos\alpha\).
b) \(cos\left(x+\dfrac{\pi}{2}\right)=sin\left[\dfrac{\pi}{2}-\left(x+\dfrac{\pi}{2}\right)\right]=sin\left(-x\right)=-sinx\).
c) \(tan\left(\alpha+\dfrac{\pi}{2}\right)=\dfrac{sin\left(\alpha+\dfrac{\pi}{2}\right)}{cos\left(\alpha+\dfrac{\pi}{2}\right)}=\dfrac{cos\alpha}{-sin\alpha}=-cot\alpha\).
d) \(cot\left(\alpha+\dfrac{\pi}{2}\right)=\dfrac{cos\left(\alpha+\dfrac{\pi}{2}\right)}{sin\left(\alpha+\dfrac{\pi}{2}\right)}=\dfrac{-sin\alpha}{cos\alpha}=-tan\alpha\).