Cho \(sin\left(x-y\right)=\frac{1}{2}\). Chứng minh: \(1-\sin^2x-\sin^2y+2\sin x.\sin y.\cos\left(x-y\right)=\frac{3}{4}\)
Những câu hỏi liên quan
Câu 1 : Dùng công thức cộng chứng minh các đẳng thức sau :
a/ sin(frac{pi}{4}+x) -sin left(frac{pi}{4}-xright)sqrt{2}sinx
b/ cos(x+y) cos(x-y)cos^2x - sin^2y
c/frac{tan^2x-tan^2y}{1-tan^2x.tan^2y}tanleft(x+yright)tanleft(x-yright)
d/ cot2xfrac{cot^2x-1}{2cotx}
e/ sin15^o + tan30^o cos15^ofrac{sqrt{6}}{3}
f/ cos^2x-sinleft(frac{pi}{6}+xright)sinleft(frac{pi}{6}-xright)frac{3}{4}
h/ frac{tanx+tany}{tanleft(x+
yright)}-frac{tanx-tany}{tanleft(x-yright)}-2tanx.tany
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Câu 1 : Dùng công thức cộng chứng minh các đẳng thức sau :
a/ sin(\(\frac{\pi}{4}+x\)) -sin \(\left(\frac{\pi}{4}-x\right)\)=\(\sqrt{2}sinx\)
b/ cos(x+y) cos(x-y)=cos\(^2\)x - sin\(^2\)y
c/\(\frac{tan^2x-tan^2y}{1-tan^2x.tan^2y}=tan\left(x+y\right)tan\left(x-y\right)\)
d/ cot2x=\(\frac{cot^2x-1}{2cotx}\)
e/ sin15\(^o\) + tan30\(^o\) cos15\(^o\)=\(\frac{\sqrt{6}}{3}\)
f/ \(cos^2x-sin\left(\frac{\pi}{6}+x\right)sin\left(\frac{\pi}{6}-x\right)=\frac{3}{4}\)
h/ \(\frac{tanx+tany}{tan\left(x+ y\right)}-\frac{tanx-tany}{tan\left(x-y\right)}=-2tanx.tany\)
Hãy chứng min rằng :
1) \(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2},\forall a,b,c,d\in R\)
2) \(\sqrt{4\cos^2x.\cos^2y+\sin^2\left(x-y\right)}+\sqrt{4\sin^2x.\sin^2y+\sin^2\left(x-y\right)}\ge2,\forall x,y\in R\)
1) Bất đẳng thức cần chứng minh
\(\Leftrightarrow\) a2 + b2 + c2 + d2 + \(2\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\ge\left(a+c\right)^2+\left(b+d\right)^2\)
\(\Leftrightarrow\) \(ac+bd\le\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\left(1\right)\)
Nếu : ac + bd < 0 : BĐT luôn đúng
Nếu : ac + bd \(\ge\) 0 : Thì (1) tương đương
( ac + bd )2 \(\le\) ( a2 + b2 )( c2 + d2 )
\(\Leftrightarrow\) \(\left(ac\right)^2+\left(bd\right)^2+2abcd\le\left(ac\right)^2+\left(ad\right)^2+\left(bc\right)^2+\left(bd\right)^2\)
\(\Leftrightarrow\) \(\left(ad\right)^2+\left(bc\right)^2-2abcd\ge0\)
\(\Leftrightarrow\) \(\left(ad-bc\right)^2\ge0\) , luôn đúng , vậy bài toán được chứng minh
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2) Chọn :\(\left\{{}\begin{matrix}a=2\cos x.\cos y\\c=2\sin x.\sin y\\b=d=\sin\left(x-y\right)\end{matrix}\right.\)
Từ câu 1) ta có :
\(\sqrt{4\cos^2x.\cos^2y+\sin^2\left(x-y\right)}+\sqrt{4\sin^2x.\sin^2y+\sin^2\left(x-y\right)}\)
\(\ge\sqrt{\left(2\cos x.\cos y+2\sin x.\sin y\right)^2+\left(2\sin\left(x-y\right)\right)^2}\)
\(\ge\sqrt{4\cos^2\left(x-y\right)+4\sin^2\left(x-y\right)}=2\)
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\(y=\dfrac{1}{\left(x^2-2x+5\right)^2}\)
y=2sin3xcos5x
\(y=\left(1+\sqrt{1-2x}\right)^3\)
\(y=x^2\sin\left(3x-1\right)\)
