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Câu a)
Từ \(\tan a=3\Leftrightarrow \frac{\sin a}{\cos a}=3\Rightarrow \sin a=3\cos a\)
Do đó:
\(\frac{\sin a\cos a+\cos ^2a}{2\sin ^2a-\cos ^2a}=\frac{3\cos a\cos a+\cos ^2a}{2(3\cos a)^2-\cos ^2a}\)
\(=\frac{\cos ^2a(3+1)}{\cos ^2a(18-1)}=\frac{4}{17}\)
Câu b)
Có: \(\cot \left(\frac{\pi}{2}-x\right)=\tan x=\frac{\sin x}{\cos x}\)
\(\cos\left(\frac{\pi}{2}+x\right)=-\sin x\)
\(\Rightarrow \cot \left(\frac{\pi}{2}-x\right)\cos \left(\frac{\pi}{2}+x\right)=\frac{-\sin ^2x}{\cos x}\)
Và:
\(\frac{\sin (\pi-x)\cot x}{1-\sin ^2x}=\frac{\sin x\cot x}{\cos^2x}=\frac{\sin x.\frac{\cos x}{\sin x}}{\cos^2x}=\frac{1}{\cos x}\)
Do đó:
\(\Rightarrow \cot \left(\frac{\pi}{2}-x\right)\cos \left(\frac{\pi}{2}+x\right)+\frac{\sin (\pi-x)\cot x}{1-\sin ^2x}=\frac{1-\sin ^2x}{\cos x}=\frac{\cos ^2x}{\cos x}=\cos x\)
Ta có đpcm.
phần chứng minh biểu thức không phụ thuộc \(x\)
ta có : \(A=\dfrac{cot^2a-cos^2a}{cot^2a}+\dfrac{sinacosa}{cota}=\dfrac{cot^2a-cos^2a}{cot^2a}+\dfrac{cos^2a}{cot^2a}\)
\(=\dfrac{cot^2a-cos^2a+cos^2a}{cot^2a}=\dfrac{cot^2a}{cot^2a}=1\left(đpcm\right)\)
ý còn lại : xem lại đề nha bn
phần chứng minh đẳng thức
ta có : \(\dfrac{sin2a-2sina}{sin2a+2sina}+tan^2\dfrac{a}{2}=\dfrac{2sinacosa-2sina}{2sinacosa+2sina}+tan^2\dfrac{a}{2}\)
\(=\dfrac{2sina\left(cosa-1\right)}{2sina\left(cosa+1\right)}+tan^2\dfrac{a}{2}=\dfrac{cosa-1}{cosa+1}+tan^2\dfrac{a}{2}\)
\(=\dfrac{1-2sin^2\dfrac{a}{2}-1}{2cos^2\dfrac{a}{2}-1+1}+tan^2\dfrac{a}{2}=\dfrac{-2sin^2\dfrac{a}{2}}{2cos^2\dfrac{a}{2}}+tan^2\dfrac{a}{2}\)
\(=-tan^2\dfrac{a}{2}+tan^2\dfrac{a}{2}=0\left(đpcm\right)\)
ta có : \(\dfrac{sina}{1+cosa}+\dfrac{1+cosa}{sina}=\dfrac{sin^2a+\left(1+cosa\right)^2}{sina\left(1+cosa\right)}\)
\(=\dfrac{sin^2a+cos^2a+2cosa+1}{sina\left(1+cosa\right)}=\dfrac{2cosa+2}{sina\left(cosa+1\right)}\)
\(=\dfrac{2\left(cosa+1\right)}{sina\left(cosa+1\right)}=\dfrac{2}{sina}\left(đpcm\right)\)
còn 2 câu kia để chừng nào rảnh mk giải cho nha
mk lm 2 câu còn lại nha
ta có : \(\dfrac{sin^2x}{sinx-cosx}-\dfrac{sinx+cosx}{tan^2x-1}=\dfrac{\left(1-cos^2x\right)\left(tan^2x-1\right)-\left(sin^2x-cos^2x\right)}{\left(sinx-cosx\right)\left(tan^2x-1\right)}\)
\(=\dfrac{tan^2x-sin^2x-sin^2x-sin^2x+cos^2x}{\left(sinx-cosx\right)\left(tan^2x-1\right)}=\dfrac{\dfrac{sin^4x}{cos^2x}-sin^2x-sin^2x+cos^2x}{\left(sinx-cosx\right)\left(tan^2-1\right)}\)
\(=\dfrac{tan^2x\left(sin^2x-cos^2x\right)-\left(sin^2x-cos^2x\right)}{\left(sinx-cosx\right)\left(tan^2x-1\right)}=\dfrac{\left(tan^2x-1\right)\left(sin^2x-cos^2x\right)}{\left(sinx-cosx\right)\left(tan^2x-1\right)}\)
\(=sinx+cosx\left(đpcm\right)\)
