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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: \(cos^2\dfrac{\Pi}{8}-sin^2\dfrac{\Pi}{8}=cos2\left(\dfrac{\Pi}{8}\right)=cos\dfrac{\Pi}{4}=\dfrac{\sqrt{2}}{2}\)
bài 1) ta có : \(G=cos\left(\alpha-5\pi\right)+sin\left(\dfrac{-3\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{3\pi}{2}-\alpha\right)\)
\(G=cos\left(\alpha-\pi\right)+sin\left(\dfrac{\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{\pi}{2}-\alpha\right)\)
\(G=cos\left(\pi-\alpha\right)+sin\left(\dfrac{\pi}{2}-\left(-\alpha\right)\right)-tan\left(\pi+\alpha-\dfrac{\pi}{2}\right).cot\left(\dfrac{\pi}{2}-\alpha\right)\) \(G=cos\left(\alpha\right)+cos\left(\alpha\right)+tan\left(\dfrac{\pi}{2}-\alpha\right).cot\left(\dfrac{\pi}{2}-\alpha\right)=2cos\alpha+1\) bài 2) ta có : \(H=cot\left(\alpha\right).cos\left(\alpha+\dfrac{\pi}{2}\right)+cos\left(\alpha\right)-2sin\left(\alpha-\pi\right)\) \(H=cot\left(\alpha\right).cos\left(\dfrac{\pi}{2}-\left(-\alpha\right)\right)+cos\left(\alpha\right)+2sin\left(\pi-\alpha\right)\) \(H=-cot\left(\alpha\right).sin\left(\alpha\right)+cos\left(\alpha\right)+2sin\left(\alpha\right)\) \(H=-cos\alpha+cos\alpha+2sin\alpha=2sin\alpha\)
a) \(A=sin\left(\dfrac{\pi}{4}+x\right)-cos\left(\dfrac{\pi}{4}-x\right)\)
\(\Leftrightarrow A=sin\dfrac{\pi}{4}.cosx+cos\dfrac{\pi}{4}.sinx-\left(cos\dfrac{\pi}{4}.cosx+sin\dfrac{\pi}{4}.sinx\right)\)
\(\Leftrightarrow A=sin\dfrac{\pi}{4}.cosx+cos\dfrac{\pi}{4}.sinx-cos\dfrac{\pi}{4}.cosx-sin\dfrac{\pi}{4}.sinx\)
\(\Leftrightarrow A=\dfrac{\sqrt{2}}{2}.cosx+\dfrac{\sqrt{2}}{2}.sinx-\dfrac{\sqrt{2}}{2}.cosx-\dfrac{\sqrt{2}}{2}.sinx\)
\(\Leftrightarrow A=0\)
b) \(B=cos\left(\dfrac{\pi}{6}-x\right)-sin\left(\dfrac{\pi}{3}+x\right)\)
\(\Leftrightarrow B=cos\dfrac{\pi}{6}.cosx+sin\dfrac{\pi}{6}.sinx-\left(sin\dfrac{\pi}{3}.cosx+cos\dfrac{\pi}{3}.sinx\right)\)
\(\Leftrightarrow B=cos\dfrac{\pi}{6}.cosx+sin\dfrac{\pi}{6}.sinx-sin\dfrac{\pi}{3}.cosx-cos\dfrac{\pi}{3}.sinx\)
\(\Leftrightarrow B=\dfrac{\sqrt{3}}{2}.cosx+\dfrac{1}{2}.sinx-\dfrac{\sqrt{3}}{2}.cosx-\dfrac{1}{2}.sinx\)
\(\Leftrightarrow B=0\)
c) \(C=sin^2x+cos\left(\dfrac{\pi}{3}-x\right).cos\left(\dfrac{\pi}{3}+x\right)\)
