\(cos^2x+\frac{1}{cos^2x}+cosx-\frac{1}{cosx}-\frac{7}{4}=0\)
mọi người ơi giúp em với :((em cảm ơn mọi người nhìu
1.\(cos\frac{4x}{3}=sin^2x\)
2.cos3x-cos2x+9sinx-4=0
mọi người ơi giải giúp em với !!!
cảm ơn mọi người nhiều ạaaa
1.
Đặt \(\frac{x}{3}=t\) pt trở thành:
\(cos4t=sin^23t\Leftrightarrow2cos4t=1-cos6t\)
\(\Leftrightarrow cos6t+2cos4t-1=0\)
\(\Leftrightarrow4cos^32t-3cos2t+2\left(2cos^22t-1\right)-1=0\)
\(\Leftrightarrow4cos^32t+2cos^22t-3cos2t-3=0\)
\(\Leftrightarrow\left(cos2t-1\right)\left(4cos^22t+6cos2t+3\right)=0\)
\(\Leftrightarrow cos2t=1\Leftrightarrow cos\frac{2x}{3}=1\)
\(\Leftrightarrow\frac{2x}{3}=k2\pi\Leftrightarrow x=k3\pi\)
2.
\(\Leftrightarrow4cos^3x-3cosx-\left(1-2sin^2x\right)+9sinx-4=0\)
\(\Leftrightarrow cosx\left(4cos^2x-3\right)+2sin^2x+9sinx-5=0\)
\(\Leftrightarrow cosx\left(4\left(1-sin^2x\right)-3\right)+\left(2sinx-1\right)\left(sinx+5\right)=0\)
\(\Leftrightarrow cosx\left(1-4sin^2x\right)+\left(2sinx-1\right)\left(sinx+5\right)=0\)
\(\Leftrightarrow\left(cosx+2sinx.cosx\right)\left(1-2sinx\right)-\left(1-2sinx\right)\left(sinx+5\right)=0\)
\(\Leftrightarrow\left(1-2sinx\right)\left(cosx-sinx+2sinx.cosx-5\right)=0\)
\(\Leftrightarrow\left(1-2sinx\right)\left(\sqrt{2}cos\left(x+\frac{\pi}{4}\right)+sin2x-5\right)=0\)
\(\Leftrightarrow1-2sinx=0\) (do \(\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\le\sqrt{2};sin2x\le1\) nên ngoặc sau luôn âm)
\(\Leftrightarrow sinx=\frac{1}{2}\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\text{2cos2x+cos^2\frac{x}{2}-10cos(\frac{5\pi}{2}-x)+\frac{7}{2}=\frac{1}{2}cosx}\)
2cos6x+sin4x+cos2x=0
MỌI NGƯỜI GIÚP MÌNH VỚI MÌNH CẢM ƠN NHIỀU
Câu đầu ko dịch được đề, lỗi kí tự rồi bạn
b/
\(\Leftrightarrow2cos^6x+sin^4x+2cos^2x-1=0\)
\(\Leftrightarrow2cos^2x\left(cos^4x+1\right)+\left(sin^2x-1\right)\left(sin^2x+1\right)=0\)
\(\Leftrightarrow cos^2x\left(2cos^4x+2\right)-cos^2x\left(sin^2x+1\right)=0\)
\(\Leftrightarrow cos^2x\left(2cos^4x+1-sin^2x=0\right)\)
\(\Leftrightarrow cos^2x\left(2cos^4x+cos^2x\right)=0\)
\(\Leftrightarrow cos^4x\left(2cos^2x+1\right)=0\)
\(\Leftrightarrow cos^4x=0\Leftrightarrow cosx=0\)
\(\Leftrightarrow x=\frac{\pi}{2}+k\pi\)
giải các pt
