Question

cho 0 < \(\alpha\)<\(\dfrac{\pi}{2}\): xác định dấu của các giá trị lượng giác :

a) sin(\(\alpha\) - \(\pi\))

b) cos(\(\dfrac{3\pi}{2}\) - \(\alpha\) )

c) tan(\(\alpha\) + \(\pi\) )

d) cot(\(\alpha\) + \(\dfrac{\pi}{2}\) )

giúp nhau nha

Answer 1

Ta có : \(0< \alpha< \dfrac{\pi}{2}\)

=> \(\sin\alpha>0,\cos\alpha>\text{0},\tan\alpha>\text{0},\cot\alpha>\text{0}\)

a, Ta có : \(\sin\left(\alpha-\pi\right)=-\sin\left(\pi-\alpha\right)=-\left[-\sin\left(\alpha\right)\right]=\sin\alpha\)

=> \(sin\left(\alpha-\pi\right)>\text{0}\)

b, \(\cos\left(\dfrac{3\pi}{2}-\alpha\right)=\cos\left(\pi+\dfrac{\pi}{2}-\alpha\right)=-\cos\left(\dfrac{\pi}{2}-\alpha\right)=-sin\alpha\)

=> \(\cos\left(\dfrac{3\pi}{2}-\alpha\right)< \text{0}\)

 

Answer 2

c, \(tan\left(\alpha+\pi\right)=tan\alpha\)

=> \(tan\left(\alpha+\pi\right)>\text{0}\)

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

=> \(cot\left(\alpha+\dfrac{\pi}{2}\right)< \text{0}\)