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\(a,\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4sin^2x.cos^2x}=-1\)
\(VT=\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4.sin^2x.cos^2x}=\left(\frac{1}{tan2x}\right)^2-\frac{1}{sin^22x}=\left(\frac{cos2x}{sin2x}\right)^2-\frac{1}{sin^22x}=\frac{cos^22x-1}{sin^22x}=\frac{-sin^22x}{sin^22x}=-1=VP\)
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
\(cot^2x-cos^2x=\frac{cos^2x}{sin^2x}-cos^2x=cos^2x\left(\frac{1}{sin^2x}-1\right)=\frac{cos^2x\left(1-sin^2x\right)}{sin^2x}\)
\(=cos^2x.\left(\frac{cos^2x}{sin^2x}\right)=cot^2x.cos^2x\)
\(\frac{cosx+sinx}{cosx-sinx}-\frac{cosx-sinx}{cosx+sinx}=\frac{\left(cosx+sinx\right)^2-\left(cosx-sinx\right)^2}{\left(cosx-sinx\right)\left(cosx+sinx\right)}\)
\(=\frac{cos^2x+sin^2x+2sinx.cosx-\left(cos^2x+sin^2x-2sinx.cosx\right)}{cos^2x-sin^2x}=\frac{4sinx.cosx}{cos2x}=\frac{2sin2x}{cos2x}=2tan2x\)
\(\frac{sin4x+cos2x}{1-cos4x+sin2x}=\frac{2sin2x.cos2x+cos2x}{1-\left(1-2sin^22x\right)+sin2x}=\frac{cos2x\left(2sin2x+1\right)}{sin2x\left(2sin2x+1\right)}=\frac{cos2x}{sin2x}=cot2x\)
\(A=sin^2x\left(sinx+cosx\right)+cos^2x\left(sinx+cosx\right)\)
\(=\left(sin^2x+cos^2x\right)\left(sinx+cosx\right)=sinx+cosx\)
\(B=\frac{sinx}{cosx}\left(\frac{1+cos^2x-sin^2x}{sinx}\right)=\frac{sinx}{cosx}\left(\frac{2cos^2x}{sinx}\right)=2cosx\)
đề bài đầy đủ: rút gọn các biểu thức lượng giác sau trên điều kiện xác định của chúng:
\(\frac{sin^2x}{cosx+cosx.\frac{sinx}{cosx}}-\frac{cos^2x}{sinx+sinx.\frac{cosx}{sinx}}=\frac{sin^2x}{sinx+cosx}-\frac{cos^2x}{sinx+cosx}=\frac{sin^2x-cos^2x}{sinx+cosx}\)
\(=\frac{\left(sinx+cosx\right)\left(sinx-cosx\right)}{sinx+cosx}=sinx-cosx\)
\(\left(\frac{sinx}{cosx}+\frac{cosx}{1+sinx}\right)\left(\frac{cosx}{sinx}+\frac{sinx}{1+cosx}\right)=\left(\frac{sinx+sin^2x+cos^2x}{cosx\left(1+sinx\right)}\right)\left(\frac{cosx+cos^2x+sin^2x}{sinx\left(1+cosx\right)}\right)\)
\(=\left(\frac{sinx+1}{cosx\left(1+sinx\right)}\right)\left(\frac{cosx+1}{sinx\left(1+cosx\right)}\right)=\frac{1}{sinx.cosx}\)
\(\frac{sin^2a+1}{2.cos^2a}+\frac{1+cos^2a}{2.sin^2a}+1=\frac{tan^2a}{2}+\frac{1}{2cos^2a}+\frac{cot^2a}{2}+\frac{1}{2sin^2a}+1\)
\(=\frac{1}{2}\left(tan^2a+1+tan^2a+cot^2a+1+cot^2a+2\right)\)
\(=\frac{1}{2}\left(2tan^2a+4+2cot^2a\right)=tan^2a+2+cot^2a=\left(tana+cota\right)^2\)
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
\(A=\frac{2sinx.cosx+sinx}{1+2cos^2x-1+cosx}=\frac{sinx\left(2cosx+1\right)}{cosx\left(2cosx+1\right)}=\frac{sinx}{cosx}=tanx\)