a) \(sin\left(\frac{3\pi}{10}-\frac{x}{2}\right)=\frac{1}{2}sin\left(\frac{\pi}{10}+\frac{3x}{2}\right)\)
b) \(4\left(sin^2x+\frac{1}{sin^2x}\right)+4\left(sinx+\frac{1}{sinx}\right)=7\)
c) \(9\left(\frac{2}{cosx}+cosx\right)+2\left(cos^2x+\frac{4}{cos^2x}\right)=1\)
d) \(2\left(cos^2x+\frac{4}{cos^2x}\right)+9\left(\frac{2}{cosx}-cosx\right)=1\)
a/
\(\Leftrightarrow cos\left(\frac{x}{2}+\frac{\pi}{5}\right)=\frac{1}{2}sin\left(\frac{3x}{2}+\frac{\pi}{10}\right)\)
Đặt \(\frac{x}{2}+\frac{\pi}{5}=a\Rightarrow\frac{x}{2}=a-\frac{\pi}{5}\Rightarrow\frac{3x}{2}=3a-\frac{3\pi}{5}\)
Pt trở thành:
\(cosa=\frac{1}{2}sin\left(3a-\frac{3\pi}{5}+\frac{\pi}{10}\right)\)
\(\Leftrightarrow cosa=\frac{1}{2}sin\left(3a-\frac{\pi}{2}\right)\)
\(\Leftrightarrow cosa=-\frac{1}{2}sin\left(\frac{\pi}{2}-3a\right)=-\frac{1}{2}cos3a\)
\(\Leftrightarrow cos3a+2cosa=0\)
\(\Leftrightarrow4cos^3a-3cosa+2cosa=0\)
\(\Leftrightarrow4cos^3a-cosa=0\)
\(\Leftrightarrow cosa\left(4cos^2a-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}cosa=0\\cosa=\frac{1}{2}\\cosa=-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}cos\left(\frac{x}{2}+\frac{\pi}{5}\right)=0\\cos\left(\frac{x}{2}+\frac{\pi}{5}\right)=\frac{1}{2}\\cos\left(\frac{x}{2}+\frac{\pi}{5}\right)=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{2}+\frac{\pi}{5}=\frac{\pi}{2}+k\pi\\\frac{x}{2}+\frac{\pi}{5}=\pm\frac{\pi}{3}+k2\pi\\\frac{x}{2}+\frac{\pi}{5}=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\) \(\Rightarrow x=...\) (5 nghiệm bạn tự biến đổi)
b/
ĐKXĐ: ...
Đặt \(sinx+\frac{1}{sinx}=a\Rightarrow sin^2x+\frac{1}{sin^2x}=a^2-2\)
Pt trở thành:
\(4\left(a^2-2\right)+4a=7\)
\(\Leftrightarrow4a^2+4a-15=0\Rightarrow\left[{}\begin{matrix}a=\frac{3}{2}\\a=-\frac{5}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sinx+\frac{1}{sinx}=\frac{3}{2}\\sinx+\frac{1}{sinx}=-\frac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin^2x-\frac{3}{2}sinx+1=0\left(vn\right)\\sin^2x+\frac{5}{2}sinx+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sinx=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
c/
ĐKXĐ: ...
Đặt \(cosx+\frac{2}{cosx}=a\Rightarrow cos^2x+\frac{4}{cos^2x}=a^2-4\)
Pt trở thành:
\(9a+2\left(a^2-4\right)=1\)
\(\Leftrightarrow2a^2+9a-9=0\)
Pt này nghiệm xấu quá bạn :(
d/ĐKXĐ: ...