\(y=\dfrac{\sin x+\cos x}{\sin x-\cos x}\)
giải giúp e câu này với ạ
1) \(\sin^2x-\sin x=2\cos^2x\)
2) \(2\sin^2x+\left(1-\sqrt{3}\right)\cos\left(\frac{5pi}{2}-x\right)-\sin\frac{pi}{3}=0\)
3) \(\cos\left(3x+\frac{pi}{4}\right)=\cos\frac{pi}{8}\)
a/
\(sin^2x-sinx=2\left(1-sin^2x\right)\)
\(\Leftrightarrow3sin^2x-sinx-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\frac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{2}+k2\pi\\x=arcsin\left(\frac{2}{3}\right)+k2\pi\\x=\pi-arcsin\left(\frac{2}{3}\right)+k2\pi\end{matrix}\right.\)
2.
\(2sin^2x+\left(1-\sqrt{3}\right)sinx-\frac{\sqrt{3}}{2}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sinx=\frac{\sqrt{3}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\\x=\frac{\pi}{3}+k2\pi\\x=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
3.
\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{\pi}{4}=\frac{\pi}{8}+k2\pi\\3x+\frac{\pi}{4}=-\frac{\pi}{8}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{24}+\frac{k2\pi}{3}\\x=-\frac{\pi}{8}+\frac{k2\pi}{3}\end{matrix}\right.\)
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\(1.\sin^2x-\sin x=2\cdot\cos^2x\)
\(\Leftrightarrow\sin^2x-\sin x=2\cdot\left(1-\sin^2x\right)\)
\(\Leftrightarrow3\cdot\sin^2x-\sin x-2=0\)
\(\orbr{\begin{cases}\sin x=1\\\sin x=\frac{-2}{3}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\pi}{2}+k2\pi\\\orbr{\begin{cases}x=arcsin\left(\frac{-2}{3}\right)+k2\pi\\x=\pi-arcsin\left(\frac{-2}{3}\right)+k2\pi\end{cases}}\end{cases}}\)
\(\hept{\begin{cases}x=\frac{\pi}{2}+k2\pi\\x=arcsin\left(\frac{-2}{3}\right)+k2\pi\\x=\pi-arcsin\left(\frac{-2}{3}\right)+k2\pi\end{cases}}\)
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 \(x+\frac{\pi}{3}=a\Rightarrow2x=2a-\frac{2\pi}{3}\Rightarrow2x-\frac{\pi}{3}=2a-\pi\)
\(\sqrt{2}sina=-sin\left(2a-\pi\right)=sin2a=2sina.cosa\)
\(\Leftrightarrow\sqrt{2}sina\left(\sqrt{2}cosa-1\right)=0\)
rút gọn biểu thức:
A= cosa.sin( b-c )+ cosb. sin(c-a) + cosc.sin( a-b)
B= \(sin^2x+cos\left(\frac{\pi}{3}-x\right).cos\left(\frac{\pi}{3}+x\right)\)
C=\(sin^2x+sin^2\left(\frac{2\pi}{3}+x\right)+sin^2\left(\frac{2\pi}{3}-x\right)\)
D=\(sin^2\left(\frac{\pi}{4}+x\right)-sin^2x-2sinx.sin\frac{\pi}{4}.cos\left(\frac{\pi}{4}+x\right)\)
\(A=cosa\left(sinb.cosc-cosb.sinc\right)+cosb\left(sinc.cosa-cosc.sina\right)+cosc\left(sinacosb-cosasinb\right)\)
\(A=cosasinbcosc-cosacosbsinc+cosacosbsinc-sinacosbcosc+sinacosbcosc-cosasinbcosc\)
\(A=0\)
\(B=sin^2x+\frac{1}{2}\left(cos\frac{2\pi}{3}+cos2x\right)\)
\(B=\frac{1}{2}-\frac{1}{2}cos2x-\frac{1}{4}+\frac{1}{2}cos2x\)
\(B=\frac{1}{4}\)
\(C=\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}+2x\right)+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}-2x\right)\)
