Vì $\frac{-π}{2}$ < ∝ < 0 ⇒ ∝ ∈ góc phần tư thứ tư
⇒ cos ∝ > 0 , sin ∝ < 0
- Ta có: cos∝ = $\sqrt{1^2-sin∝^2}$ = $\sqrt{1^2-(\frac{-2}{5})²}$ = $\frac{√21}{5}$
⇒ cos2 ∝ = 1- 2.sin²∝ = 1 - 2.($\frac{-2}{5}$)² = $\frac{17}{25}$
- tan ∝ = $\frac{sin∝}{cos∝}$ = $\frac{\frac{-2}{5}}{\frac{√21}{5} }$ = $\frac{-2√21}{21}$
⇒ tan2∝ = $\frac{2tan∝}{1-tan²∝}$ = $\frac{2.\frac{-2√21}{21}}{1-(\frac{-2√21}{21})²}$ = $\frac{-4√21}{17}$