Đặt \(\frac{2}{cosx}-cosx=a\Rightarrow cos^2x+\frac{4}{cos^2x}=a^2+4\)
Pt trở thành:
\(2\left(a^2+4\right)+9a-1=0\)
\(\Leftrightarrow2a^2+9a+7=0\Rightarrow\left[{}\begin{matrix}a=-1\\a=-\frac{7}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{2}{cosx}-cosx=-1\\\frac{2}{cosx}-cosx=-\frac{7}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-cos^2x+cosx+2=0\\-cos^2x+\frac{7}{2}cosx+2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}cosx=-1\\cosx=2\left(l\right)\\cosx=4\left(l\right)\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pi+k2\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
Chứng minh các đẳng thức sau:
a.\(\frac{1+sin^2x}{1-sin^{2^{ }}x}=1+2tan^2x\)
b.\(\frac{sin^3a-cos^3a}{sina-cosa}-sina.cosa=1\)
c.\(\frac{1+cosx+cos2x+cos3x}{2cos^2x+cosx-1}=2cosx\)
e.\(\frac{1-2sin^2a}{cosa+sina}+\frac{2cos^2a-1}{cosa-sina}=2cosa\)
d.\(\frac{1-cosx+cos2x}{sin2x-sinx}=cotx\)
MỌI NGƯỜI GIÚP MÌNH VỚI .MÌNH CẢM ƠN RẤT NHIỀU
\(\frac{1+sin^2x}{1-sin^2x}=\frac{1+sin^2x}{cos^2x}=\frac{1}{cos^2x}+\frac{sin^2x}{cos^2x}=1+tan^2x+tan^2x=1+2tan^2x\)
\(\frac{sin^3a-cos^3a}{sina-cosa}-sina.cosa=\frac{\left(sina-cosa\right)\left(sin^2a+cos^2a+sina.cosa\right)}{sina-cosa}-sina.cosa\)
\(=sin^2a+cos^2a+sina.cosa-sina.cosa=1\)
\(\frac{1+cos2x+cosx+cos3x}{2cos^2x-1+cosx}=\frac{1+2cos^2x-1+2cosx.cos2x}{cos2x+cosx}=\frac{2cosx\left(cosx+cos2x\right)}{cos2x+cosx}=2cosx\)
\(\frac{1-2sin^2a}{cosa+sina}+\frac{2cos^2a-1}{cosa-sina}=\frac{cos^2a-sin^2a}{cosa+sina}+\frac{cos^2a-sin^2a}{cosa-sina}\)
\(=\frac{\left(cosa+sina\right)\left(cosa-sina\right)}{cosa+sina}+\frac{\left(cosa+sina\right)\left(cosa-sina\right)}{cosa-sina}=cosa-sina+cosa+sina=2cosa\)
\(\frac{1-cosx+cos2x}{sin2x-sinx}=\frac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}=\frac{cosx\left(2cosx-1\right)}{sinx\left(2cosx-1\right)}=\frac{cosx}{sinx}=cotx\)
Mọi người giúp em với, em cảm ơn ạ
Bài tập quy về dạng phương trình cơ bản:
\(1.\sin\left(x-\frac{\pi}{3}\right)+2cos\left(x-\frac{\pi}{6}\right)=0\);
\(2.\sin^23x=cos^2x\);
\(3.sin\left(2x-\frac{7\pi}{2}\right)+cos2x=1\)
\(4.\sqrt{2}cos\left(x-\frac{3\pi}{4}\right)=1+sinx\)
\(5.\sin\left(2x-\frac{7\pi}{2}\right)+cós2x=1\)
Câu 1:
\(\Leftrightarrow sinx.cos\frac{\pi}{3}-cosx.sin\frac{\pi}{3}+2\left(cosx.cos\frac{\pi}{6}+sinx.sin\frac{\pi}{6}\right)=0\)
\(\Leftrightarrow sinx+\frac{1}{\sqrt{3}}cosx=0\)
Nhận thấy \(cosx=0\) không phải nghiệm, chia 2 vế cho \(cosx\)
\(tanx+\frac{1}{\sqrt{3}}=0\Rightarrow tanx=-\frac{1}{\sqrt{3}}\Rightarrow x=\frac{\pi}{6}+k\pi\)
Câu 2:
\(\Leftrightarrow1-cos6x=1+cos2x\)
\(\Leftrightarrow-cos6x=cos2x\)
