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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/
\(cosA+cosB+cosC=2cos\frac{A+B}{2}cos\frac{A-B}{2}+1-2sin^2\frac{C}{2}\)
\(=1+2sin\frac{C}{2}.cos\frac{A-B}{2}-2sin^2\frac{C}{2}\)
\(=1+2sin\frac{C}{2}\left(cos\frac{A-B}{2}-sin\frac{C}{2}\right)\)
\(=1+2sin\frac{C}{2}\left(cos\frac{A-B}{2}-cos\frac{A+B}{2}\right)\)
\(=1+4sin\frac{C}{2}.sin\frac{A}{2}sin\frac{B}{2}\)
a, \(\left(1-sin^2x\right)cot^2x+1-cot^2x\)
\(=cot^2x-sin^2x.cot^2x+1-cot^2x\)
\(=1-sin^2x.\frac{\text{cos}^2x}{sin^2x}=1-\text{cos}^2x=sin^2x\)
b,\(\left(tanx+cotx\right)^2-\left(tanx-cotx\right)2\)
\(=tan^2x2.tanx.cotx+cot^2x-tan^2x+2tanx.cotx-cot^2x\)
\(=4tanxcotx=4\)
c,\(\left(xsina-y\text{cos}a\right)^2+\left(x\text{cos}a+ysina\right)^2\)
\(=x^2sin^2a=2xysina\text{cos}a+y^2\text{cos}^2a+2xysina\text{cos}a+y^2sin^2a\)
\(=x^2\left(sin^2a+\text{cos}^2a\right)+y^2\left(sin^2a+\text{cos}^2a\right)\)
\(=x^2+y^2\)
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/ \(cot^2x-cos^2x=\frac{cos^2x}{sin^2x}-cos^2x=cos^2x\left(\frac{1}{sin^2x}-1\right)=cos^2x\left(\frac{1-sin^2x}{sin^2x}\right)\)
\(=cos^2x.\frac{cos^2x}{sin^2x}=cos^2x.cot^2x\)
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: \(\tan \frac{{B + C}}{2} = \cot \left( {{{90}^o} - \frac{{B + C}}{2}} \right) = \cot \frac{A}{2}\)
\(a)sin^4x+cos^4x=1-2sin^2x\cdot cos^2x\)
\(\Leftrightarrow sin^4x+2sin^2x\cdot cos^2x+cos^4x=1\)
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2=1\)(luôn đúng)
a)
\(\sin ^4a-\cos ^4a+1=(\sin ^2a-\cos ^2a)(\sin ^2a+\cos^2a)+1\)
\(=(\sin ^2a-\cos ^2a).1+1=\sin ^2a-\cos ^2a+\sin ^2a+\cos ^2a\)
\(=2\sin ^2a\)
b) \(\sin ^2a+2\cos ^2a-1=(\sin ^2a+\cos^2a)+\cos ^2a-1\)
\(=1+\cos ^2a-1=\cos ^2a\)
\(\Rightarrow \frac{\sin ^2a+2\cos ^2a-1}{\cot ^2a}=\frac{\cos ^2a}{\cot ^2a}=\frac{\cos ^2a}{\frac{\cos ^2a}{\sin ^2a}}=\sin ^2a\)
c)
\(\frac{1-\sin ^2a\cos ^2a}{\cos ^2a}-\cos ^2a=\frac{1}{\cos ^2a}-\sin ^2a-\cos ^2a\)
\(=\frac{1}{\cos ^2a}-(\sin ^2a+\cos ^2a)=\frac{1}{\cos ^2a}-1\)
\(=\frac{1-\cos ^2a}{\cos ^2a}=\frac{\sin ^2a}{\cos ^2a}=\tan ^2a\)
d)
\(\frac{\sin ^2a-\tan ^2a}{\cos ^2a-\cot ^2a}=\frac{\sin ^2a-\frac{\sin ^2a}{\cos ^2a}}{\cos ^2a-\frac{\cos ^2a}{\sin ^2a}}\) \(=\frac{\sin ^2a(1-\frac{1}{\cos ^2a})}{\cos ^2a(1-\frac{1}{\sin ^2a})}\)
\(=\frac{\sin ^2a.\frac{\cos ^2a-1}{\cos ^2a}}{\cos ^2a.\frac{\sin ^2a-1}{\sin ^2a}}\) \(=\frac{\sin ^2a.\frac{-\sin ^2a}{\cos ^2a}}{\cos ^2a.\frac{-\cos ^2a}{\sin ^2a}}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)
f)
\(\frac{(\sin a+\cos a)^2-1}{\cot a-\sin a\cos a}=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\frac{\cos a}{\sin a}-\sin a\cos a}\)
\(=\sin a.\frac{1+2\sin a\cos a-1}{\cos a-\cos a\sin ^2a}\)
\(=\sin a. \frac{2\sin a\cos a}{\cos a(1-\sin ^2a)}=\sin a. \frac{2\sin a\cos a}{\cos a. \cos^2 a}=\frac{2\sin ^2a}{\cos ^2a}=2\tan ^2a\)
\(\frac{sinA}{cosA}+\frac{sinB}{cosB}=\frac{2cos\frac{C}{2}}{sin\frac{C}{2}}\Leftrightarrow\frac{sinA.cosB+cosA.sinB}{cosA.cosB}=\frac{2sin\frac{C}{2}.cos\frac{C}{2}}{sin^2\frac{C}{2}}\)
\(\Leftrightarrow\frac{sin\left(A+B\right)}{cosA.cosB}=\frac{2sinC}{1-cosC}\Leftrightarrow\frac{sinC}{cosA.cosB}=\frac{2sinC}{1-cosC}\)
\(\Leftrightarrow1-cosC=2cosA.cosB=cos\left(A+B\right)+cos\left(A-B\right)\)
\(\Leftrightarrow1-cosC=-cosC+cos\left(A-B\right)\)
\(\Leftrightarrow cos\left(A-B\right)=1\Rightarrow A-B=0\Rightarrow A=B\)
\(\Rightarrow\) Tam giác ABC cân tại C
\(\frac{cos^2A+cos^2B}{sin^2A+sin^2B}=\frac{1}{2}\left(cot^2A+cot^2B\right)\)
\(\Leftrightarrow2cos^2A+2cos^2B=\left(sin^2A+sin^2B\right)\left(cot^2A+cot^2B\right)\)
\(\Leftrightarrow2cos^2A+2cos^2B=cos^2A+cos^2B+sin^2A.cot^2B+sin^2B.cot^2A\)
\(\Leftrightarrow cos^2A+cos^2B=\frac{sin^2A.cos^2B}{sin^2B}+\frac{sin^2B.cos^2A}{sin^2A}\)
\(\Leftrightarrow cos^2A\left(\frac{sin^2B}{sin^2A}-1\right)=cos^2B\left(1-\frac{sin^2A}{sin^2B}\right)\)
\(\Leftrightarrow\frac{cos^2A\left(sin^2B-sin^2A\right)}{sin^2A}=\frac{cos^2B\left(sin^2B-sin^2A\right)}{sin^2B}\)
\(\Leftrightarrow cot^2A\left(sin^2B-sin^2A\right)=cot^2B\left(sin^2B-sin^2A\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}sin^2B=sin^2A\\cot^2A=cot^2B\end{matrix}\right.\) \(\Rightarrow A=B\)
Chọn C.
Ta có :
+ tanA + tanB + tanC = tanA.tanB.tanC ⇔ -tanA(1 – tanBtanC) = tanB + tanC
+ cotA + cotB + cotC = cotA.cotB.cotC ⇔ cotA(cotBcotC – 1) = cotB + cotC