\(B=\frac{cosa}{sina}\left(\frac{1+sin^2a}{cosa}-cosa\right)=\frac{cosa}{sina}\left(\frac{1+sin^2a-cos^2a}{cosa}\right)=\frac{cosa}{sina}.\frac{2sin^2a}{cosa}=2sina\)
\(C=\frac{1+cos2x+cosx+cos3x}{2cos^2x-1+cosx}=\frac{1+2cos^2x-1+2cos2x.cosx}{cos2x+cosx}=\frac{2cosx\left(cosx+cos2x\right)}{cos2x+cosx}=2cosx\)
\(D=\frac{2sinx.cosx.\left(-tanx\right)}{-tanx.sinx}-2cosx=2cosx-2cosx=0\)
\(E=cos^2x.cot^2x-cot^2x+cos^2x+2cos^2x+2sin^2x\)
\(E=cot^2x\left(cos^2x-1\right)+cos^2x+2=\frac{cos^2x}{sin^2x}\left(-sin^2x\right)+cos^2x+2=2\)
\(F=\frac{sin^2x\left(1+tan^2x\right)}{cos^2x\left(1+tan^2x\right)}=\frac{sin^2x}{cos^2x}=tan^2x\)
Câu G mẫu số có gì đó sai sai, sao lại là \(2sina-sina?\)
\(H=sin^4\left(\frac{\pi}{2}+a\right)-cos^4\left(\frac{3\pi}{2}-a\right)+1=cos^4a-sin^4a+1\)
\(=\left(cos^2a-sin^2a\right)\left(cos^2a+sin^2a\right)+1=cos^2a-\left(1-cos^2a\right)+1=2cos^2a\)
\(\frac{1+sin^2x}{1-sin^2x}=\frac{1+sin^2x}{cos^2x}=\frac{1}{cos^2x}+\frac{sin^2x}{cos^2x}=1+tan^2x+tan^2x=1+2tan^2x\)
\(\frac{sin^3a-cos^3a}{sina-cosa}-sina.cosa=\frac{\left(sina-cosa\right)\left(sin^2a+cos^2a+sina.cosa\right)}{sina-cosa}-sina.cosa\)
\(=sin^2a+cos^2a+sina.cosa-sina.cosa=1\)
\(\frac{1+cos2x+cosx+cos3x}{2cos^2x-1+cosx}=\frac{1+2cos^2x-1+2cosx.cos2x}{cos2x+cosx}=\frac{2cosx\left(cosx+cos2x\right)}{cos2x+cosx}=2cosx\)
\(\frac{1-2sin^2a}{cosa+sina}+\frac{2cos^2a-1}{cosa-sina}=\frac{cos^2a-sin^2a}{cosa+sina}+\frac{cos^2a-sin^2a}{cosa-sina}\)
\(=\frac{\left(cosa+sina\right)\left(cosa-sina\right)}{cosa+sina}+\frac{\left(cosa+sina\right)\left(cosa-sina\right)}{cosa-sina}=cosa-sina+cosa+sina=2cosa\)
\(\frac{1-cosx+cos2x}{sin2x-sinx}=\frac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}=\frac{cosx\left(2cosx-1\right)}{sinx\left(2cosx-1\right)}=\frac{cosx}{sinx}=cotx\)
\(sina+cosa=\sqrt{2}\Leftrightarrow\left(sina+cosa\right)^2=2\\ \)
\(\Leftrightarrow\sin^2a+2\sin a.cosa+cos^2a=2\)
\(\Leftrightarrow1+2.sina.cosa=2\)
\(\Leftrightarrow2.sina.cosa=2-1=1\)
\(\Leftrightarrow\sin a.cosa=\frac{1}{2}\)
Vậy P=sina.cosa=\(\frac{1}{2}\)
\(Q=\sin^4a+cos^4a\)
\(\Leftrightarrow\left(sin^2a\right)^2+\left(cos^2a\right)^2\)
\(\Leftrightarrow\left(sin^2a+cos^2a\right)^2-2.sin^2a.cos^2a\)
\(\Leftrightarrow1^2-2.sin^2a.cos^2a\) tách tiếp rồi thế vào là được .tương tự phàn P ý
còn R thì tách sin^3a=sin^2a+sina tương tự cos mũ 3 a cụng vậy
theo tớ là như thế còn có sai thì đừng có ném đá ném gạch na
\(\frac{1-cosx+cos2x}{sin2x-sinx}=\frac{1-cosx+2cos^2x-1}{2sinx.cosx-sinx}=\frac{cosx\left(2cosx-1\right)}{sinx\left(2cosx-1\right)}=\frac{cosx}{sinx}=cotx\)
