Trong tam giác ABC.Chứng minh rằng:
\(\frac{b^2-c^2}{c\text{os}B+c\text{os}C}\)+\(\frac{c^2-a^2}{c\text{os}C+c\text{os}A}\)+\(\frac{a^2-b^2}{c\text{os}A+c\text{os}B}\)=0
Những câu hỏi liên quan
chứng minh rằng:
a)\(\frac{c\text{os}a.cot\text{a}-sin\text{a}.t\text{ana}}{\frac{1}{sin\text{a}}-\frac{1}{c\text{os}a}}=1+sin\text{a}.c\text{os}a\)
b)\(\frac{c\text{os}a+sin\text{a}-1}{c\text{os}a-sin\text{a}+1}=\frac{sin\text{a}}{1+c\text{os}a}\)
c)\(\frac{sin\text{a}}{1+c\text{os}a}+\frac{1+c\text{os}a}{sin\text{a}}=\frac{2}{sin\text{a}}\)
chứng minh rằng
a)
frac{sintext{a}}{1+ctext{os}a}+cottext{a}frac{1}{sintext{a}}
b)frac{1}{ctext{os}a}-frac{ctext{os}a}{1+sintext{a}}ttext{ana}
c) frac{ttext{ana}-sintext{a}}{sin^3a}frac{1}{ctext{os}aleft(1+ctext{os}aright)}
d) frac{sintext{a}+ctext{os}a-1}{sintext{a}-ctext{os}a+1}frac{ctext{os}a}{1+sintext{a}}
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chứng minh rằng
a)
\(\frac{sin\text{a}}{1+c\text{os}a}+cot\text{a}=\frac{1}{sin\text{a}}\)
b)\(\frac{1}{c\text{os}a}-\frac{c\text{os}a}{1+sin\text{a}}=t\text{ana}\)
c) \(\frac{t\text{ana}-sin\text{a}}{sin^3a}=\frac{1}{c\text{os}a\left(1+c\text{os}a\right)}\)
d) \(\frac{sin\text{a}+c\text{os}a-1}{sin\text{a}-c\text{os}a+1}=\frac{c\text{os}a}{1+sin\text{a}}\)
Lời giải:
a)
\(\frac{\sin a}{1+\cos a}+\cot a=\frac{\sin a}{1+\cos a}+\frac{\cos a}{\sin a}=\frac{\sin ^2a+\cos^2a+\cos a}{\sin a(1+\cos a)}\)
\(=\frac{1+\cos a}{\sin a(1+\cos a)}=\frac{1}{\sin a}\) (đpcm)
b)
\(\frac{1}{\cos a}-\frac{\cos a}{1+\sin a}=\frac{1+\sin a-\cos ^2a}{\cos a(1+\sin a)}=\frac{(1-\cos ^2a)+\sin a}{\cos a(\sin a+1)}\)
\(=\frac{\sin^2a+\sin a}{\cos a(\sin a+1)}=\frac{\sin a(\sin a+1)}{\cos a(\sin a+1)}=\frac{\sin a}{\cos a}=\tan a\) (đpcm)
c)
\(\frac{\tan a-\sin a}{\sin ^3a}=\frac{\frac{\sin a}{\cos a}-\sin a}{\sin ^3a}=\frac{\frac{1}{\cos a}-1}{\sin ^2a}=\frac{1-\cos a}{\cos a\sin ^2a}=\frac{1-\cos a}{\cos a(1-\cos ^2a)}=\frac{1}{\cos a(1+\cos a)}\)
d)
\(\frac{\sin a+\cos a-1}{\sin a-\cos a+1}=\frac{(\sin a+\cos a-1)(\sin a+\cos a+1)}{(\sin a-\cos a+1)(\sin a+\cos a+1)}=\frac{(\sin a+\cos a)^2-1}{(\sin a+1)^2-\cos ^2a}\)
\(=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\sin ^2a+1+2\sin a-\cos ^2a}=\frac{1+2\sin a\cos a-1}{\sin ^2a+1+2\sin a-(1-\sin ^2a)}\)
\(=\frac{2\sin a\cos a}{2\sin ^2a+2\sin a}=\frac{2\sin a\cos a}{2\sin a(\sin a+1)}=\frac{\cos a}{1+\sin a}\) (đpcm)
