chứng minh \(\frac{2\sin^2\frac{x}{2}+\sin2x-1}{2\sin x-1}+\sin x=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)\)
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Những câu hỏi liên quan
giải phương trình
sin xsqrt{1+2sin x}cos2x
sinleft(frac{5x}{2}-frac{pi}{4}right)-cosleft(frac{x}{2}-frac{pi}{4}right)sqrt{2}cosfrac{3x}{2}
3sqrt{tan x+1}left(sin x+2cos xright)5left(sin x+3cos xright)
sqrt{2}left(sin x+sqrt{3}cos xright)sqrt{3}cos2x-sin2x
sin2xsin4x+2left(3sin x-4sin^2x+1right)0
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giải phương trình
\(\sin x\sqrt{1+2\sin x}=\cos2x\)
\(\sin\left(\frac{5x}{2}-\frac{\pi}{4}\right)-\cos\left(\frac{x}{2}-\frac{\pi}{4}\right)=\sqrt{2}\cos\frac{3x}{2}\)
\(3\sqrt{\tan x+1}\left(\sin x+2\cos x\right)=5\left(\sin x+3\cos x\right)\)
\(\sqrt{2}\left(\sin x+\sqrt{3}\cos x\right)=\sqrt{3}\cos2x-\sin2x\)
\(\sin2x\sin4x+2\left(3\sin x-4\sin^2x+1\right)=0\)
a/ Hmm, bạn có nhầm lẫn chỗ nào ko nhỉ, nghiệm của pt này xấu khủng khiếp
b/ \(\Leftrightarrow sin\frac{5x}{2}-cos\frac{5x}{2}-sin\frac{x}{2}-cos\frac{x}{2}=cos\frac{3x}{2}\)
\(\Leftrightarrow2cos\frac{3x}{2}.sinx-2cos\frac{3x}{2}cosx=cos\frac{3x}{2}\)
\(\Leftrightarrow cos\frac{3x}{2}\left(2sinx-2cosx-1\right)=0\)
\(\Leftrightarrow cos\frac{3x}{2}\left(\sqrt{2}sin\left(x-\frac{\pi}{4}\right)-1\right)=0\)
c/ Do \(cosx\ne0\), chia 2 vế cho cosx ta được:
\(3\sqrt{tanx+1}\left(tanx+2\right)=5\left(tanx+3\right)\)
Đặt \(\sqrt{tanx+1}=t\ge0\)
\(\Leftrightarrow3t\left(t^2+1\right)=5\left(t^2+2\right)\)
\(\Leftrightarrow3t^3-5t^2+3t-10=0\)
\(\Leftrightarrow\left(t-2\right)\left(3t^2+t+5\right)=0\)
d/ \(\Leftrightarrow\sqrt{2}\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{3}\right)=-sin\left(2x-\frac{\pi}{3}\right)\)
Đặt \(x+\frac{\pi}{3}=a\Rightarrow2x=2a-\frac{2\pi}{3}\Rightarrow2x-\frac{\pi}{3}=2a-\pi\)
\(\sqrt{2}sina=-sin\left(2a-\pi\right)=sin2a=2sina.cosa\)
\(\Leftrightarrow\sqrt{2}sina\left(\sqrt{2}cosa-1\right)=0\)
Chứng minh|
a) \(\frac{1+sin2x}{sinx+cosx}-\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}=sinx\)
b) \(sin^4x+cos^4\left(x+\frac{\pi}{4}\right)=\frac{3}{4}-\frac{\sqrt{2}}{2}sin\left(2x+\frac{\pi}{4}\right)\)
\(\frac{sin^2x+cos^2x+2sinx.cosx}{sinx+cosx}-\left(1-tan^2\frac{x}{2}\right).cos^2\frac{x}{2}\)
\(=\frac{\left(sinx+cosx\right)^2}{sinx+cosx}-\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)\)
\(=sinx+cosx-cosx=sinx\)
\(sin^4x+cos^4\left(x+\frac{\pi}{4}\right)=\left(\frac{1}{2}-\frac{1}{2}cos2x\right)^2+\left(\frac{1}{2}+\frac{1}{2}cos\left(2x+\frac{\pi}{2}\right)\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\left(\frac{1}{2}-\frac{1}{2}sin2x\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\frac{1}{4}-\frac{1}{2}sin2x+\frac{1}{4}sin^22x\)
\(=\frac{1}{4}-\frac{1}{2}\left(cos2x+sin2x\right)+\frac{1}{4}\left(cos^22x+sin^22x\right)\)
\(=\frac{3}{4}-\frac{\sqrt{2}}{2}sin\left(2x+\frac{\pi}{4}\right)\)
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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) \(tan\frac{x}{2}\left(\frac{1}{cosx}+1\right)=tanx\)
Chứng minh rằng:
a) \(\sin x - \cos x = \sqrt 2 \sin \left( {x - \frac{\pi }{4}} \right)\);
b) \(\tan \left( {\frac{\pi }{4} - x} \right) = \frac{{1 - \tan x}}{{1 + \tan x}}\;\left( {x \ne \frac{\pi }{2} + k\pi ,\;x \ne \frac{{3\pi }}{4} + k\pi ,\;k \in \mathbb{Z}} \right)\;\).
