Tính : sin anfa.cosanfa biết tan anfa+ cos anfa =3 12/08/2021 Bởi Delilah Tính : sin anfa.cosanfa biết tan anfa+ cos anfa =3
$\tan\alpha+\cot\alpha$ $=\dfrac{\sin\alpha}{\cos\alpha}+\dfrac{\cos\alpha}{\sin\alpha}$ $=\dfrac{1}{\sin\alpha\cos\alpha}=3$ $\Rightarrow \sin\alpha\cos\alpha=\dfrac{1}{3}$ Bình luận
Ta có: $tana + cota = 3$ $\Leftrightarrow \dfrac{1}{sinacosa} = 3$ $\Leftrightarrow sina = \dfrac{1}{3cosa}$ Ta lại có: $sin^2a + cos^2a = 1$ $\Leftrightarrow \left(\dfrac{1}{3cosa}\right)^2 + cos^2a = 1$ $\Leftrightarrow 9cos^4a – 9cos^2a + 1 = 0$ $\Leftrightarrow \left[\begin{array}{l}cosa = \pm \sqrt{\dfrac{1}{2} – \dfrac{\sqrt5}{6}}\\cosa = \pm \sqrt{\dfrac{1}{2} + \dfrac{\sqrt5}{6}}\end{array}\right.$ $\Rightarrow \left[\begin{array}{l}sina = \pm \sqrt{\dfrac{1}{2} + \dfrac{\sqrt5}{6}}\\sina = \pm \sqrt{\dfrac{1}{2} – \dfrac{\sqrt5}{6}}\end{array}\right.$ Bình luận
$\tan\alpha+\cot\alpha$
$=\dfrac{\sin\alpha}{\cos\alpha}+\dfrac{\cos\alpha}{\sin\alpha}$
$=\dfrac{1}{\sin\alpha\cos\alpha}=3$
$\Rightarrow \sin\alpha\cos\alpha=\dfrac{1}{3}$
Ta có:
$tana + cota = 3$
$\Leftrightarrow \dfrac{1}{sinacosa} = 3$
$\Leftrightarrow sina = \dfrac{1}{3cosa}$
Ta lại có:
$sin^2a + cos^2a = 1$
$\Leftrightarrow \left(\dfrac{1}{3cosa}\right)^2 + cos^2a = 1$
$\Leftrightarrow 9cos^4a – 9cos^2a + 1 = 0$
$\Leftrightarrow \left[\begin{array}{l}cosa = \pm \sqrt{\dfrac{1}{2} – \dfrac{\sqrt5}{6}}\\cosa = \pm \sqrt{\dfrac{1}{2} + \dfrac{\sqrt5}{6}}\end{array}\right.$
$\Rightarrow \left[\begin{array}{l}sina = \pm \sqrt{\dfrac{1}{2} + \dfrac{\sqrt5}{6}}\\sina = \pm \sqrt{\dfrac{1}{2} – \dfrac{\sqrt5}{6}}\end{array}\right.$