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\(A+B+C=180^0\Rightarrow\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=90^0\Rightarrow\frac{A}{2}+\frac{B}{2}=90^0-\frac{C}{2}\)
\(\Rightarrow tan\left(\frac{A}{2}+\frac{B}{2}\right)=tan\left(90^0-\frac{C}{2}\right)\)
\(\Leftrightarrow\frac{tan\frac{A}{2}+tan\frac{B}{2}}{1-tan\frac{A}{2}.tan\frac{B}{2}}=cot\frac{C}{2}=\frac{1}{tan\frac{C}{2}}\)
\(\Leftrightarrow tan\frac{C}{2}\left(tan\frac{A}{2}+tan\frac{B}{2}\right)=1-tan\frac{A}{2}.tan\frac{B}{2}\)
\(\Leftrightarrow tan\frac{A}{2}tan\frac{C}{2}+tan\frac{B}{2}tan\frac{C}{2}+tan\frac{A}{2}.tan\frac{B}{2}=1\)
b/\(A+B+C=180^0\Rightarrow A+B=180^0-C\)
\(\Rightarrow cot\left(A+B\right)=cot\left(180^0-C\right)\)
\(\Leftrightarrow\frac{cotA.cotB-1}{cotA+cotB}=-cotC\)
\(\Leftrightarrow cotA.cotB-1=-cotA.cotC-cotB.cotC\)
\(\Leftrightarrow cotA.cotB+cotB.cotC+cotA.cotC=1\)
a/ \(\frac{A}{2}+\left(\frac{B}{2}+\frac{C}{2}\right)=90^0\)
\(\Rightarrow sin\frac{A}{2}=cos\left(\frac{B}{2}+\frac{C}{2}\right)=cos\frac{B}{2}cos\frac{C}{2}-sin\frac{B}{2}.sin\frac{C}{2}\)
b/ \(\frac{tan^2A-tan^2B}{1-tan^2A.tan^2B}=\frac{\left(tanA-tanB\right)}{\left(1+tanA.tanB\right)}.\frac{\left(tanA+tanB\right)}{\left(1-tanA.tanB\right)}=tan\left(A-B\right).tan\left(A+B\right)\)
\(=tan\left(A-B\right).tan\left(180^0-C\right)=-tan\left(A-B\right).tanC\)
c/
\(A+B+C=180^0\Rightarrow cot\left(A+B\right)=-cotC\)
\(\Leftrightarrow\frac{cotA.cotB-1}{cotA+cotB}=-cotC\)
\(\Leftrightarrow cotA.cotB-1=-cotA.cotC-cotB.cotC\)
\(\Leftrightarrow cotA.cotB+cotB.cotC+cotA.cotC=1\)
f/
\(sin2A+sin2B+sin2C=2sin\left(A+B\right).cos\left(A-B\right)+2sinC.cosC\)
\(=2sinC.cos\left(A-B\right)+2sinC.cosC\)
\(=2sinC\left(cos\left(A-B\right)+cosC\right)\)
\(=2sinC\left[cos\left(A-B\right)-cos\left(A+B\right)\right]\)
\(=4sinC.sinA.sinB\)
g/
\(cos^2A+cos^2B+cos^2C=\frac{1}{2}+\frac{1}{2}cos2A+\frac{1}{2}+\frac{1}{2}cos2B+cos^2C\)
\(=1+\frac{1}{2}\left(cos2A+cos2B\right)+cos^2C\)
\(=1+cos\left(A+B\right).cos\left(A-B\right)+cos^2C\)
\(=1-cosC.cos\left(A-B\right)+cos^2C\)
\(=1-cosC\left(cos\left(A-B\right)-cosC\right)\)
\(=1-cosC\left[cos\left(A-B\right)+cos\left(A+B\right)\right]\)
\(=1-2cosC.cosA.cosB\)
d/ \(sinA+sinB+sinC=2sin\frac{A+B}{2}cos\frac{A-B}{2}+2sin\frac{C}{2}.cos\frac{C}{2}\)
\(=2cos\frac{C}{2}.cos\frac{A-B}{2}+2sin\frac{C}{2}.cos\frac{C}{2}\)
\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+sin\frac{C}{2}\right)\)
\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+cos\frac{A+B}{2}\right)\)
\(=4cos\frac{C}{2}.cos\frac{A}{2}.cos\frac{B}{2}\)
e/
\(cosA+cosB+cosC=2cos\frac{A+B}{2}cos\frac{A-B}{2}+1-2sin^2\frac{C}{2}\)
\(=1+2sin\frac{C}{2}.cos\frac{A-B}{2}-2sin^2\frac{C}{2}\)
\(=1+2sin\frac{C}{2}\left(cos\frac{A-B}{2}-sin\frac{C}{2}\right)\)
\(=1+2sin\frac{C}{2}\left(cos\frac{A-B}{2}-cos\frac{A+B}{2}\right)\)
\(=1+4sin\frac{C}{2}.sin\frac{A}{2}sin\frac{B}{2}\)
Xét tam giác ABC, ta có:
\(\widehat A + \widehat B + \widehat C = {180^o} \Rightarrow \frac{{\widehat A}}{2} + \frac{{\widehat B + \widehat C}}{2} = {90^o}\)
Do đó \(\frac{{\widehat A}}{2}\) và \(\frac{{\widehat B + \widehat C}}{2}\) là hai góc phụ nhau.
a) Ta có: \(\sin \frac{A}{2} = \cos \left( {{{90}^o} - \frac{A}{2}} \right) = \cos \frac{{B + C}}{2}\)
b) Ta có: \(\tan \frac{{B + C}}{2} = \cot \left( {{{90}^o} - \frac{{B + C}}{2}} \right) = \cot \frac{A}{2}\)