a) Ta có:
\(\sqrt 2 \sin \left( {x - \frac{\pi }{4}} \right) = \sqrt 2 \left( {\sin x\cos \frac{\pi }{4} + \cos x\sin \frac{\pi }{4}} \right) = \sqrt 2 \left( {\sin x.\frac{{\sqrt 2 }}{2} + \cos x.\frac{{\sqrt 2 }}{2}} \right) = \sin x + \cos x\)
b) Ta có:
\(\tan \left( {\frac{\pi }{4} - x} \right) = \frac{{\tan \frac{\pi }{4} - \tan x}}{{1 + \tan \frac{\pi }{4}\tan x}} = \frac{{1 - \tan x}}{{1 + \tan x}}\;\)
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Giải các phương trình :
1) frac{sin^4x+cos^4x}{sin2x}frac{1}{2}left(tan x+cot2xright)
2) frac{1}{sin x}+frac{1}{sinleft(x-frac{3pi}{2}right)}4sinleft(frac{7pi}{4}-xright)
3)2left(cos^42x-sin^42xright)+cos8x-cos4x0
4)frac{sin^4x+cos^4x}{5sin2x}frac{1}{2}cot2x-frac{1}{8sin2x}
5)sin^4x+cos^4x-3sin2x+frac{5}{2}sin^22x0
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Giải các phương trình :
1) \(\frac{\sin^4x+\cos^4x}{\sin2x}=\frac{1}{2}\left(\tan x+\cot2x\right)\)
2) \(\frac{1}{\sin x}+\frac{1}{\sin\left(x-\frac{3\pi}{2}\right)}=4\sin\left(\frac{7\pi}{4}-x\right)\)
3)\(2\left(\cos^42x-\sin^42x\right)+\cos8x-\cos4x=0\)
4)\(\frac{\sin^4x+\cos^4x}{5\sin2x}=\frac{1}{2}\cot2x-\frac{1}{8\sin2x}\)
5)\(\sin^4x+\cos^4x-3\sin2x+\frac{5}{2}\sin^22x=0\)
Chứng minh rằng: (Pls help me)
a, frac{1}{sin x}+cot xcotfrac{x}{2}
b, frac{1-cos x}{sin x}tanfrac{x}{2}
c,tanfrac{x}{2}left(frac{1}{cos x}+1right)tan x
d,frac{sin2a}{2cos aleft(1+cos aright)}tanfrac{a}{2}
e,cot x+tanfrac{x}{2}frac{1}{sin x}
f,3-4cos2x+cos4x8sin^4x
g,frac{1-cos x}{sin x}frac{sin x}{1+cos x}
h,sin x+cos xsqrt{2}sinleft(x+frac{pi}{4}right)
i,sin x-cos xsqrt{2}sinleft(x-frac{pi}{4}right)
l,cos x-sin xsqrt{2}cosleft(x+frac{pi}{4}right)
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Chứng minh rằng: (Pls help me)
a, \(\frac{1}{\sin x}+\cot x=\cot\frac{x}{2}\)
b, \(\frac{1-\cos x}{\sin x}=\tan\frac{x}{2}\)
c,\(\tan\frac{x}{2}\left(\frac{1}{\cos x}+1\right)=\tan x\)
d,\(\frac{\sin2a}{2\cos a\left(1+\cos a\right)}=\tan\frac{a}{2}\)
e,\(\cot x+\tan\frac{x}{2}=\frac{1}{\sin x}\)
f,\(3-4\cos2x+\cos4x=8\sin^4x\)
g,\(\frac{1-\cos x}{\sin x}=\frac{\sin x}{1+\cos x}\)
h,\(\sin x+\cos x=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)\)
i,\(\sin x-\cos x=\sqrt{2}\sin\left(x-\frac{\pi}{4}\right)\)
l,\(\cos x-\sin x=\sqrt{2}\cos\left(x+\frac{\pi}{4}\right)\)
a/
\(\frac{1}{sinx}+\frac{cosx}{sinx}=\frac{1+cosx}{sinx}=\frac{1+2cos^2\frac{x}{2}-1}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{2cos^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{cos\frac{x}{2}}{sin\frac{x}{2}}=cot\frac{x}{2}\)
