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Ai giúp em 3 câu này với ạ e cảm ơn rất nhìu ạ

Nguyễn Hoàng Minh · Accepted Answer

$26,\
a,\sin45^0=\cos45^0< \sin50^025'< \sin57^048'=\cos32^012'< \sin72^0=\cos18^0< \sin75^0\
b,\tan37^026'< \tan47^0< \tan58^0=\cot32^0< \tan63^0< \tan66^019'=\cot23^041'\
27,\
A=\dfrac{\left(\sin^226^0+\sin^264^0\right)+2\left(\cos^215^0+\cos^275^0\right)}{\left(\sin^255^0+\cos^255^0\right)+\left(\sin^242^0+\cos^242^0\right)}-\dfrac{\tan81^0}{2\tan81^0}\
A=\dfrac{\left(\sin^226^0+\cos^226^0\right)+2\left(\sin^215^0+\cos^215^0\right)}{1+1}-\dfrac{1}{2}\
A=\dfrac{1+2}{2}-\dfrac{1}{2}=2-\dfrac{1}{2}=\dfrac{3}{2}$$28,\
\sin^2\alpha=1-\cos^2\alpha=1-\dfrac{1}{2}=\dfrac{1}{2}\
\Leftrightarrow\sin\alpha=\dfrac{\sqrt{2}}{2}$Câu trả lời: $26,\
a,\sin45^0=\cos45^0< \sin50^025'< \sin57^048'=\cos32^012'< \sin72^0=\cos18^0< \sin75^0\
b,\tan37^026'< \tan47^0< \tan58^0=\cot32^0< \tan63^0< \tan66^019'=\cot23^041'\
27,\
A=\dfrac{\left(\sin^226^0+\sin^264^0\right)+2\left(\cos^215^0+\cos^275^0\right)}{\left(\sin^255^0+\cos^255^0\right)+\left(\sin^242^0+\cos^242^0\right)}-\dfrac{\tan81^0}{2\tan81^0}\
A=\dfrac{\left(\sin^226^0+\cos^226^0\right)+2\left(\sin^215^0+\cos^215^0\right)}{1+1}-\dfrac{1}{2}\
A=\dfrac{1+2}{2}-\dfrac{1}{2}=2-\dfrac{1}{2}=\dfrac{3}{2}$$28,\
\sin^2\alpha=1-\cos^2\alpha=1-\dfrac{1}{2}=\dfrac{1}{2}\
\Leftrightarrow\sin\alpha=\dfrac{\sqrt{2}}{2}$

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