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Mấu chốt trong các bài này là việc sử dụng công thức $\sin ^2a+\cos ^2a=1$
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Lời giải:
a)
\(\frac{\sin a}{1+\cos a}+\cot a=\frac{\sin a}{1+\cos a}+\frac{\cos a}{\sin a}=\frac{\sin ^2a+\cos^2a+\cos a}{\sin a(1+\cos a)}\)
\(=\frac{1+\cos a}{\sin a(1+\cos a)}=\frac{1}{\sin a}\) (đpcm)
b)
\(\frac{1}{\cos a}-\frac{\cos a}{1+\sin a}=\frac{1+\sin a-\cos ^2a}{\cos a(1+\sin a)}=\frac{(1-\cos ^2a)+\sin a}{\cos a(\sin a+1)}\)
\(=\frac{\sin^2a+\sin a}{\cos a(\sin a+1)}=\frac{\sin a(\sin a+1)}{\cos a(\sin a+1)}=\frac{\sin a}{\cos a}=\tan a\) (đpcm)
c)
\(\frac{\tan a-\sin a}{\sin ^3a}=\frac{\frac{\sin a}{\cos a}-\sin a}{\sin ^3a}=\frac{\frac{1}{\cos a}-1}{\sin ^2a}=\frac{1-\cos a}{\cos a\sin ^2a}=\frac{1-\cos a}{\cos a(1-\cos ^2a)}=\frac{1}{\cos a(1+\cos a)}\)
d)
\(\frac{\sin a+\cos a-1}{\sin a-\cos a+1}=\frac{(\sin a+\cos a-1)(\sin a+\cos a+1)}{(\sin a-\cos a+1)(\sin a+\cos a+1)}=\frac{(\sin a+\cos a)^2-1}{(\sin a+1)^2-\cos ^2a}\)
\(=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\sin ^2a+1+2\sin a-\cos ^2a}=\frac{1+2\sin a\cos a-1}{\sin ^2a+1+2\sin a-(1-\sin ^2a)}\)
\(=\frac{2\sin a\cos a}{2\sin ^2a+2\sin a}=\frac{2\sin a\cos a}{2\sin a(\sin a+1)}=\frac{\cos a}{1+\sin a}\) (đpcm)
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chứng minh rằng
a)
\(\frac{1-2\text{s}in^2x}{2cot\left(\frac{\pi}{4}+\alpha\right).c\text{os}^2\left(\frac{\pi}{4}-\alpha\right)}=1\)
b)
\(\frac{\frac{\sqrt{3}}{2}c\text{os}2\text{a}-\frac{1}{2}sin2\text{a}}{1-\frac{1}{2}c\text{os}2\text{a}-\frac{\sqrt{3}}{2}sin2\text{a}}=tan\left(a+\frac{\pi}{4}\right)\)
1) \(c\text{os}x+c\text{os}2x+c\text{os}3x=0\)
2) \(c\text{os}3x+c\text{os}4x+c\text{os}5x=0\)
3) \(c\text{os^2}x+c\text{os^2}2x+c\text{os^2}3x=0\)
4) \(c\text{os^2}2x+c\text{os^2}3x+c\text{os^2}4x=0\)
1.
\(cosx+cos3x+cos2x=0\)
\(\Leftrightarrow2cos2x.cosx+cos2x=0\)
\(\Leftrightarrow cos2x\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
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2.
\(cos3x+cos5x+cos4x=0\)
\(\Leftrightarrow2cos4x.cosx+cos4x=0\)
\(\Leftrightarrow cos4x\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+k\pi\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
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3.
