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Vì 0 < α < π/2 nên sin α > 0, cos α > 0, tan α > 0, cot α > 0.
\(sin^2\text{α}=1-cos^2\text{α}=1-\left(\dfrac{1}{3}\right)^2=\dfrac{8}{9}\)
vì π<α<\(\dfrac{\Pi}{2}\)⇒sin α=\(\dfrac{2\sqrt{2}}{3}\)
\(sin^2\alpha=1-sin^2\alpha=1-\left(\dfrac{-4}{5}\right)^2=\dfrac{9}{25}\)
vì π<α<\(\dfrac{3\Pi}{2}\)⇒cos α =\(\dfrac{-3}{5}\)
cos2a =1- sin2a =1-\(\left(\dfrac{-4}{5}\right)^2\)=\(\dfrac{3}{5}\)
Vì π<a<\(\dfrac{3\pi}{2}\)
=>cos a =\(\dfrac{-3}{5}\)
\(\dfrac{3\pi}{2}\le a\le2\pi\Rightarrow3\pi\le2a\le4\pi\)
\(\Rightarrow sin2a\le0\)
\(cos^2a-sin^2a=\dfrac{1}{2}\Leftrightarrow cos2a=\dfrac{1}{2}\)
\(\Rightarrow sin2a=-\sqrt{1-cos^22a}=-\dfrac{\sqrt{3}}{2}\)
\(sin\left(\text{α}-\dfrac{\Pi}{4}\right)-cos\left(\text{α}-\dfrac{\Pi}{4}\right)\)
\(=sin\text{α}.cos\dfrac{\Pi}{4}-cos\text{α}-sin\dfrac{\Pi}{4}-\left(cos\text{α}.cos\dfrac{\Pi}{4}+sin\text{α}.sin\dfrac{\Pi}{4}\right)\)
\(=sin\text{α}.\dfrac{\sqrt{2}}{2}-\dfrac{1}{3}.\dfrac{\sqrt{2}}{2}-\dfrac{1}{3}.\dfrac{\sqrt{2}}{2}-sin\text{α}.\dfrac{\sqrt{2}}{2}\)
\(=\dfrac{-2\sqrt{2}}{6}\)
\(=\dfrac{-\sqrt{2}}{3}\)
Câu 1
Ta có vế trái:
\[ VT=\sin\left(\frac{\pi}{2}-\alpha\right) \cos(\pi-\alpha) \] \[ =\cos\alpha.(-\cos\alpha) \] \[ =-\cos^2\alpha \]Ta có vế phải:
\[ VP=-\frac{1}{1+\tan^2(\pi-\alpha)} \] \[ =-\frac{1}{1+\tan^2\alpha} \] \[ =-\frac{1}{\frac{1}{\cos^2\alpha}} \] \[ =-\cos^2\alpha \]Vậy VT = VP (đpcm).
Câu 2
Ta có vế trái:
\[ VT=\sin^2\left(\frac{\pi}{2}-\alpha\right) \] \[ =\cos^2\alpha \]Ta có vế phải:
\[ VP=\frac{1}{1+\tan^2\alpha} \] \[ =\frac{1}{1+\frac{\sin^2\alpha}{\cos^2\alpha}} \] \[ =\frac{1}{\frac{\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha}} \] \[ =\frac{1}{\frac{1}{\cos^2\alpha}} \] \[ =\cos^2\alpha \]Vậy VT = VP (đpcm).
do a ∈ \(\left(0;\dfrac{\pi}{2}\right)\)⇒ \(\left\{{}\begin{matrix}sinx>0\\cosx>0\end{matrix}\right.\)
Mà tanx = 3 ⇒ \(\dfrac{sinx}{cosx}=3\Leftrightarrow\dfrac{sin^2x}{cos^2x}=9\Rightarrow10sin^2x=9\)
⇒ sinx = \(\dfrac{3}{\sqrt{10}}\)
⇒ sin (x + π) = -sinx = -\(\dfrac{3}{\sqrt{10}}\)