Chứng minh:
x.tan34o = (350−x)tan40o
\(\Leftrightarrow\)x≈\(\frac{350\tan40^o}{\tan34^o+\tan40^o}\)
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Biểu thức này chỉ rút gọn được khi mẫu là \(1-2sin^210^0\)
a/ \(\tan40.\cot40+\frac{\sin50}{\cos40}\)
\(=1+\frac{\cos40}{\cos40}=1+1=2\)
\(A=\frac{sinx}{cosx}+\frac{cosx}{sinx}+\frac{sin3x}{cos3x}+\frac{cos3x}{sin3x}\)
\(=\frac{sin^2x+cos^2x}{sinx.cosx}+\frac{sin^23x+cos^23x}{sin3x.cos3x}=\frac{2}{2sinx.cosx}+\frac{2}{2sin3x.cos3x}\)
\(=\frac{2}{sin2x}+\frac{2}{sin6x}=\frac{2\left(sin2x+sin6x\right)}{sin2x.sin6x}=\frac{4sin4x.cos2x}{sin2x.sin6x}\)
\(=\frac{8sin2x.cos^22x}{sin2x.sin6x}=\frac{8cos^22x}{sin6x}\)
\(B=\frac{sin30}{cos30}+\frac{sin60}{cos60}+\frac{sin40}{cos40}+\frac{sin50}{cos50}=\frac{sin30.cos60+cos30.sin60}{cos30.cos60}+\frac{sin40.cos50+sin50.cos40}{cos40.cos50}\)
\(=\frac{sin90}{cos30.cos60}+\frac{sin90}{cos40.cos50}=\frac{1}{\frac{1}{2}.\frac{\sqrt{3}}{2}}+\frac{1}{\frac{1}{2}cos90+\frac{1}{2}cos10}\)
\(=\frac{4\sqrt{3}}{3}+\frac{2}{cos10}=\frac{4\sqrt{3}\left(cos10+\frac{\sqrt{3}}{2}\right)}{3cos10}=\frac{4\sqrt{3}\left(cos10+cos30\right)}{3cos10}\)
\(=\frac{8\sqrt{3}cos20.cos10}{3cos10}=\frac{8\sqrt{3}}{3}cos20\)
c/
\(\Leftrightarrow tan\left(60^0-x\right)=-\frac{1}{\sqrt{3}}\)
\(\Rightarrow60^0-x=-30^0+k180^0\)
\(\Rightarrow x=90^0+k180^0\)
d/
\(\Leftrightarrow tan\left(3x+\frac{2\pi}{5}\right)=-tan\left(\frac{\pi}{5}\right)\)
\(\Leftrightarrow tan\left(3x+\frac{2\pi}{5}\right)=tan\left(-\frac{\pi}{5}\right)\)
\(\Rightarrow3x+\frac{2\pi}{5}=-\frac{\pi}{5}+k\pi\)
\(\Rightarrow x=-\frac{\pi}{5}+\frac{k\pi}{3}\)
a/
\(\Leftrightarrow tan2x=-tan40^0\)
\(\Leftrightarrow tan2x=tan\left(-40^0\right)\)
\(\Rightarrow2x=-40^0+k180^0\)
\(\Rightarrow x=-20^0+k90^0\)
b/
\(\Leftrightarrow tan\left(2x-15^0\right)=1\)
\(\Rightarrow2x-15^0=45^0+k180^0\)
\(\Rightarrow x=30^0+k90^0\)
\(\frac{sin20}{cos20}+\frac{sin40}{cos40}+\frac{\sqrt{3}sin20.sin40}{cos20.cos40}=\frac{sin20cos40+cos40sin20}{cos20cos40}+\frac{-\frac{\sqrt{3}}{2}\left(cos60-cos20\right)}{cos20cos40}\)
\(=\frac{sin60}{cos20cos40}-\frac{\frac{\sqrt{3}}{2}\left(\frac{1}{2}-cos20\right)}{cos20cos40}=\frac{\sqrt{3}}{2}\left(\frac{1-\frac{1}{2}+cos20}{cos20cos40}\right)=\frac{\sqrt{3}}{2}\left(\frac{\frac{1}{2}+cos20}{\frac{1}{2}\left(cos60+cos20\right)}\right)\)
\(=\sqrt{3}\left(\frac{\frac{1}{2}+cos20}{\frac{1}{2}+cos20}\right)=\sqrt{3}\)
\(=\left(1+tan^220\right).cos^220-tan40.cot\left(90-50\right)\)
\(=\left(1+\frac{sin^220}{cos^220}\right).cos^220-tan40.cot40\)
\(=cos^220+sin^220-1\)
\(=1-1=0\)
a) Ta có: \(sin\alpha=cos\left(90-\alpha\right)\Rightarrow sin42=cos48\)
\(\Rightarrow sin42-cos48=0\)
b) Ta có: \(sin\alpha=cos\left(90-\alpha\right)\Rightarrow sin61=cos29\Rightarrow sin^261=cos^229\)
\(\Rightarrow sin^261+sin^229=sin^229+cos^229=1\)
c) Ta có: \(tan\alpha=\dfrac{1}{tan\left(90-\alpha\right)}\Rightarrow tan40=\dfrac{1}{tan50}\)
\(\Rightarrow tan40.tan50=1\) mà \(tan45=1\Rightarrow tan40.tan45.tan50=1\)
\(sin42^0-cos48^0=sin42^0-sin\left(90^0-48^0\right)=sin42^0-sin42^0=0\)
\(sin^261^0+sin^229^0=sin^261^0+cos^2\left(90^0-29^0\right)=sin^261^0+cos^261^0=1\)
\(tan40^0.tan50^0.tan45^0=tan40^0.cot\left(90^0-50^0\right).1=tan40^0.cot40^0=1\)
Sử dụng các công thức:
\(cosa=sin\left(90^0-a\right)\) ; \(sina=cos\left(90^0-a\right)\) ; \(tana=cot\left(90^0-a\right)\) ; \(tana.cota=1\)