\(\Leftrightarrow cos\left(\pi-6x\right)=cos2x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\pi-6x+k2\pi\\2x=6x-\pi+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\frac{\pi}{4}+\frac{k\pi}{2}\end{matrix}\right.\)
Câu 3:
\(\Leftrightarrow sin\left(2x+\frac{\pi}{2}-4\pi\right)+cos2x=1\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{2}\right)+cos2x=1\)
\(\Leftrightarrow cos2x+cos2x=1\)
\(\Leftrightarrow cos2x=\frac{1}{2}\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{3}+k2\pi\\2x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)
Câu 4:
\(\sqrt{2}\left(cosx.cos\frac{3\pi}{4}+sinx.sin\frac{3\pi}{4}\right)=1+sinx\)
\(\Leftrightarrow-cosx+sinx=1+sinx\)
\(\Leftrightarrow cosx=-1\Rightarrow x=\pi+k\pi2\)
Câu 5:
Giống câu 3, chắc bạn ghi nhầm đề
giải các phường trình sau:
a/\(sin^3x+cos^3x=sinx+cosx\)
b/\(sin^3x+2sin^2xcosx-3cos^3x=0\)
c/\(3cos^4x-4cos^2xsin^2x-sin^4x=0\)
d/\(sinx-4sin^3x+cosx=0\)
mọi người giúp em với em cảm ơn mọi người nhìu
\(a\text{) }sin^3x+cos^3x=sinx+cosx\\ \Leftrightarrow\left(sinx+cosx\right)\left(sin^2x-sinx\cdot cosx+cos^2x\right)=sinx+cosx\\ \Leftrightarrow-\frac{1}{2}sin2x\left(sinx+cosx\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}sinx=-cosx=sin\left(x-\frac{\pi}{2}\right)\\sin2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3\pi}{2}-x+a2\pi\\2x=b\pi\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\frac{3\pi}{4}+a\pi\\x=\frac{b\pi}{2}\end{matrix}\right.\)
\(\text{b) }sin^3x+2sin^2x\cdot cosx-3cos^3x=0\\ \Leftrightarrow\left(sin^3x-cos^3x\right)+2cosx\cdot\left(sin^2x-cos^2x\right)=0\\ \Leftrightarrow\left(sinx-cosx\right)\left(sinx\cdot cosx+1\right)+\left(sinx-cosx\right)\left(2sinx\cdot cosx+2cos^2x\right)=0\\ \Leftrightarrow\left(sinx-cosx\right)\left(3sinx\cdot cosx+1+2cos^2x\right)=0\\ \Leftrightarrow\left(sinx-cosx\right)\left(\frac{3}{2}sin2x+2+cos2x\right)=0\)
Với \(sinx-cosx=0\)
\(\Leftrightarrow sinx=cosx=sin\left(\frac{\pi}{2}-x\right)\\ \Leftrightarrow x=\frac{\pi}{2}-x+a2\pi\\ \Leftrightarrow x=\frac{\pi}{4}+a\pi\)
Với \(\frac{3}{2}sin2x+2+cos2x=0\)
\(\Leftrightarrow sin^22x+\left(\frac{3}{2}sin2x+2\right)^2=1\left(VN\right)\)
\(\text{c) }3cos^4x-4cos^2x\cdot sin^2x-sin^4x=0\)
Nhận thấy sinx=0 không là nghiệm pt.
Chia cả 2 vế cho sin4x ta được
\(pt\Leftrightarrow\frac{3cos^4x}{sin^4x}-\frac{4cos^2x}{sin^2x}-1=0\\ \Leftrightarrow3cot^4x-4cot^2x-1=0\\ \Leftrightarrow cot^2x=\frac{2+\sqrt{7}}{3}\\ \Leftrightarrow cotx=\pm\sqrt{\frac{2+\sqrt{7}}{3}}\\ \Leftrightarrow x=arccot\left(\pm\sqrt{\frac{2+\sqrt{7}}{3}}\right)+k2\pi\)
d) kiểm tra đề.
