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\(\pi< a< \dfrac{3\pi}{2}\Rightarrow\left\{{}\begin{matrix}sina< 0\\cosa< 0\end{matrix}\right.\)
\(sin\left(\dfrac{7\pi}{2}+a\right)=sin\left(4\pi-\dfrac{\pi}{2}+a\right)=sin\left(-\dfrac{\pi}{2}+a\right)=-sin\left(\dfrac{\pi}{2}-a\right)=-cosa>0\)
Đáp án A
Ta có \(F=sin^2\dfrac{\pi}{6}+...+sin^2\pi=\left(sin^2\dfrac{\pi}{6}+sin^2\dfrac{5\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+sin^2\dfrac{4\pi}{6}\right)+\left(sin^2\dfrac{3\pi}{6}+sin^2\pi\right)=\left(sin^2\dfrac{\pi}{6}+cos^2\dfrac{\pi}{6}\right)+\left(sin^2\dfrac{2\pi}{6}+cos^2\dfrac{2\pi}{6}\right)+\left(1+0\right)=1+1+1=3\)
a)\(sin^4\dfrac{\pi}{16}+sin^4\dfrac{3\pi}{16}+sin^4\dfrac{5\pi}{16}+sin^4\dfrac{7\pi}{16}\)
\(=\left(sin^4\dfrac{\pi}{16}+sin^4\dfrac{7\pi}{16}\right)+\left(sin^4\dfrac{3\pi}{16}+sin^4\dfrac{5\pi}{16}\right)\)
\(=\left(sin^4\dfrac{\pi}{16}+cos^4\dfrac{\pi}{16}\right)+\left(sin^4\dfrac{3\pi}{16}+cos^4\dfrac{3\pi}{16}\right)\)
\(=1-2sin^2\dfrac{\pi}{16}cos^2\dfrac{\pi}{16}+1-2sin^2\dfrac{3\pi}{16}cos^2\dfrac{3\pi}{16}\)
\(=2-\dfrac{1}{2}sin^2\dfrac{\pi}{8}-\dfrac{1}{2}sin^2\dfrac{3\pi}{8}\)
\(=2-\dfrac{1}{2}\left(sin^2\dfrac{\pi}{8}+sin^2\dfrac{3\pi}{8}\right)\)
\(=2-\dfrac{1}{2}\left(sin^2\dfrac{\pi}{8}+cos^2\dfrac{\pi}{8}\right)\)
\(=2-\dfrac{1}{2}=\dfrac{3}{2}\).
Có: \(cotx-tanx=\dfrac{cosx}{sinx}-\dfrac{sinx}{cosx}=\dfrac{cos^2x-sin^2x}{sinxcosx}=\dfrac{2cos2x}{sin2x}\)
Vì vậy:
\(cot7,5^o+tan67,5^o-tan7,5^o-cot67,5^o\)
\(=\left(cot7,5^o-tan7,5^o\right)-\left(cot67,5^o-tan67,5^o\right)\)
\(=\dfrac{2cos15^o}{sin15^o}-\dfrac{2cos135^o}{sin135^o}\)
\(=2\left(\dfrac{cos15^osin135^o-sin15^ocos135^o}{sin15^osin135^o}\right)\)
\(=2.\dfrac{sin120^o}{\dfrac{1}{2}\left(cos120^o-cos150^o\right)}\)
\(=\dfrac{4.\dfrac{\sqrt{3}}{2}}{\dfrac{-1}{2}+\dfrac{\sqrt{3}}{2}}=\dfrac{4\sqrt{3}}{\sqrt{3}-1}\)
1,\(VT=\dfrac{sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}{cos\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}+\dfrac{cos\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}{sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}\)\(=\dfrac{sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)^2+cos^2\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}{cos\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right).sin\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)}\)
\(=\dfrac{1}{\dfrac{1}{2}.sin\left(\dfrac{\pi}{2}+x\right)}=\dfrac{2}{cosx}=VP\)
2,\(VT=\left(sin^4x-cos^4x\right)\left(sin^4x+cos^4x\right)=\left(sin^2x+cos^2x\right)\left(sin^2x-cos^2x\right)\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)
\(=\left(sin^2-cos^2x\right)\left(1-2sin^2x.cos^2x\right)\)\(=-cos2x\left(1-\dfrac{1}{2}sin^22x\right)\)\(=-\dfrac{cos2x\left(2-sin^22x\right)}{2}=-\dfrac{cos2x\left(1+cos^22x\right)}{2}\)
\(VP=-\left(\dfrac{7}{8}cos2x+\dfrac{1}{8}cos6x\right)=-\dfrac{7}{8}cos2x-\dfrac{1}{8}\left[4cos^32x-3cos2x\right]=-\dfrac{7}{8}.cos2x-\dfrac{1}{2}cos^32x+\dfrac{3}{8}cos2x\)
\(=-\dfrac{1}{2}cos2x-\dfrac{1}{2}cos^32x=\dfrac{-cos2x\left(1+cos^22x\right)}{2}\)
\(\Rightarrow VT=VP\)(đpcm)
3, \(VT=3-4\left(1-2sin^2x\right)+1-2sin^22x=8sin^2x-2sin^22x=8sin^2x-8.sin^2x.cos^2x=8sin^2x\left(1-cos^2x\right)=8sin^4x=VP\)
4,\(VP=\dfrac{1}{2}\left[sin\left(x+\dfrac{\pi}{2}\right)+sin\left(3x+\dfrac{\pi}{6}\right)\right]-\dfrac{1}{2}\left[cos\left(3x-\dfrac{\pi}{3}\right)+cos\left(x+\pi\right)\right]\)
\(=\dfrac{1}{2}\left(cosx+sin3x.\dfrac{\sqrt{3}}{2}+\dfrac{cos3x}{2}\right)-\dfrac{1}{2}\left(\dfrac{cos3x}{2}+sin3x.\dfrac{\sqrt{3}}{2}-cosx\right)\)
\(=\dfrac{1}{2}.2cosx=cosx=VP\)
5, \(VP=4cos\left(2x-\dfrac{\pi}{6}\right).\left(sinx.\dfrac{\sqrt{3}}{2}+\dfrac{cosx}{2}\right)^2\)\(=cos\left(2x-\dfrac{\pi}{6}\right).\left(sinx.\sqrt{3}+cosx\right)^2\)
\(VT=2.cos\left(2x-\dfrac{\pi}{6}\right)+2.sin\left(2x-\dfrac{\pi}{6}\right).cos\left(2x-\dfrac{\pi}{6}\right)=2cos\left(2x-\dfrac{\pi}{6}\right)\left[1+sin\left(2x-\dfrac{\pi}{6}\right)\right]\)
\(=2cos\left(2x-\dfrac{\pi}{6}\right)\left(1+\dfrac{sin2x.\sqrt{3}}{2}-\dfrac{cos2x}{2}\right)\)\(=2cos\left(2x-\dfrac{\pi}{6}\right)\left(sin^2x+cos^2x+sinx.cosx.\sqrt{3}-\dfrac{cos^2x-sin^2x}{2}\right)\)
\(=2cos\left(2x-\dfrac{\pi}{6}\right)\left(sin^2x.\dfrac{3}{2}+sinx.cosx.\sqrt{3}+\dfrac{cos^2x}{2}\right)\)\(=cos\left(2x-\dfrac{\pi}{6}\right)\left(sin^2x.3+2sinx.cosx.\sqrt{3}+cos^2x\right)\)
\(=cos\left(2x-\dfrac{\pi}{6}\right)\left(sinx.\sqrt{3}+cosx\right)^2\)
\(\Rightarrow VT=VP\) (dpcm)