\(C=\frac{3}{2}-\frac{1}{2}cos2x-\frac{1}{2}\left(cos\left(\frac{4\pi}{3}+2x\right)+cos\left(\frac{4\pi}{3}-2x\right)\right)\)
\(C=\frac{3}{2}-\frac{1}{2}cos2x-cos\frac{4\pi}{3}.cos2x\)
\(C=\frac{3}{2}-\frac{1}{2}cos2x+\frac{1}{2}cos2x\)
\(C=\frac{3}{2}\)
\(D=\frac{1}{2}\left[\sqrt{2}sin\left(\frac{\pi}{4}+x\right)\right]^2-sin^2x-sinx.\sqrt{2}cos\left(\frac{\pi}{4}+x\right)\)
\(D=\frac{1}{2}\left(sinx+cosx\right)^2-sin^2x-sinx\left(sinx-cosx\right)\)
\(D=\frac{1}{2}\left(1+2sinx.cosx\right)-sin^2x-sin^2x+sinx.cosx\)
\(D=\frac{1}{2}+sinxcosx+sinxcosx=\frac{1}{2}+sin2x\)
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Góc độ cao của thang dựa vào tường là 60º và chân thang cách tường 4,6 m. Chiều dài của thang là
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Xét tình chẵn lẻ của hàm số
1) \(y=x.\sin2x+cot\left|x\right|+\frac{\left|1-\sin x\right|+\left|1+sinx\right|}{1+\sin^2x}\)
2) \(y=\frac{x^3}{\cos3x}+\frac{2\sin^3x}{1+\left|\cos x\right|}\)
3) \(y=\frac{x+\sin x}{\tan x+\cot x}\)
Giải phương trình:
1.cos^3x.cos3x+sin^3x.sin3xfrac{sqrt{2}}{4}
2.cos^34xcos^3x.cos3x+sin^3x.sin3x
3.cos^2x-4sin^2left(frac{x}{2}-frac{pi}{4}right)+20
4.sin^4x+sin^4left(x+frac{pi}{4}right)frac{1}{4}
5.sin^6x+cos^6xfrac{5}{6}left(sin^4x+cos^4xright)
6.sin^6x+cos^6x+frac{1}{2}sinx.cosx0
7.frac{1}{2}left(sin^4x+cos^4xright)sin^2x.cos^2x+sinx.cosx
8.sin^6x+cos^6x-3cos8x+20
9.sin^4x+cos^4x-2left(sin^6frac{x}{2}+cos^6frac{x}{2}right)+10
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Giải phương trình:
1.\(cos^3x.cos3x+sin^3x.sin3x=\frac{\sqrt{2}}{4}\)
2.\(cos^34x=cos^3x.cos3x+sin^3x.sin3x\)
3.\(cos^2x-4sin^2\left(\frac{x}{2}-\frac{\pi}{4}\right)+2=0\)
4.\(sin^4x+sin^4\left(x+\frac{\pi}{4}\right)=\frac{1}{4}\)
5.\(sin^6x+cos^6x=\frac{5}{6}\left(sin^4x+cos^4x\right)\)
6.\(sin^6x+cos^6x+\frac{1}{2}sinx.cosx=0\)
7.\(\frac{1}{2}\left(sin^4x+cos^4x\right)=sin^2x.cos^2x+sinx.cosx\)
8.\(sin^6x+cos^6x-3cos8x+2=0\)
9.\(sin^4x+cos^4x-2\left(sin^6\frac{x}{2}+cos^6\frac{x}{2}\right)+1=0\)
5.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\frac{5}{6}\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)
\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)
\(\Leftrightarrow\frac{1}{3}sin^22x=\frac{1}{6}\)
\(\Leftrightarrow sin^22x=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{2}}{2}\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)
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6.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-3sin^2x.cos^2x+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x+\frac{1}{4}sin2x=0\)
\(\Leftrightarrow-3sin^22x+sin2x+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow2x=-\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)
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1.