giải các phương trình sau
a/ \(^{tan^2x-\frac{4}{cotx}+=0}\)
b/\(cos2\left(x+\frac{\text{π}}{3}\right)+4cos\left(\frac{\text{π}}{6}-x\right)=\frac{5}{2}\)
c/\(\frac{1}{cos^2x}-1+tanx-\sqrt{3}\left(tanx+1\right)=0\)
d/tanx-2cotx+1=0
Mọi người ơi giúp mình với <3 cảm ơn mọi người nhìu ạ
Bạn ghi đề chính xác ra đi, câu a và câu b đó bạn
Câu a sau \(\frac{4}{cotx}\) còn dấu + nhưng không biết cộng với cái gì
Câu b biểu thức cos đầu tiên là \(cos^2\left(x+\frac{\pi}{3}\right)\) hay \(cos\left(2x+\frac{2\pi}{3}\right)\)
a) Đề thiếu
b)
PT $\Leftrightarrow 1-2\sin^2(x+\frac{\pi}{3})+4\cos (\frac{\pi}{6}-x)-\frac{5}{2}=0$
$\Leftrightarrow 1-2\sin ^2[\frac{\pi}{2}-(\frac{\pi}{6}-x)]+4\cos (\frac{\pi}{6}-x)-\frac{5}{2}=0$
$\Leftrightarrow -2\cos ^2(\frac{\pi}{6}-x)+4\cos (\frac{\pi}{6}-x)-\frac{3}{2}=0$
$\Leftrightarrow -2t^2+4t-\frac{3}{2}=0$ với $t=\cos (\frac{\pi}{6}-x)$
Đến đây bạn giải pt bậc 2 thu được $\cos (\frac{\pi}{6}-x)=\frac{1}{2}$
$\Rightarrow x=2k\pi +\frac{\pi}{2}$ hoặc $x=2k\pi -\frac{\pi}{6}$ với $k$ nguyên
c)
ĐK:.............
PT $\Leftrightarrow 1+\frac{\sin ^2x}{\cos ^2x}-1+\tan x-\sqrt{3}(\tan x+1)=0$
$\Leftrightarrow \tan ^2x+\tan x-\sqrt{3}(\tan x+1)=0$
$\Leftrightarrow \tan ^2x+(1-\sqrt{3})\tan x-\sqrt{3}=0$
$\Rightarrow \tan x=\sqrt{3}$ hoặc $\tan x=-1$
$\Rightarrow x=\pi (k-\frac{1}{4})$ hoặc $x=\pi (k+\frac{1}{3})$ với $k$ nguyên
d)
ĐK:.......
PT $\Leftrightarrow \tan x-\frac{2}{\tan x}+1=0$
$\Leftrightarrow \tan ^2x+\tan x-2=0$
$\Leftrightarrow (\tan x-1)(\tan x+2)=0$
$\Rightarrow \tan x=1$ hoặc $\tan x=-2$
$\Rightarrow x=k\pi +\frac{\pi}{4}$ hoặc $x=k\pi +\tan ^{-2}(-2)$ với $k$ nguyên.
giải phương trình
1.\(sin^3x+2cosx-2+sin^2x=0\)
\(2.\frac{\sqrt{3}}{2}sin2x+\sqrt{2}cos^2x+\sqrt{6}cosx=0\)
3.\(2sin2x-cos2x=7sinx+2cosx-4\)
4.\(2cos2x-8cosx+7=\frac{1}{cosx}\)
5.\(cos^8x+sin^8x=2\left(cos^{10}x+sin^{10}x\right)+\frac{5}{4}cos2x\)
6.\(1+sinx+cos3x=cosx+sin2x+cos2x\)
7.\(1+sinx+cosx+sin2x+cos2x=0\)
1.
\(\Leftrightarrow sin^2x\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cos^2x\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cosx\right)\left(1+cosx\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cosx\right)\left(sinx+cosx+sinx.cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\Leftrightarrow...\\sinx+cosx+sinx.cosx-1=0\left(1\right)\end{matrix}\right.\)
Xét (1):
Đặt \(sinx+cosx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
\(\Leftrightarrow t+\frac{t^2-1}{2}-1=0\)
\(\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
2.