b/
\(\frac{1-cosx}{sinx}=\frac{1-\left(1-2sin^2\frac{x}{2}\right)}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{2sin^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=tan\frac{x}{2}\)
c/
\(tan\frac{x}{2}\left(\frac{1}{cosx}+1\right)=\left(\frac{1-cosx}{sinx}\right)\left(\frac{1}{cosx}+1\right)=\frac{\left(1-cosx\right)\left(1+cosx\right)}{sinx.cosx}=\frac{1-cos^2x}{sinx.cosx}\)
\(=\frac{sin^2x}{sinx.cosx}=\frac{sinx}{cosx}=tanx\)
d/
\(\frac{sin2a}{2cosa\left(1+cosa\right)}=\frac{2sina.cosa}{2cosa\left(1+2cos^2\frac{a}{2}-1\right)}=\frac{sina}{2cos^2\frac{a}{2}}=\frac{2sin\frac{a}{2}cos\frac{a}{2}}{2cos^2\frac{a}{2}}=tan\frac{a}{2}\)
e/
\(cotx+tan\frac{x}{2}=\frac{cosx}{sin}+\frac{1-cosx}{sinx}=\frac{cosx+1-cosx}{sinx}=\frac{1}{sinx}\)
Các câu c, e đều sử dụng kết quả từ câu b
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f/
\(3-4cos2x+cos4x=3-4cos2x+2cos^22x-1\)
\(=2cos^22x-4cos2x+2=2\left(cos^22x-2cos2x+1\right)\)
\(=2\left(cos2x-1\right)^2=2\left(1-2sin^2x-1\right)^2\)
\(=2.\left(-2sin^2x\right)^2=8sin^4x\)
g/
\(\frac{1-cosx}{sinx}=\frac{sinx\left(1-cosx\right)}{sin^2x}=\frac{sinx\left(1-cosx\right)}{1-cos^2x}=\frac{sinx\left(1-cosx\right)}{\left(1-cosx\right)\left(1+cosx\right)}=\frac{sinx}{1+cosx}\)
h/
\(sinx+cosx=\sqrt{2}\left(sinx.\frac{\sqrt{2}}{2}+cosx.\frac{\sqrt{2}}{2}\right)\)
\(=\sqrt{2}\left(sinx.cos\frac{\pi}{4}+cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
i/
\(sinx-cosx=\sqrt{2}\left(sinx.\frac{\sqrt{2}}{2}-cosx.\frac{\sqrt{2}}{2}\right)\)
\(=\sqrt{2}\left(sinx.cos\frac{\pi}{4}-cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\)
j/
\(cosx-sinx=\sqrt{2}\left(cosx.\frac{\sqrt{2}}{2}-sinx\frac{\sqrt{2}}{2}\right)\)
\(=\sqrt{2}\left(cosx.cos\frac{\pi}{4}-sinx.sin\frac{\pi}{4}\right)=\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)
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1) frac{1}{cos x}+frac{1}{sin2x}frac{2}{sin4x}
2) cos3xcdottan5xsin7x
3) tan5xcdottan2x1
4) 4cos x-2cos2x-cos4x1
5) sinleft(2x+frac{5pi}{2}right)-2cosleft(x-frac{7pi}{2}right)1+2sin x
6) sin^22x-cos^28xsinleft(frac{17pi}{2}+10xright)
7) 8cos xfrac{sqrt{3}}{sin x}+frac{1}{cos x}
Đọc tiếp
1) \(\frac{1}{\cos x}+\frac{1}{\sin2x}=\frac{2}{\sin4x}\)
2) \(\cos3x\cdot\tan5x=\sin7x\)
3) \(\tan5x\cdot\tan2x=1\)
4) \(4\cos x-2\cos2x-\cos4x=1\)
5) \(\sin\left(2x+\frac{5\pi}{2}\right)-2\cos\left(x-\frac{7\pi}{2}\right)=1+2\sin x\)
6) \(\sin^22x-\cos^28x=\sin\left(\frac{17\pi}{2}+10x\right)\)
7) \(8\cos x=\frac{\sqrt{3}}{\sin x}+\frac{1}{\cos x}\)
1.