\(\Rightarrow4cos^3x.cos3x+4sin^3x.sin3x=\sqrt{2}\)
\(\Leftrightarrow\left(3cosx+cos3x\right)cos3x+\left(3sinx-sin3x\right)sin3x=\sqrt{2}\)
\(\Leftrightarrow3\left(cos3x.cosx+sin3x.sinx\right)+cos^23x-sin^23x=\sqrt{2}\)
\(\Leftrightarrow3cos2x+cos6x=\sqrt{2}\)
\(\Leftrightarrow3cos2x+4cos^32x-3cos2x=\sqrt{2}\)
\(\Leftrightarrow4cos^32x=\sqrt{2}\)
\(\Leftrightarrow cos2x=\frac{\sqrt{2}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{4}+k2\pi\\2x=-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\end{matrix}\right.\)
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Xem thêm câu trả lời
16 tháng 9 2020 lúc 14:50
GPT
a) \(sin\left(2x+1\right)+cos\left(3x-1\right)=0\)
b) \(sin\left(2x-\frac{\pi}{6}\right)=-sin\left(x-\frac{\pi}{4}\right)\)
c) \(sin\left(3x+\frac{2\pi}{3}\right)+sin\left(x-\frac{7\pi}{5}\right)=0\)
d) \(cos\left(4x+\frac{\pi}{3}\right)+sin\left(x-\frac{\pi}{4}\right)=0\)
a.
\(sin\left(2x+1\right)=-cos\left(3x-1\right)\)
\(\Leftrightarrow sin\left(2x+1\right)=sin\left(3x-1-\frac{\pi}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1-\frac{\pi}{2}=2x+1+k2\pi\\3x-1-\frac{\pi}{2}=\pi-2x-1+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+2+k2\pi\\x=\frac{3\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)
b.
\(sin\left(2x-\frac{\pi}{6}\right)=sin\left(\frac{\pi}{4}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=\frac{\pi}{4}-x+k2\pi\\2x-\frac{\pi}{6}=\frac{3\pi}{4}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{12}+k2\pi\end{matrix}\right.\)
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c.
\(\Leftrightarrow sin\left(3x+\frac{2\pi}{3}\right)=-sin\left(x-\frac{2\pi}{5}-\pi\right)\)
\(\Leftrightarrow sin\left(3x+\frac{2\pi}{3}\right)=sin\left(x-\frac{2\pi}{5}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{2\pi}{3}=x-\frac{2\pi}{5}+k2\pi\\3x+\frac{2\pi}{3}=\frac{7\pi}{5}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{8\pi}{15}+k\pi\\x=\frac{11\pi}{60}+\frac{k\pi}{2}\end{matrix}\right.\)
d.
\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=sin\left(\frac{\pi}{4}-x\right)\)
\(\Leftrightarrow cos\left(4x+\frac{\pi}{3}\right)=cos\left(\frac{\pi}{4}+x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{3}=\frac{\pi}{4}+x+k2\pi\\4x+\frac{\pi}{3}=-\frac{\pi}{4}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{36}+\frac{k2\pi}{3}\\x=-\frac{7\pi}{60}+\frac{k2\pi}{5}\end{matrix}\right.\)
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