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a) \(\dfrac{\sin2\text{a}+\cos a}{1+\cos2\text{a}+\cos a}=2\tan a\)
a) \(\dfrac{sin2\alpha+sin\alpha}{1+cos2\alpha+cos\alpha}=\dfrac{2sin\alpha cos\alpha+sin\alpha}{2cos^2\alpha+cos\alpha}\)\(=\dfrac{sin\alpha\left(2cos\alpha+1\right)}{cos\alpha\left(2cos\alpha+1\right)}=\dfrac{sin\alpha}{cos\alpha}=tan\alpha\).
rút gọn biểu thức:
E=cos(\(\dfrac{3\pi}{3}-\alpha\))-sin(\(\dfrac{3\pi}{2}-\alpha\))+sin(\(\alpha+4\pi\))
Ta có : A+B+C= 180
=>sin(A+B)/2 = sin(180/2 - C/2) = cosC/2
ttcó: sinC/2 = cos(A+B)/2
=> sA+sB+sC =2cosC/2*cos(A-B)/2 + 2cos(A+B)/2*cosC/2
=2cosC/2
=4cosA/2cosB/2cosC/2
a)
\(\cos\dfrac{22\pi}{3}=\cos\left(8\pi-\dfrac{2\pi}{3}\right)\\ =\cos\left(-\dfrac{2\pi}{3}\right)\\ =\cos\left(\dfrac{2\pi}{3}\right)\\ =-\cos\dfrac{\pi}{3}\\ =-\dfrac{1}{2}\)
b)
\(\sin\dfrac{23\pi}{4}=\sin\left(6\pi-\dfrac{\pi}{4}\right)\\ =\sin\left(-\dfrac{\pi}{4}\right)\\ =-\dfrac{\sqrt{2}}{2}\)
c)
\(\sin\dfrac{25\pi}{3}-\tan\dfrac{10\pi}{3}\\ =\sin\left(8\pi+\dfrac{\pi}{3}\right)-\tan\left(3\pi+\dfrac{\pi}{3}\right)\\ =\sin\dfrac{\pi}{3}-\tan\dfrac{\pi}{3}\\ =\dfrac{\sqrt{3}}{2}-\sqrt{3}\\ =\dfrac{-\sqrt{3}}{2}\)
d)
\(\cos^2\dfrac{\pi}{8}-\sin^2\dfrac{\pi}{8}\\ =\cos\dfrac{\pi}{4}\\ =\dfrac{\sqrt{2}}{2}\)
cau a: \(cos\dfrac{22\Pi}{3}=cos\dfrac{24\Pi-2\Pi}{3}=cos\left(8\Pi-\dfrac{2\Pi}{3}\right)=cos\dfrac{2\Pi}{3}=-\dfrac{1}{2}\)
câu b: \(sin\dfrac{23\Pi}{4}=sin\dfrac{24\Pi-\Pi}{4}=sin\left(6\Pi-\dfrac{\Pi}{4}\right)=-sin\dfrac{\Pi}{4}=-\dfrac{\sqrt{2}}{2}\)
cau c: \(=sin\left(8\Pi-\dfrac{\Pi}{3}\right)-tan\left(3\Pi+\dfrac{\Pi}{3}\right)=-sin\dfrac{\Pi}{3}-tan\dfrac{\Pi}{3}=-\dfrac{\sqrt{3}}{2}-\sqrt{3}=\dfrac{-3\sqrt{3}}{2}\)
cau d: \(cos^2\dfrac{\Pi}{8}-sin^2\dfrac{\Pi}{8}=cos2\left(\dfrac{\Pi}{8}\right)=cos\dfrac{\Pi}{4}=\dfrac{\sqrt{2}}{2}\)