Chứng minh các đẳng thức sau:
sinx(1+cos2x)=sin2x.cosx
\(tanx-\frac{1}{tanx}=-\frac{2}{tan2x}\)
\(tan\frac{x}{2}\left(\frac{1}{cosx}+1\right)=tanx\)
Chứng minh đẳng thức sau :
a, \(\left(\frac{tan^2x-1}{2tanx}\right)^2\) - \(\frac{1}{4sin^2x.cos^2x}\) = -1
b, \(\frac{cos^2x-sin^2x}{sin^4x+cos^4x-sin^2x}\) = 1 + tan2x
c, \(\frac{sin^2x}{cosx.\left(1+tanx\right)}-\frac{cos^2x}{sinx.\left(1+cotx\right)}=sinx-cosx\)
d, \(\left(\frac{cosx}{1+sinx}+tanx\right).\left(\frac{sinx}{1+cosx}+cotx\right)=\frac{1}{sinx.cosx}\)
e, cos2x.(cos2x + 2sin2x + sin2x.tan2x) = 1
\(a,\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4sin^2x.cos^2x}=-1\)
\(VT=\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4.sin^2x.cos^2x}=\left(\frac{1}{tan2x}\right)^2-\frac{1}{sin^22x}=\left(\frac{cos2x}{sin2x}\right)^2-\frac{1}{sin^22x}=\frac{cos^22x-1}{sin^22x}=\frac{-sin^22x}{sin^22x}=-1=VP\)
b, \(VT=\frac{cos^2x-sin^2x}{sin^4x+cos^4x-sin^2x}=\frac{cos2x}{\left(sin^2x+cos^2x\right)^2-sin^2x-2.sin^2x.cos^2x}=\frac{cos2x}{1-sin^2x-2.sin^2x.cos^2x}=\frac{cos2x}{cos^2x-2.sin^2x.cos^2x}\)
=\(\frac{cos2x}{cos^2x.\left(1-2.sin^2x\right)}=\frac{cos2x}{cos^2x.cos2x}=\frac{1}{cos^2x}=1+tan^2x=VP\)
d, \(VT=\left(\frac{cosx}{1+sinx}+tanx\right).\left(\frac{sinx}{1+cosx}+cotx\right)=\left(\frac{cosx}{1+sinx}+\frac{sinx}{cosx}\right).\left(\frac{sinx}{1+cosx}+\frac{cosx}{sinx}\right)\)
\(=\left(\frac{cos^2x+sinx.\left(1+sinx\right)}{cosx.\left(1+sinx\right)}\right).\left(\frac{sin^2x+cosx.\left(1+cosx\right)}{sinx.\left(1+cosx\right)}\right)=\left(\frac{cos^2x+sinx+sin^2x}{cosx.\left(1+sinx\right)}\right).\left(\frac{sin^2x+cosx+cos^2x}{sinx.\left(1+cosx\right)}\right)\)
=\(\frac{1}{cosx.sinx}=VP\)
e, \(VT=cos^2x.\left(cos^2x+2sin^2x+sin^2x.tan^2x\right)=cos^2x.\left(1+sin^2x.\left(1+tan^2x\right)\right)=cos^2x.\left(1+tan^2x\right)=cos^2x.\frac{1}{cos^2x}=1=VP\)
c, \(VT=\frac{sin^2x}{cosx.\left(1+tanx\right)}-\frac{cos^2x}{sinx.\left(1+cosx\right)}=\frac{sin^3x.\left(1+cosx\right)-cos^3x.\left(1+tanx\right)}{sinx.cosx.\left(1+tanx\right).\left(1+cosx\right)}\)
=\(\frac{sin^3x+sin^3x.cotx-cos^3x-cos^3.tanx}{\left(sinx+cosx\right)^2}=\frac{sin^3x+sin^2xcosx-cos^3x-cos^2sinx}{\left(sinx+cosx\right)^2}=\frac{sin^2x.\left(sinx+cosx\right)-cos^2x.\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}\)
\(=\frac{\left(sin^2x-cos^2x\right).\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}=\frac{\left(sinx-cosx\right).\left(sinx+cosx\right).\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}=sinx-cosx=VP\)
Đây nha bạn
giải phương trình sau:
a,\(\frac{sin2x+2cosx-sinx-1}{tanx+\sqrt{3}}=0\)
b,\(\frac{\left(1+sinx+cos2x\right)sinx\left(x+\frac{\pi}{4}\right)}{1+tanx}=\frac{1}{\sqrt{2}}cosx\)
c,\(\frac{\left(1-sin2x\right)cosx}{\left(1+sin2x\right)\left(1-sinx\right)}=\sqrt{3}\)
d,\(\frac{1}{sinx}+\frac{1}{sin\left(x-\frac{3\pi}{2}\right)}=4sin\left(\frac{7\pi}{4}-x\right)\)
Chứng minh đẳng thức sau :
