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Câu 1:
\(a.sin\left(B-C\right)=a.sinBcosC-a.cosB.sinC\)
\(bsin\left(C-A\right)=bsinC.cosA-bcosC.sinA\)
\(csin\left(A-B\right)=csinAcosB-csinB.cosA\)
Cộng lại:
\(VT=cosA\left(bsinC-c.sinB\right)+cosB\left(c.sinA-a.sinC\right)+cosC\left(a.sinB-bsinA\right)\)
\(=cosA\left(\frac{b.c}{2R}-\frac{bc}{2R}\right)+cosB\left(\frac{ac}{2R}-\frac{ac}{2R}\right)+cosC\left(\frac{ab}{2R}-\frac{ab}{2R}\right)=0\)
Câu 2:
\(sin^2A+sin^2B+sin^2C=\frac{1}{2}-\frac{1}{2}cos2A+\frac{1}{2}-\frac{1}{2}cos2B+1-cos^2C\)
\(=2-\frac{1}{2}\left(cos2A+cos2B\right)-cosC.cosC\)
\(=2-cos\left(A+B\right)cos\left(A-B\right)+cosC.cos\left(A+B\right)\)
\(=2+cosC.cos\left(A-B\right)+cosC.cos\left(A+B\right)\)
\(=2+cosC\left[cos\left(A-B\right)+cos\left(A+B\right)\right]\)
\(=2+2cosA.cosB.cosC\)
Câu 3:
Ta có \(sin^2\frac{A}{2}=\frac{1-cosA}{2}=\frac{1-\frac{b^2+c^2-a^2}{2bc}}{2}=\frac{a^2-b^2-c^2+2bc}{4bc}=\frac{a^2-\left(b-c\right)^2}{4bc}\)
\(=\frac{\left(a+b-c\right)\left(a+c-b\right)}{4bc}=\frac{\left(p-c\right)\left(p-b\right)}{bc}\Rightarrow sin\frac{A}{2}=\sqrt{\frac{\left(p-b\right)\left(p-c\right)}{bc}}\)
Tương tự ta có \(sin\frac{B}{2}=\sqrt{\frac{\left(p-a\right)\left(p-c\right)}{ac}}\) ; \(sin\frac{C}{2}=\sqrt{\frac{\left(p-a\right)\left(p-b\right)}{ab}}\)
\(\Rightarrow4Rsin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}=4\left(\frac{abc}{4S}\right)\sqrt{\frac{\left(p-a\right)^2\left(p-b\right)^2\left(p-c\right)^2}{a^2b^2c^2}}\)
\(=\frac{abc.\left(p-a\right)\left(p-b\right)\left(p-c\right)}{S.abc}=\frac{\left(p-a\right)\left(p-b\right)\left(p-c\right)}{S}=\frac{\left(p-a\right)\left(p-b\right)\left(p-c\right)}{\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}}=\sqrt{\frac{\left(p-a\right)\left(p-b\right)\left(p-c\right)}{p}}=r\)
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
\(2\left[\left(sinx+cosx+1\right)\left(sinx+cosx-1\right)\right]^2\)
\(=2\left[\left(sinx+cosx\right)^2-1\right]^2=2\left(sin^2x+cos^2x+2sinx.cosx-1\right)^2\)
\(=2\left(2sinx.cosx\right)^2=2sin^22x=1-cos4x\)
b/ \(\frac{3-4cos2a+2cos^22a-1}{3+4cos2a+2cos^22a-1}=\frac{2\left(cos^22a-2cos2a+1\right)}{2\left(cos^22a+2cos2a+1\right)}=\frac{\left(cos2a-1\right)^2}{\left(cos2a+1\right)^2}\)
\(\frac{\left(1-2sin^2a-1\right)^2}{\left(2cos^2a-1+1\right)^2}=\frac{4sin^4a}{4cos^4a}=tan^4a\)
c/ \(cos^22x+sin^22x-2sin2x.cos2x+2sin3x.cosx-2sinx.cosx-sin^2x\)
\(=1-sin4x+sin4x+sin2x-sin2x-sin^2x\)
\(=1-sin^2x=cos^2x\)