\(A=sin\left(\frac{\pi}{4}+x\right)-sin\left(\frac{\pi}{2}-\frac{\pi}{4}+x\right)=sin\left(\frac{\pi}{4}+x\right)-sin\left(\frac{\pi}{4}+x\right)=0\)
cos đó bạn
Lời giải:
a)
\(\cos 2a=\frac{2}{5}\Rightarrow \sin ^22a=1-(\cos 2a)^2=1-(\frac{2}{5})^2=\frac{21}{25}\)
Vì $a\in (0; \frac{\pi}{4})\Rightarrow 2a\in (0; \frac{\pi}{2})$
$\Rightarrow \sin 2a>0\Rightarrow \sin 2a=\frac{\sqrt{21}}{5}$
$\tan 2a=\frac{\sin 2a}{\cos 2a}=\frac{\sqrt{21}}{5.\frac{2}{5}}=\frac{\sqrt{21}}{2}$
$\cot 2a=\frac{1}{\tan 2a}=\frac{2}{\sqrt{21}}$
-------------------------
$\sin 2a=\frac{24}{25}\Rightarrow \cos ^22a=1-(\sin 2a)^2=\frac{49}{625}$
$a\in [\frac{-3}{4}\pi; \frac{-\pi}{2}]\Rightarrow 2a\in [\frac{-3}{2}\pi ; -\pi]\Rightarrow \cos 2a< 0$
$\Rightarrow \cos 2a=\frac{-7}{25}$
$\Rightarrow \tan 2a=\frac{\sin 2a}{\cos 2a}=\frac{24}{25.\frac{-7}{25}}=\frac{-24}{7}$
$\Rightarrow \cot 2a=\frac{-7}{24}$
Bài 4:
\(\sin x=\frac{3}{5}\Rightarrow \sin ^2x=\frac{9}{25}\)
\(A=\frac{\cot x}{\cot x-\tan x}=\frac{\frac{\cos x}{\sin x}}{\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}}=\frac{\cos ^2x}{\cos ^2x-\sin ^2x}=\frac{1-\sin ^2x}{1-\sin ^2x-\sin ^2x}\)
\(=\frac{1-\sin ^2x}{1-2\sin ^2x}=\frac{1-\frac{9}{25}}{1-\frac{18}{25}}=\frac{16}{7}\)
--------------------
\(\cos a=\frac{1}{2}\Rightarrow \sin ^2a=1-\cos ^2a=\frac{3}{4}\)
Vì $a\in (1,5\pi; 2\pi)\Rightarrow \sin a< 0\Rightarrow \sin a=\frac{-\sqrt{3}}{2}$
\(B=\sin a+\cos a\tan a=\sin a+\cos a. \frac{\sin a}{\cos a}=2\sin a=2.\frac{-\sqrt{3}}{2}=-\sqrt{3}\)
Bài 3:
ĐK:..............
\(\tan x=2\Leftrightarrow \frac{\sin x}{\cos x}=2\Rightarrow \sin x=2\cos x\)
\(\Rightarrow \sin ^2x=4\cos ^2x\Rightarrow 5\cos ^2x=\sin ^2x+\cos ^2x=1\Rightarrow \cos x=\pm \sqrt{\frac{1}{5}}\)
\(A=\frac{5\cos x+6\tan x}{5\cos x-6\tan x}=\frac{5\cos x+12}{5\cos x-12}=1+\frac{24}{5\cos x-12}=1+\frac{24}{5.\pm \sqrt{\frac{1}{5}}-12}\)
\(=1+\frac{24}{\pm \sqrt{5}-12}\)
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ĐK:...........
\(\cot x=-6\Rightarrow \cos x=-6\sin x\)
\(B=\frac{4\sin x\cos x-3\cos ^2x}{1+3\sin ^2x}=\frac{4\sin x.(-6\sin x)-3(-6\sin x)^2}{\sin ^2x+\cos ^2x+3\sin ^2x}\)
\(=\frac{-132\sin ^2x}{4\sin ^2x+\cos ^2x}=\frac{-132\sin ^2x}{4\sin ^2x+(-6\sin x)^2}=\frac{-132\sin ^2x}{40\sin ^2x}=\frac{-33}{10}\)
Bài 2: ĐK:..............
\(A=\frac{\cot ^2x-\cos ^2x}{\cot ^2x}+\frac{\sin x\cos x}{\cot x}=1-(\frac{\cos x}{\cot x})^2+\frac{\sin x\cos x}{\cot x}\)
\(=1-(\frac{\cos x}{\frac{\cos x}{\sin x}})^2+\frac{\sin x\cos x}{\frac{\cos x}{\sin x}}=1-\sin ^2x+\sin ^2x=1\) là giá trị không phụ thuộc vào biến $x$
Ta có đpcm.