1) \(sin^2\left(\frac{\pi}{8}+x\right)-sin^2\left(\frac{\pi}{8}-x\right)=\frac{\sqrt{2}}{2}sin2x\)
2) \(tan\frac{x}{2}\left(\frac{1}{cosx}+1\right)=tanx\)
rút gọn các biểu thức lượng giác sau:
\(\frac{sin^2x}{cosx\left(1+tanx\right)}-\frac{cos^2x}{sinx\left(1+cotx\right)}=sinx-cosx\)
\(\left(tanx+\frac{cosx}{1+sinx}\right)\left(cotx+\frac{sinx}{1+cosx}\right)=\frac{1}{sinx.cosx}\)
đề bài đầy đủ: rút gọn các biểu thức lượng giác sau trên điều kiện xác định của chúng:
\(\frac{sin^2x}{cosx+cosx.\frac{sinx}{cosx}}-\frac{cos^2x}{sinx+sinx.\frac{cosx}{sinx}}=\frac{sin^2x}{sinx+cosx}-\frac{cos^2x}{sinx+cosx}=\frac{sin^2x-cos^2x}{sinx+cosx}\)
\(=\frac{\left(sinx+cosx\right)\left(sinx-cosx\right)}{sinx+cosx}=sinx-cosx\)
\(\left(\frac{sinx}{cosx}+\frac{cosx}{1+sinx}\right)\left(\frac{cosx}{sinx}+\frac{sinx}{1+cosx}\right)=\left(\frac{sinx+sin^2x+cos^2x}{cosx\left(1+sinx\right)}\right)\left(\frac{cosx+cos^2x+sin^2x}{sinx\left(1+cosx\right)}\right)\)
\(=\left(\frac{sinx+1}{cosx\left(1+sinx\right)}\right)\left(\frac{cosx+1}{sinx\left(1+cosx\right)}\right)=\frac{1}{sinx.cosx}\)
giải các pt
a) \(tanx+tan\left(\frac{2\pi}{3}-3x\right)=0\)
b) \(tan\left(2x-15^o\right)-tanx=0\)
c) \(\frac{tan2x-2}{2tan2x+1}=3\)
d) \(\frac{sinx+\sqrt{3}cosx}{3sinx-\sqrt{3}cosx}=1\)
a/
\(\Leftrightarrow tanx=-tan\left(\frac{2\pi}{3}-3x\right)\)
\(\Leftrightarrow tanx=tan\left(3x-\frac{2\pi}{3}\right)\)
\(\Rightarrow x=3x-\frac{2\pi}{3}+k\pi\)
\(\Rightarrow x=\frac{\pi}{3}+\frac{k\pi}{2}\)
b/
\(tan\left(2x-15^0\right)=tanx\)
\(\Rightarrow2x-15^0=x+k180^0\)
\(\Rightarrow x=15^0+k180^0\)
c/
ĐKXĐ: ...
\(\Leftrightarrow tan2x-2=3\left(2tan2x+1\right)\)
\(\Leftrightarrow5tan2x=-5\)
\(\Rightarrow tan2x=-1\)
\(\Rightarrow2x=-\frac{\pi}{4}+k\pi\)
\(\Rightarrow x=-\frac{\pi}{8}+\frac{k\pi}{2}\)
d/
ĐKXĐ: ...
\(\Leftrightarrow sinx+\sqrt{3}cosx=3sinx-\sqrt{3}cosx\)
\(\Leftrightarrow2sinx=2\sqrt{3}cosx\)
\(\Rightarrow tanx=\sqrt{3}\Rightarrow x=\frac{\pi}{3}+k\pi\)
Trong các hệ thức sau , hệ thức nào sai ?Nếu sai hãy sửa lại cho đúng và chứng minh các hệ thức đúng còn lại ?
\(A.\frac{sin^2\alpha+1}{2\left(1-sin^2\alpha\right)}+\frac{1+cos^2\alpha}{2\left(1-cos^2\alpha\right)}+1=\left(tan\alpha+cot\alpha\right)^2\)
\(B.\frac{1-4sin^2x.cos^2x}{4sin^2x.cos^2x}=\frac{1+tan^4x-2tan^2x}{4tan^2x}\)
\(C.\frac{sinx+tanx}{tanx}=1+sinx+cotx\)
\(D.tanx+\frac{cosx}{1+sinx}=\frac{1}{cosx}\)
\(\frac{sin^2a+1}{2.cos^2a}+\frac{1+cos^2a}{2.sin^2a}+1=\frac{tan^2a}{2}+\frac{1}{2cos^2a}+\frac{cot^2a}{2}+\frac{1}{2sin^2a}+1\)
\(=\frac{1}{2}\left(tan^2a+1+tan^2a+cot^2a+1+cot^2a+2\right)\)
\(=\frac{1}{2}\left(2tan^2a+4+2cot^2a\right)=tan^2a+2+cot^2a=\left(tana+cota\right)^2\)
B.
\(\frac{1-4sin^2a.cos^2a}{4sin^2a.cos^2a}=\frac{\frac{1}{cos^4a}-\frac{4sin^2a}{cos^2a}}{\frac{4sin^2a}{cos^2a}}=\frac{\left(\frac{1}{cos^2a}\right)^2-4tan^2a}{4tan^2a}=\frac{\left(1+tan^2a\right)^2-4tan^2a}{4tan^2a}\)
\(=\frac{tan^4a-2tan^2a+1}{4tan^2a}\)
C.