DKXĐ: \(sin4x\ne0\)
\(\Leftrightarrow\frac{4sinx.cos2x}{sin4x}+\frac{2cos2x}{sin4x}=\frac{2}{sin4x}\)
\(\Leftrightarrow2sinx.cos2x+cos2x=1\)
\(\Leftrightarrow2sinx\left(1-2sin^2x\right)+1-2sin^2x=1\)
\(\Leftrightarrow sinx\left(1-2sin^2x-sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(l\right)\\-2sin^2x-sinx+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}sinx=-1\left(l\right)\\sinx=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
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2.
ĐKXĐ: ...
\(\Leftrightarrow\frac{cos3x.sin5x}{cos5x}=sin7x\)
\(\Leftrightarrow cos3x.sin5x=sin7x.cos5x\)
\(\Leftrightarrow sin8x+sin2x=sin12x+sin2x\)
\(\Leftrightarrow sin8x=sin12x\)
\(\Leftrightarrow\left[{}\begin{matrix}12x=8x+k2\pi\\12x=\pi-8x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{\pi}{20}+\frac{k\pi}{10}\end{matrix}\right.\)
Ở nghiệm đầu tiên loại các giá trị k lẻ do đó nghiệm của pt là:
\(\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{20}+\frac{k\pi}{10}\end{matrix}\right.\)
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3.
ĐKXĐ: ...
\(\Leftrightarrow tan5x=\frac{1}{tan2x}\)
\(\Leftrightarrow tan5x=cot2x\)
\(\Leftrightarrow tan5x=tan\left(\frac{\pi}{2}-2x\right)\)
\(\Leftrightarrow5x=\frac{\pi}{2}-2x+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{14}+\frac{k\pi}{7}\)
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Xem thêm câu trả lời
giải phương trình
a, \(2\sin\frac{x}{2}\left(\sin\frac{3x}{2}+\cos\frac{3x}{2}\right)=3-4\cos x\)
b, \(\frac{2\cos^2x+\sqrt{3}\sin2x+3}{2\cos^2x.\sin\left(x+\frac{\pi}{3}\right)}=3\left(\tan^2x+1\right)\)
a/ \(cosx-cos2x+sin2x-sinx=3-4cosx\)
\(\Leftrightarrow2sinx.cosx-sinx-2cos^2x+5cosx-2=0\)
\(\Leftrightarrow sinx\left(2cosx-1\right)-\left(2cosx-1\right)\left(cosx-2\right)=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(sinx-cosx+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2cosx-1=0\\sinx-cosx=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin\left(x-\frac{\pi}{4}\right)=-\sqrt{2}< -1\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=\pm\frac{\pi}{3}+k2\pi\)
b/ ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\sin\left(x+\frac{\pi}{3}\right)\ne0\end{matrix}\right.\) \(\Rightarrow...\)
\(\frac{2cos^2x+\sqrt{3}sin2x+3}{2cos^2x.sin\left(x+\frac{\pi}{3}\right)}=\frac{3}{cos^2x}\)
\(\Leftrightarrow2cos^2x+2\sqrt{3}sinx.cosx+3=3\left(sinx+\sqrt{3}cosx\right)\)
\(\Leftrightarrow2cos^2x-3\sqrt{3}cosx+3+2\sqrt{3}sinx.cosx-3sinx=0\)
\(\Leftrightarrow\left(2cosx-\sqrt{3}\right)\left(cosx-\sqrt{3}\right)+\sqrt{3}sinx\left(2cosx-\sqrt{3}\right)=0\)
\(\Leftrightarrow\left(2cosx-\sqrt{3}\right)\left(cosx+\sqrt{3}sinx-\sqrt{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{\sqrt{3}}{2}\\sin\left(x+\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\end{matrix}\right.\) \(\Rightarrow...\)
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Chứng minh đẳng thức lượng giác:
\(\frac{2sin^2\frac{x}{2}+sin2x-1}{2sinx-1}+sinx=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
\(\frac{2sin^2\frac{x}{2}+sin2x-1}{2sinx-1}+sinx=\frac{1-cosx+2sin2x.cosx-1}{2sinx-1}+sinx\)
\(=\frac{cosx\left(2sinx-1\right)}{2sinx-1}+sinx=cosx+sinx\)
\(=\sqrt{2}\left(\frac{\sqrt{2}}{2}sinx+\frac{\sqrt{2}}{2}cosx\right)=\sqrt{2}\left(sinx.cos\frac{\pi}{4}+cosx.sin\frac{\pi}{4}\right)\)
\(=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
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