\(B=\cos ^4x+\sin ^2x\cos ^2x+\sin ^2x=\cos ^2x(\cos ^2x+\sin ^2x)+\sin ^2x\)
\(=\cos ^2x.1+\sin ^2x=\cos ^2x+\sin ^2x=1\) là giá trị không phụ thuộc vào biến $x$
Ta có đpcm.
Bài 1:
a)
ĐK: $\cos x\neq 0$ $\Rightarrow \sin x\neq -1$. Ta có:
\(\frac{1-\sin x}{\cos x}=\frac{(1-\sin x)(1+\sin x)}{\cos x(1+\sin x)}=\frac{1-\sin ^2x}{\cos x(1+\sin x)}=\frac{\cos ^2x}{\cos x(1+\sin x)}=\frac{\cos x}{1+\sin x}\)
(đpcm)
b) ĐK: $\sin x; \cos x\neq 0$
\(\frac{\tan x}{\sin x}-\frac{\sin x}{\cot x}=\frac{\tan x\cot x-\sin ^2x}{\sin x\cot x}=\frac{1-\sin ^2x}{\sin x.\cot x}=\frac{\cos ^2x}{\sin x.\frac{\cos x}{\sin x}}=\frac{\cos ^2x}{\cos x}=\cos x\) (đpcm)
Bài 4:
\(0< x< 90^0\Rightarrow cosx>0\)
\(\Rightarrow cosx=\sqrt{1-sin^2x}=\frac{4}{5}\)
\(tanx=\frac{sinx}{cosx}=\frac{3}{4}\) ; \(cotx=\frac{1}{tanx}=\frac{4}{3}\)
\(\Rightarrow A=\frac{\frac{4}{3}}{\frac{4}{3}-\frac{3}{4}}=\frac{16}{7}\)
\(\frac{3\pi}{2}< a< 2\pi\Rightarrow sina< 0\)
\(\Rightarrow sina=-\sqrt{1-cos^2a}=-\frac{\sqrt{3}}{2}\)
\(B=sina+cosa=\frac{1-\sqrt{3}}{2}\)
Bài 5:
\(0< a< \frac{\pi}{4}\Rightarrow0< 2a< \frac{\pi}{2}\Rightarrow sin2a>0\)
\(\Rightarrow sin2a=\sqrt{1-cos^2a}=\frac{\sqrt{21}}{5}\)
\(tan2a=\frac{sin2a}{cos2a}=\frac{\sqrt{21}}{2}\)
\(cot2a=\frac{1}{tan2a}=\frac{2\sqrt{21}}{21}\)
b/ \(-\frac{3\pi}{4}\le a\le-\frac{\pi}{2}\Rightarrow-\frac{3\pi}{2}\le2a\le-\pi\Rightarrow cos2a< 0\)
\(\Rightarrow cos2a=-\sqrt{1-sin^2a}=-\frac{7}{25}\)
\(tan2a=\frac{sin2a}{cos2a}=-\frac{24}{7}\)
\(cot2a=\frac{1}{tan2a}=-\frac{7}{24}\)
Bài 1:
\(\frac{1-sinx}{cosx}=\frac{cosx\left(1-sinx\right)}{cos^2x}=\frac{cosx\left(1-sinx\right)}{1-sin^2x}=\frac{cosx\left(1-sinx\right)}{\left(1-sinx\right)\left(1+sinx\right)}=\frac{cosx}{1+sinx}\)
\(\frac{tanx}{sinx}-\frac{sinx}{cotx}=\frac{tanx.cotx-sin^2x}{sinx.cotx}=\frac{1-sin^2x}{cosx}=\frac{cos^2x}{cosx}=cosx\)
Bài 2:
\(A=\frac{cot^2x-cos^2x}{cot^2x}+\frac{sinx.cosx}{cotx}=\frac{sin^2x\left(cot^2x-cos^2x\right)}{cos^2x}+\frac{sinx.sinx.cosx}{cosx}\)
\(=\frac{cos^2x-cos^2x.sin^2x}{cos^2x}+sin^2x=1-sin^2x+sin^2x=1\)
\(B=cos^2x\left(cos^2x+sin^2x\right)+sin^2x=cos^2x+sin^2x=1\)
Bài 3:
A đề đúng chứ bạn? Là cos hay cot?
\(B=\frac{\frac{4sinx.cosx}{sin^2x}-\frac{3cos^2x}{sin^2x}}{\frac{1}{sin^2x}+3}=\frac{4cotx-3cot^2x}{1+cot^2x+3}=\frac{4.\left(-6\right)-3.\left(-6\right)^2}{1+\left(-6\right)^2+3}=...\)