\(\sin^4x.\sin^2x+\cos^4x.\cos^2x-\left(\sin^4x+\cos^4x+\dfrac{1}{2}\sin^4x+\dfrac{1}{2}\cos^4x-\dfrac{3}{2}\right)-1=-\sin^4x.\left(1-\sin^2x\right)-cos^4x.\left(1-\cos^2x\right)-\dfrac{1}{2}\left(\sin^4x+\cos^4x\right)+\dfrac{1}{2}=-\left(\sin^4x.\cos^2x+\cos^4x.\sin^2x\right)-\dfrac{1}{2}\left(\left(\sin^2x+\cos^2x\right)^2-2\sin^2x.\cos^2x\right)+\dfrac{1}{2}=-\left(\sin^2x.\cos^2x.\left(\sin^2x+\cos^2x\right)\right)-\dfrac{1}{2}.\left(1-2\sin^2x.\cos^2x\right)+\dfrac{1}{2}=-\sin^2x.\cos^2x+\sin^2x.\cos^2x-\dfrac{1}{2}+\dfrac{1}{2}=0\)
Lời giải:
a)
\(\frac{1-\cos x}{\sin x}=\frac{(1-\cos x)(1+\cos x)}{\sin x(1+\cos x)}=\frac{1-\cos ^2x}{\sin x(1+\cos x)}=\frac{\sin ^2x}{\sin x(1+\cos x)}=\frac{\sin x}{1+\cos x}\)
b)
\((\sin x+\cos x-1)(\sin x+\cos x+1)=(\sin x+\cos x)^2-1^2\)
\(=\sin ^2x+\cos ^2x+2\sin x\cos x-1=1+2\sin x\cos x-1=2\sin x\cos x\)
c)
\(\frac{\sin ^2x+2\cos x-1}{2+\cos x-\cos ^2x}=\frac{1-\cos ^2x+2\cos x-1}{2+\cos x-\cos ^2x}=\frac{-\cos ^2x+2\cos x}{2+\cos x-\cos ^2x}\)
\(=\frac{\cos x(2-\cos x)}{(2-\cos x)(\cos x+1)}=\frac{\cos x}{\cos x+1}\)
d)
\(\frac{\cos ^2x-\sin ^2x}{\cot ^2x-\tan ^2x}=\frac{\cos ^2x-\sin ^2x}{\frac{\cos ^2x}{\sin ^2x}-\frac{\sin ^2x}{\cos ^2x}}=\frac{\sin ^2x\cos ^2x(\cos ^2x-\sin ^2x)}{\cos ^4x-\sin ^4x}\)
\(=\frac{\sin ^2x\cos ^2x(\cos ^2x-\sin ^2x)}{(\cos ^2x-\sin ^2x)(\cos ^2x+\sin ^2x)}=\frac{\sin ^2x\cos ^2x}{\sin ^2x+\cos ^2x}=\sin ^2x\cos ^2x\)
e)
\(1-\cot ^4x=1-\frac{\cos ^4x}{\sin ^4x}=\frac{\sin ^4x-\cos ^4x}{\sin ^4x}=\frac{(\sin ^2x-\cos ^2x)(\sin ^2x+\cos ^2x)}{\sin ^4x}\)
\(=\frac{\sin ^2x-\cos ^2x}{\sin ^4x}=\frac{\sin ^2x-(1-\sin ^2x)}{\sin ^4x}=\frac{2\sin ^2x-1}{\sin ^4x}=\frac{2}{\sin ^2x}-\frac{1}{\sin ^4x}\)
Ta có ddpcm.
Giả sử các biểu thức đều xác định
a/
\(sinx.cotx+cosx.tanx=sinx.\frac{cosx}{sinx}+cosx.\frac{sinx}{cosx}=sinx+cosx\)
b/
\(\left(1+cosx\right)\left(sin^2x+cos^2x-cosx\right)=\left(1+cosx\right)\left(1-cosx\right)=1-cos^2x=sin^2x\)
c/
\(\frac{sinx+cosx}{cos^3x}=\frac{1}{cos^2x}\left(\frac{sinx+cosx}{cosx}\right)=\left(1+tan^2x\right)\left(tanx+1\right)=tan^3x+tan^2x+tanx+1\)
d/
\(tan^2x-sin^2x=\frac{sin^2x}{cos^2x}-sin^2x=sin^2x\left(\frac{1}{cos^2x}-1\right)\)
\(=sin^2x\left(\frac{1-cos^2x}{cos^2x}\right)=sin^2x.\frac{sin^2x}{cos^2x}=sin^2x.tan^2x\)
e/ \(cot^2x-cos^2x=\frac{cos^2x}{sin^2x}-cos^2x=cos^2x\left(\frac{1}{sin^2x}-1\right)=cos^2x\left(\frac{1-sin^2x}{sin^2x}\right)\)
\(=cos^2x.\frac{cos^2x}{sin^2x}=cos^2x.cot^2x\)