Ta có: \(\left\{{}\begin{matrix}cos^2x\ge0\\cos^22x\ge0\\cos^23x\ge0\end{matrix}\right.\) với mọi x
\(\Rightarrow cos^2x+cos^22x+cos^23x\ge0\) với mọi x
Dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}cosx=0\\cos2x=0\\cos3x=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}cosx=0\\2cos^2x-1=0\\cos3x=0\end{matrix}\right.\)
Pt vô nghiệm (do nghiệm của pt thứ nhất ko thể là nghiệm của pt thứ 2)
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Xem thêm câu trả lời
Cho tam giác nhọn ABC , diện tích bằng 1 . Vẽ ba đường cao AD , BE , CF . Chứng minh rằng :
a ) _{S_{AEF}+S_{BFD}+S_{CDE}} ctext{os}^2 A + ctext{os}^2 B + ctext{os}^2 C
b ) _{S_{Dtext{EF}}} sin^2 A - ctext{os}^2 B - ctext{os}^2C
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Cho tam giác nhọn ABC , diện tích bằng 1 . Vẽ ba đường cao AD , BE , CF . Chứng minh rằng :
a ) \(_{S_{AEF}+S_{BFD}+S_{CDE}}\) = \(c\text{os}^2\) A + \(c\text{os}^2\) B + \(c\text{os}^2\) C
b ) \(_{S_{D\text{EF}}}\) = \(sin^2\) A - \(c\text{os}^2\) B - \(c\text{os}^2\)C
\(c\text{os}3a=4cosa.c\text{os}\left(\frac{\pi}{3}-a\right).c\text{os}\left(\frac{\pi}{3}+a\right)\)
\(4cosa.cos\left(\frac{\pi}{3}-a\right).cos\left(\frac{\pi}{3}+a\right)=2cosa.\left(cos\frac{2\pi}{3}+cos2a\right)\)
\(=2cosa\left(cos2a-\frac{1}{2}\right)=2cosa.cos2a-cosa\)
\(=cos3a+cosa-cosa=cos3a\)
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Thu gọn các biểu thức sau:
a. \(sin^6a+c\text{os}^6a+3sin^2a.c\text{os}^2a\)
b.\(sin^4a-c\text{os}^4a-\left(sina+c\text{os}a\right)\left(sina-c\text{os}a\right)\)
c.\(c\text{os}^2a+tan^2a.c\text{os}^2a\)
d.\(c\text{os}^2a+tan^2a.c\text{os}^2a\)
a) \(sin^6x+cos^6x+3sin^2x.cos^2x\)
\(=\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cox^2x+cos^4x\right)+3sin^2x.cos^2x\)
\(=sin^4x-sin^2x.cox^2x+cos^4x+3sin^2x.cos^2x\)
\(=sin^4x+2sin^2x.cox^2x+cos^4x=\left(sin^2x+cos^2x\right)^2=1\text{}\text{}\)
b) \(sin^4x-cos^4x-\left(sinx+cosx\right)\left(sinx-cosx\right)\)
\(=\left(sin^2x+cos^2x\right)\left(sin^2x-cos^2x\right)-\left(sin^2x-cos^2x\right)\)
\(=1\left(sin^2x-cos^2x\right)-\left(sin^2x-cos^2x\right)=0\)
c) \(cos^2x+tan^2x.cos^2x\)
\(=cos^2x+\dfrac{sin^2x}{cos^2x}.cos^2x=sin^2x+cos^2x=1\)
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\(\frac{1+c\text{os}x-s\text{inx}}{1-c\text{os}x-s\text{inx}}=-cot\frac{x}{2}\)
\(\frac{1+cosx-sinx}{1-cosx-sinx}=\frac{1+2cos^2\frac{x}{2}-1-2sin\frac{x}{2}.cos\frac{x}{2}}{1-1+2sin^2\frac{x}{2}-2sin\frac{x}{2}.cos\frac{x}{2}}=\frac{2cos^2\frac{x}{2}-2sin\frac{x}{2}.cos\frac{x}{2}}{2sin^2\frac{x}{2}-2sin\frac{x}{2}.cos\frac{x}{2}}\)
\(=\frac{-2cos\frac{x}{2}\left(sin\frac{x}{2}-cos\frac{x}{2}\right)}{2sin\frac{x}{2}\left(sin\frac{x}{2}-cos\frac{x}{2}\right)}=\frac{-cos\frac{x}{2}}{sin\frac{x}{2}}=-cot\frac{x}{2}\)
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Cho tam giác ABC có C<45 độ AB=c ; Bc = a ; AC=b. đường cao AH=h. Trung tuyến AM=m
CMR
a) \(c\text{os}2\alpha=c\text{os}^2\alpha-sin^2\alpha\)
b) \(tan2\alpha=\frac{2tan\alpha}{1tan^2\alpha}\)