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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/
\(cosA+cosB+cosC=2cos\frac{A+B}{2}cos\frac{A-B}{2}+1-2sin^2\frac{C}{2}\)
\(=1+2sin\frac{C}{2}.cos\frac{A-B}{2}-2sin^2\frac{C}{2}\)
\(=1+2sin\frac{C}{2}\left(cos\frac{A-B}{2}-sin\frac{C}{2}\right)\)
\(=1+2sin\frac{C}{2}\left(cos\frac{A-B}{2}-cos\frac{A+B}{2}\right)\)
\(=1+4sin\frac{C}{2}.sin\frac{A}{2}sin\frac{B}{2}\)
\(\frac{cos\left(a-b\right)}{sin\left(a+b\right)}=\frac{cosa.cosb+sina.sinb}{sina.cosb+cosa.sinb}=\frac{\frac{cosa.cosb}{sina.sinb}+1}{\frac{sina.cosb}{sina.sinb}+\frac{cosa.sinb}{sina.sinb}}=\frac{cota.cotb+1}{cota+cotb}\)
Bạn ghi đề ko đúng
\(sin\left(a+b\right)sin\left(a-b\right)=\frac{1}{2}\left[cos2b-cos2a\right]\)
\(=\frac{1}{2}\left[1-2sin^2b-1+2sin^2a\right]\)
\(=sin^2a-sin^2b\)
\(=1-cos^2a-1+cos^2b=cos^2b-cos^2a\)
Câu này bạn cũng ghi đề ko đúng
\(cos\left(a+b\right)cos\left(a-b\right)=\frac{1}{2}\left[cos2a+cos2b\right]\)
\(=\frac{1}{2}\left[2cos^2a-1+1-2sin^2b\right]=cos^2a-sin^2b\)
\(=1-sin^2a-1+cos^2b=cos^2b-sin^2a\)
\(\frac{sin2a-2sina}{sin2a+2sina}=\frac{2sina.cosa-2sina}{2sina.cosa+2sina}=\frac{2sina\left(cosa-1\right)}{2sina\left(cosa+1\right)}=\frac{cosa-1}{cosa+1}\)
\(=\frac{1-2sin^2\frac{a}{2}-1}{2cos^2\frac{a}{2}-1+1}=\frac{-sin^2\frac{a}{2}}{cos^2\frac{a}{2}}=-tan^2\frac{a}{2}\)
\(\frac{sin^4x-sin^2x+cos^2x}{cos^4x-cos^2x+sin^2x}=\frac{sin^2x\left(sin^2x-1\right)+cos^2x}{cos^2x\left(cos^2x-1\right)+sin^2x}=\frac{-sin^2x.cos^2x+cos^2x}{-cos^2x.sin^2x+sin^2x}\)
\(=\frac{cos^2x\left(1-sin^2x\right)}{sin^2x\left(1-cos^2x\right)}=\frac{cos^4x}{sin^4x}=cot^4x\)
\(\frac{sin^3a-cos^3a}{sina-cosa}=\frac{\left(sina-cosa\right)\left[sin^2a+cos^2a+sina.cosa\right]}{sina-cosa}=1+sina.cosa=1+\frac{1}{2}sin2a\)
@Akai Haruma @Nguyễn Việt Lâm @Nguyễn Việt Lâm @Lightning Farron giúp em
\(A+B+C=180^0\Rightarrow\frac{A+B}{2}+\frac{C}{2}=90^0\)
\(\Rightarrow sin\left(\frac{A+B}{2}\right)=cos\left(90^0-\frac{A+B}{2}\right)=cos\frac{C}{2}\)
\(cos\left(A+B\right)=-cos\left(180^0-\left(A+B\right)\right)=-cosC\)
\(cos\left(\frac{A+B}{2}\right)=sin\left(90-\frac{A+B}{2}\right)=sin\frac{C}{2}\)
\(sinA=sin\left(180^0-A\right)=sin\left(B+C\right)\)
\(sin\left(A+B\right)=sin\left(180^0-\left(A+B\right)\right)=sinC\)
\(cosA=-cos\left(180^0-A\right)=-cos\left(B+C\right)\)
1) \(sin\left(A+2B+C\right)=sin\left(\pi-B+2B\right)\)
=\(sin\left(\pi+B\right)=sin\left(-B\right)=-sinB\)
2) \(sinBsinC-cosBcosC=-cos\left(B+C\right)\)
\(=-cos\left(\pi-A\right)=cosA\)
4) bạn ơi +2 vào vế phải mới đúng nhé
2+ \(2cosAcosBcosC=\left[cos\left(A+B\right)+cos\left(A-B\right)\right]cosC+2\)
\(=cos\left(\pi-C\right)cosC+cos\left(A-B\right)cos\left(\pi-\left(A+B\right)\right)+2\)
=\(-cos^2C-cos\left(A-B\right)cos\left(A+B\right)+2\)
\(=-cos^2C-\frac{1}{2}\left(cos2A+cos2B\right)+2\)
\(=-cos^2C-\frac{1}{2}\left(2cos^2A-1\right)-\frac{1}{2}\left(2cos^2B-1\right)+2\)
\(=-cos^2C-cos^2A+\frac{1}{2}-cos^2C+\frac{1}{2}+2\)
= sin2C - 1 + sin2A - 1 + sin2C - 1 + 3
= sin2A + sin2B + sin2C