\(\Leftrightarrow\sqrt{3}sinx.cosx+\sqrt{2}cos^2x+\sqrt{6}cosx=0\)
\(\Leftrightarrow cosx\left(\sqrt{3}sinx+\sqrt{2}cosx+\sqrt{6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Leftrightarrow...\\\sqrt{3}sinx+\sqrt{2}cosx=-\sqrt{6}\left(1\right)\end{matrix}\right.\)
Xét (1):
Do \(\sqrt{3}^2+\sqrt{2}^2< \left(-\sqrt{6}\right)^2\) nên (1) vô nghiệm
3.
\(\Leftrightarrow4sinx.cosx-\left(1-2sin^2x\right)=7sinx+2cosx-4\)
\(\Leftrightarrow2cosx\left(2sinx-1\right)+2sin^2x-7sinx+3=0\)
\(\Leftrightarrow2cosx\left(2sinx-1\right)+\left(sinx-3\right)\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left(2cosx+sinx-3\right)\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\Leftrightarrow...\\2cosx+sinx=3\left(1\right)\end{matrix}\right.\)
Xét (1), do \(2^2+1^2< 3^2\) nên (1) vô nghiệm
Mọi người giúp em giải bài này ạ, em cảm ơn
Bài 1: Rút gọn biểu thức:
A=\(\frac{\sin2x+\sin x}{1+\cos2x+\cos x}\)
B=\(cota\left(\frac{1+\sin^2a}{\cos a}-cosa\right)\)
C=\(\frac{1+\cos x+\cos2x+\cos3x}{2\cos^2x+\cos x-1}\)
D=\(\frac{2\cos\left(\frac{\pi}{2}-x\right)\cdot\sin\left(\frac{\pi}{2}+x\right)\cdot\tan\left(\pi-x\right)}{\cot\left(\frac{\pi}{2}+x\right)\cdot\sin\left(\pi-x\right)}-2\cos x\)
E=\(\cos^2x\cdot\cot^2x+3\cos^2x-\cot^2x+2\sin^2x\)
\(F=\frac{\sin^2x+\sin^2x\tan^2x}{\cos^2x+\cos^2x\tan^2x}\)
\(G=\frac{1+cos2a-cosa}{2sina-sina}\)
H=\(sin^{^{ }4}\left(\frac{\pi}{2}+\alpha\right)-cos^4\left(\frac{3\pi}{2}-\alpha\right)+1\)
Bài 2: chứng minh
a) cho \(\Delta ABCchứngminhsin\frac{A+B}{2}=cos\frac{C}{2}\)
b) chứng minh biểu thức sau độc lập với biến x:
A=\(cosx+cos\left(x+\frac{2\pi}{3}\right)+cos\left(x+\frac{4\pi}{3}\right)\)
c)cho \(\Delta\) ABC chứng minh : sin A+sin B+ sin C= \(4cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}\)
d)CMR: \(\frac{cos2a}{1+sin2a}=\frac{cosa-sina}{cosa+sina}\)
e) CMR:\(E=\frac{sin\alpha+cos\alpha}{cos^3\alpha}=1+tan\alpha+tan^2\alpha+tan^3\alpha\)
f) CMR \(\Delta\)ABC cân khi và chỉ khi \(sinB=2cosAsinC\)
g) CM: \(\frac{1-cosx+cos2x}{sin2x-sinx}=cotx\)
h)CM: \(\left(cos3x-cosx\right)^2+\left(sin3x-sinx\right)^2=4sin^2x\)
k) CMR trong tam giac ABC ta có: \(sin2A+sin2B+sin2C=4sinA\cdot sinB\cdot sinC\)
Bài 3: giải bất phương trình:
a)\(\frac{\left(1-3x\right)\left(2x^2+1\right)}{-2x^2-3x+5}>0\)