\(\frac{sina+tana}{tana}=\frac{sina}{tana}+1=1+sina.\frac{cosa}{sina}=1+cosa\)
D.
\(tana+\frac{cosa}{1+sina}=\frac{sina}{cosa}+\frac{cosa\left(1-sina\right)}{1-sin^2a}=\frac{sina.cosa}{cos^2a}+\frac{cosa-cosa.sina}{cos^2a}\)
\(=\frac{sina.cosa+cosa-sina.cosa}{cos^2a}=\frac{cosa}{cos^2a}=\frac{1}{cosa}\)
Câu C sai
chứng minh rằng
3) \(\frac{sin2x-sinx}{1-cosx+cos2x}=tanx\)
4) \(\left(\frac{sinx+cotx}{1+sinx.tanx}\right)^{2014}=\frac{sin^{2014}x+cot^{2014}x}{1+sin^{2014}x.tan^{2014}x}\)
3/
\(\frac{sin2x-sinx}{1-cosx+cos2x}=\frac{2sinxcosx-sinx}{1-cosx+2cos^2x-1}=\frac{sinx\left(2cosx-1\right)}{cosx\left(2cosx-1\right)}=\frac{sinx}{cosx}=tanx\)
4/
\(\left(\frac{sinx+cotx}{1+sinx.tanx}\right)^{2014}=\left(\frac{sinx+\frac{1}{tanx}}{1+sinxtanx}\right)^{2014}=\left(\frac{sinxtanx+1}{tanx\left(sinxtanx+1\right)}\right)^{2014}\)
\(=\left(\frac{1}{tanx}\right)^{2014}=cot^{2014}x\)
\(\frac{sin^{2014}x+cot^{2014}x}{1+\left(sinx.tanx\right)^{2014}}=\frac{sin^{2014}x+\frac{1}{tan^{2014}x}}{1+\left(sinx.tanx\right)^{2014}}=\frac{\left(sinxtanx\right)^{2014}+1}{tan^{2014}x\left[\left(sinxtanx\right)^{2014}+1\right]}\)
\(=\frac{1}{tan^{2014}x}=\left(\frac{1}{tanx}\right)^{2014}=cot^{2014}x\)
\(\Rightarrow\left(\frac{sinx+cotx}{1+sinx.tanx}\right)^{2014}=\frac{sin^{2014}x+cot^{2014}x}{1+\left(sinx.tanx\right)^{2014}}\)
a) \(1-cot^4x=\frac{2}{sin^2x}-\frac{1}{sin^4x}\)
b)\(\frac{1-2sinx.cosx}{cos^2-sin^2}\)\(=\frac{1-tanx}{1+tanx}\)\(\)
c)\(\frac{sin^2x}{sinx-cosx}+\frac{sinx+cosx}{1-tanx}=sinx+cosx\)
d)\(\sqrt{\frac{1+cosx}{1-cosx}}-\sqrt{\frac{1-cosx}{1+cosx}}=\frac{2.cosx}{|sin|}\)
e)\(tan^3x+tan^2x+tanx+1=\frac{sinx+cosx}{cos^3x}\)
Giải phương trình
1.\(\frac{\left(2sinx+1\right).\left(3cos4x+2sinx\right)+4cos^2x+1}{1+sinx}=8\)
2.\(\frac{\left(1+sinx+cos2x\right).sin\left(x+\frac{\pi}{4}\right)}{1+tanx}=\frac{1}{\sqrt{2}}cosx\)
a/ ĐKXĐ: \(sinx\ne-1\)
\(\Leftrightarrow\left(2sinx+1\right)\left(3cos4x+2sinx\right)+4cos^2x+1=8+8sinx\)
\(\Leftrightarrow6sinx.cos4x+4sin^2x+3cos4x+2sinx+4cos^2x+1=8+8sinx\)
\(\Leftrightarrow6sinx.cos4x+3cos4x-6sinx-3=0\)
\(\Leftrightarrow6sinx\left(cos4x-1\right)+3\left(cos4x-1\right)=0\)
\(\Leftrightarrow\left(6sinx+3\right)\left(cos4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\cos4x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\1-2sin^22x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sin^2x\left(1-sin^2x\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sin^2x\left(1+sinx\right)\left(1-sinx\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sinx=0\\sinx=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\\x=k\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
b/ ĐKXĐ: \(\left\{{}\begin{matrix}tanx\ne-1\\cosx\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left(1+sinx+cos2x\right).\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=cosx\left(1+\frac{sinx}{cosx}\right)\)
\(\Leftrightarrow\left(1+sinx+cos2x\right)\left(sinx+cosx\right)=cosx+sinx\)
\(\Leftrightarrow\left(cosx+sinx\right)\left(sinx+cos2x\right)=0\)
\(\Leftrightarrow sinx+cos2x=0\)
\(\Leftrightarrow-2sin^2x+sinx+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\left(l\right)\\sinx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)