\(\Leftrightarrow C=sin^2x+\left(cos\dfrac{\pi}{3}.cosx+sin\dfrac{\pi}{3}.sinx\right).\left(cos\dfrac{\pi}{3}.cosx-sin\dfrac{\pi}{3}.sinx\right)\)
\(\Leftrightarrow C=sin^2x+\left(\dfrac{1}{2}.cosx+\dfrac{\sqrt{3}}{2}.sinx\right).\left(\dfrac{1}{2}.cosx-\dfrac{\sqrt{3}}{2}.sinx\right)\)
\(\Leftrightarrow C=sin^2x+\dfrac{1}{4}.cos^2x-\dfrac{3}{4}.sin^2x\)
\(\Leftrightarrow C=\dfrac{1}{4}.sin^2x+\dfrac{1}{4}.cos^2x\)
\(\Leftrightarrow C=\dfrac{1}{4}\left(sin^2x+cos^2x\right)\)
\(\Leftrightarrow C=\dfrac{1}{4}\)
d) \(D=\dfrac{1-cos2x+sin2x}{1+cos2x+sin2x}.cotx\)
\(\Leftrightarrow D=\dfrac{1-\left(1-2sin^2x\right)+2sinx.cosx}{1+2cos^2a-1+2sinx.cosx}.cotx\)
\(\Leftrightarrow D=\dfrac{2sin^2x+2sinx.cosx}{2cos^2x+2sinx.cosx}.cotx\)
\(\Leftrightarrow D=\dfrac{2sinx\left(sinx+cosx\right)}{2cosx\left(cosx+sinx\right)}.cotx\)
\(\Leftrightarrow D=\dfrac{sinx}{cosx}.cotx\)
\(\Leftrightarrow D=tanx.cotx\)
\(\Leftrightarrow D=1\)
đề sai nhỉ? sina/2; cos a/2; tana/2; cota/2 chứ?
ta có:
\(sin^2\dfrac{a}{2}=\dfrac{1-cosa}{2}=\dfrac{1-\dfrac{5}{13}}{2}=\dfrac{4}{13}\)
\(\dfrac{3\pi}{2}< a< 2\pi\Leftrightarrow\dfrac{3\pi}{4}< \dfrac{a}{2}< \pi\)
=> sina/2 > 0 => sina/2 = \(\dfrac{2}{\sqrt{13}}\)
ta có:
\(cos^2\left(\dfrac{a}{2}\right)=1-sin^2\left(\dfrac{a}{2}\right)=1-\dfrac{4}{13}=\dfrac{9}{13}\)
\(\dfrac{3\pi}{2}< a< 2\pi\Leftrightarrow\dfrac{3\pi}{4}< \dfrac{a}{2}< \pi\) (cung2)
=> cosa/2 < 0 => cosa/2 = \(\dfrac{-3}{\sqrt{13}}\)
\(tan\left(\dfrac{a}{2}\right)=\dfrac{sin\left(\dfrac{a}{2}\right)}{cos\left(\dfrac{a}{2}\right)}=\dfrac{\dfrac{2}{\sqrt{13}}}{-\dfrac{3}{\sqrt{13}}}=-\dfrac{2}{3}\)
\(cot\left(\dfrac{a}{2}\right)=\dfrac{1}{tan\left(\dfrac{a}{2}\right)}=\dfrac{1}{-\dfrac{2}{3}}=-\dfrac{3}{2}\)
\(H.sin\dfrac{\pi}{19}=sin\dfrac{\pi}{19}.cos\dfrac{\pi}{19}+sin\dfrac{\pi}{19}cos\dfrac{3\pi}{19}+...+sin\dfrac{\pi}{19}cos\dfrac{17\pi}{19}\)
\(=\dfrac{1}{2}sin\dfrac{2\pi}{19}+\dfrac{1}{2}sin\dfrac{4\pi}{19}-\dfrac{1}{2}sin\dfrac{2\pi}{19}+...+\dfrac{1}{2}sin\dfrac{18\pi}{19}-\dfrac{1}{2}sin\dfrac{16\pi}{19}\)
\(=\dfrac{1}{2}sin\dfrac{18\pi}{19}=\dfrac{1}{2}sin\left(\pi-\dfrac{\pi}{19}\right)=\dfrac{1}{2}sin\dfrac{\pi}{19}\)
\(\Rightarrow H=\dfrac{1}{2}\)