b)\(\frac{2x+1}{\left(x-1\right)\left(x+2\right)}\ge0\)
c)\(\frac{\left(3x-2\right)\left(x^2-9\right)}{x^2-4x+4}\le0\)
d)\(\frac{\left(2x^2+3x\right)\left(3-2x\right)}{1-x^2}\ge0\)
e)\(\frac{\left(x^2+2x+1\right)\left(x-1\right)}{3-x^2}\)
f)\(\frac{2x+1}{-x^2+x+6}\ge0\)
\(A=\frac{2sinx.cosx+sinx}{1+2cos^2x-1+cosx}=\frac{sinx\left(2cosx+1\right)}{cosx\left(2cosx+1\right)}=\frac{sinx}{cosx}=tanx\)
\(B=\frac{cosa}{sina}\left(\frac{1+sin^2a}{cosa}-cosa\right)=\frac{cosa}{sina}\left(\frac{1+sin^2a-cos^2a}{cosa}\right)=\frac{cosa}{sina}.\frac{2sin^2a}{cosa}=2sina\)
\(C=\frac{1+cos2x+cosx+cos3x}{2cos^2x-1+cosx}=\frac{1+2cos^2x-1+2cos2x.cosx}{cos2x+cosx}=\frac{2cosx\left(cosx+cos2x\right)}{cos2x+cosx}=2cosx\)
\(D=\frac{2sinx.cosx.\left(-tanx\right)}{-tanx.sinx}-2cosx=2cosx-2cosx=0\)
\(E=cos^2x.cot^2x-cot^2x+cos^2x+2cos^2x+2sin^2x\)
\(E=cot^2x\left(cos^2x-1\right)+cos^2x+2=\frac{cos^2x}{sin^2x}\left(-sin^2x\right)+cos^2x+2=2\)
\(F=\frac{sin^2x\left(1+tan^2x\right)}{cos^2x\left(1+tan^2x\right)}=\frac{sin^2x}{cos^2x}=tan^2x\)
Câu G mẫu số có gì đó sai sai, sao lại là \(2sina-sina?\)
\(H=sin^4\left(\frac{\pi}{2}+a\right)-cos^4\left(\frac{3\pi}{2}-a\right)+1=cos^4a-sin^4a+1\)
\(=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+1=cos^2a-\left(1-cos^2a\right)+1=2cos^2a\)
Bài 2:
\(sin\frac{A+B}{2}=sin\left(\frac{180^0-C}{2}\right)=sin\left(90^0-\frac{C}{2}\right)=cos\frac{C}{2}\)
b/
\(A=cosx+cos\left(x+\frac{2\pi}{3}\right)+cos\left(x+\frac{4\pi}{3}\right)=cosx+2cos\left(x+\pi\right).cos\frac{\pi}{3}\)
\(=cosx-2cosx.\frac{1}{2}=0\)
c/
\(sinA+sinB+sinC=2sin\frac{A+B}{2}cos\frac{A-B}{2}+2sin\frac{C}{2}cos\frac{C}{2}=2cos\frac{C}{2}.cos\frac{A-B}{2}+2sin\frac{C}{2}cos\frac{C}{2}\)
\(=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+sin\frac{C}{2}\right)=2cos\frac{C}{2}\left(cos\frac{A-B}{2}+cos\frac{A+B}{2}\right)=4cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}\)
d/ \(\frac{cos2a}{1+sin2a}=\frac{cos^2a-sin^2a}{cos^2a+sin^2a+2sina.cosa}=\frac{\left(cosa-sina\right)\left(cosa+sina\right)}{\left(cosa+sina\right)^2}=\frac{cosa-sina}{cosa+sina}\)
e/
\(E=\frac{sina+cosa}{cos^3a}=\frac{1}{cos^2a}\left(tana+1\right)=\left(1+tan^2a\right)\left(tana+1\right)\)
\(E=tan^3a+tan^2a+tana+1\)