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a)
\(\cos225^0=\cos\left(180^0+45^0\right)=-\cos45^0=-\dfrac{\sqrt{2}}{2}\)
\(\sin240^0=\sin\left(180^0+60^0\right)=-\sin60^0=-\dfrac{\sqrt{3}}{2}\)
\(\cos\left(-15^0\right)=-\cot15^0=-\tan75^0=-\tan\left(30^0+45^0\right)\)
\(=\dfrac{-\tan30^0-\tan45^0}{1-\tan30^0\tan45^0}=\dfrac{-\dfrac{1}{\sqrt{3}}-1}{1-\dfrac{1}{\sqrt{3}}}=-\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(=-\dfrac{\left(\sqrt{3}+1\right)^2}{2}=-2-\sqrt{3}\)
\(\tan75^0=\cot15^0=2+\sqrt{3}\)
b)
\(\sin\dfrac{7\pi}{12}=\sin\left(\dfrac{\pi}{3}+\dfrac{\pi}{4}\right)=\sin\dfrac{\pi}{3}\cos\dfrac{\pi}{4}+\cos\dfrac{\pi}{3}\sin\dfrac{\pi}{4}\)
\(=\dfrac{\sqrt{2}}{2}\left(\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\right)=\dfrac{\sqrt{6}+\sqrt{2}}{4}\)
\(\cos\left(-\dfrac{\pi}{12}\right)=\cos\left(\dfrac{\pi}{4}-\dfrac{\pi}{3}\right)=\cos\dfrac{\pi}{4}\cos\dfrac{\pi}{3}+\sin\dfrac{\pi}{3}\sin\dfrac{\pi}{4}\)
\(=\dfrac{\sqrt{2}}{2}\left(\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\right)=0,9659\dfrac{\sqrt{2}}{2}\left(\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\right)=0,9659\)
\(\tan\dfrac{13\pi}{12}=\tan\left(\pi+\dfrac{\pi}{12}\right)=\tan\dfrac{\pi}{12}=\tan\left(\dfrac{\pi}{3}-\dfrac{\pi}{4}\right)\)
\(=\dfrac{\tan\dfrac{\pi}{3}-\tan\dfrac{\pi}{4}}{1+\tan\dfrac{\pi}{3}\tan\dfrac{\pi}{4}}=\dfrac{\sqrt{3}-1}{1+\sqrt{3}}=2-\sqrt{3}\)
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) \(D=\dfrac{1-cos2x+sin2x}{1+cos2x+sin2x}.cotx\)
\(\Leftrightarrow D=\dfrac{1-\left(1-2sin^2x\right)+2sinx.cosx}{1+2cos^2a-1+2sinx.cosx}.cotx\)
\(\Leftrightarrow D=\dfrac{2sin^2x+2sinx.cosx}{2cos^2x+2sinx.cosx}.cotx\)
\(\Leftrightarrow D=\dfrac{2sinx\left(sinx+cosx\right)}{2cosx\left(cosx+sinx\right)}.cotx\)
\(\Leftrightarrow D=\dfrac{sinx}{cosx}.cotx\)
\(\Leftrightarrow D=tanx.cotx\)
\(\Leftrightarrow D=1\)
Lời giải:
Do \(0< a< \frac{\pi}{2}\Rightarrow \sin a>0\)
Ta có:
\(\sqrt{\frac{1+\cos a}{1-\cos a}}-\sqrt{\frac{1-\cos a}{1+\cos a}}=\frac{(1+\cos a)-(1-\cos a)}{\sqrt{(1-\cos a)(1+\cos a)}}\)
\(=\frac{2\cos a}{\sqrt{1-\cos ^2a}}=\frac{2\cos a}{\sqrt{\sin ^2a}}=\frac{2\cos a}{\sin a}\)
\(=2.\frac{\cos a}{\sin a}=2\cot a\)
Ta có đpcm.
rút gọn biểu thức:
E=cos(\(\dfrac{3\pi}{3}-\alpha\))-sin(\(\dfrac{3\pi}{2}-\alpha\))+sin(\(\alpha+4\pi\))
Ta có \(F=sin^2\dfrac{\pi}{6}+...+sin^2\pi=\left(sin^2\dfrac{\pi}{6}+sin^2\dfrac{5\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+sin^2\dfrac{4\pi}{6}\right)+\left(sin^2\dfrac{3\pi}{6}+sin^2\pi\right)=\left(sin^2\dfrac{\pi}{6}+cos^2\dfrac{\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+cos^2\dfrac{2\pi}{6}\right)+\left(1+0\right)=1+1+1=3\)