ta có : \(\dfrac{sin\left(a+b\right)sin\left(a-b\right)}{1-tan^2a.cot^2b}=\dfrac{sin\left(a+b\right)sin\left(a-b\right)}{1-\dfrac{sin^2a.cos^2b}{cos^2a.sin^2b}}\)
\(=\dfrac{sin\left(a+b\right)sin\left(a-b\right)}{\dfrac{cos^2a.sin^2b-sin^2a.cos^2b}{cos^2a.sin^2b}}=\dfrac{sin\left(a+b\right)sin\left(a-b\right).cos^2a.sin^2b}{-\left(sin^2a.cos^2b-cos^2a.sin^2b\right)}\)
\(=\dfrac{sin\left(a+b\right)sin\left(a-b\right).cos^2a.sin^2b}{-\left(\left(sina.cosb-cosa.sinb\right)\left(sina.cosb+cosa.sinb\right)\right)}\)
\(=\dfrac{sin\left(a+b\right)sin\left(a-b\right).cos^2a.sin^2b}{-sin\left(a-b\right)sin\left(a+b\right)}=-cos^2a.sin^2b\left(đpcm\right)\)
mk lm hơi tắc ! do tối rồi , mà mk lại đang ở quán nek nên không tiện làm dài . bạn thông cảm
Bài 1:
a)
\(\sin ^2x+\sin ^2x\cot^2x=\sin ^2x(1+\cot^2x)=\sin ^2x(1+\frac{\cos ^2x}{\sin ^2x})\)
\(=\sin ^2x.\frac{\sin ^2x+\cos^2x}{\sin ^2x}=\sin ^2x+\cos^2x=1\)
b)
\((1-\tan ^2x)\cot^2x+1-\cot^2x\)
\(=\cot^2x(1-\tan^2x-1)+1=\cot^2x(-\tan ^2x)+1=-(\tan x\cot x)^2+1\)
\(=-1^2+1=0\)
c)
\(\sin ^2x\tan x+\cos^2x\cot x+2\sin x\cos x=\sin ^2x.\frac{\sin x}{\cos x}+\cos ^2x.\frac{\cos x}{\sin x}+2\sin x\cos x\)
\(=\frac{\sin ^3x}{\cos x}+\frac{\cos ^3x}{\sin x}+2\sin x\cos x=\frac{\sin ^4x+\cos ^4x+2\sin ^2x\cos ^2x}{\sin x\cos x}=\frac{(\sin ^2x+\cos ^2x)^2}{\sin x\cos x}=\frac{1}{\sin x\cos x}\)
\(=\frac{1}{\frac{\sin 2x}{2}}=\frac{2}{\sin 2x}\)
Bài 2:
Áp dụng BĐT Cauchy Schwarz ta có:
\(P=\frac{a^2}{\sqrt{a(2c+a+b)}}+\frac{b^2}{\sqrt{b(2a+b+c)}}+\frac{c^2}{\sqrt{c(2b+c+a)}}\)
\(\geq \frac{(a+b+c)^2}{\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)}}(*)\)
Tiếp tục áp dụng BĐT Cauchy-Schwarz:
\((\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)})^2\leq (a+b+c)(2c+a+b+2a+b+c+2b+c+a)\)
\(\Leftrightarrow (\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)})^2\leq 4(a+b+c)^2\)
\(\Rightarrow \sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)}\leq 2(a+b+c)(**)\)
Từ \((*); (**)\Rightarrow P\geq \frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}=\frac{3}{2}\)
Vậy \(P_{\min}=\frac{3}{2}\)
Dấu "=" xảy ra khi $a=b=c=1$
Câu 1 đề sai, chắc chắn 1 trong 2 cái \(cot^2x\) phải có 1 cái là \(cos^2x\)
2.
\(\dfrac{1-sinx}{cosx}-\dfrac{cosx}{1+sinx}=\dfrac{\left(1-sinx\right)\left(1+sinx\right)-cos^2x}{cosx\left(1+sinx\right)}=\dfrac{1-sin^2x-cos^2x}{cosx\left(1+sinx\right)}\)
\(=\dfrac{1-\left(sin^2x+cos^2x\right)}{cosx\left(1+sinx\right)}=\dfrac{1-1}{cosx\left(1+sinx\right)}=0\)
3.
\(\dfrac{tanx}{sinx}-\dfrac{sinx}{cotx}=\dfrac{tanx.cotx-sin^2x}{sinx.cotx}=\dfrac{1-sin^2x}{sinx.\dfrac{cosx}{sinx}}=\dfrac{cos^2x}{cosx}=cosx\)
4.
\(\dfrac{tanx}{1-tan^2x}.\dfrac{cot^2x-1}{cotx}=\dfrac{tanx}{1-tan^2x}.\dfrac{\dfrac{1}{tan^2x}-1}{\dfrac{1}{tanx}}=\dfrac{tanx}{1-tan^2x}.\dfrac{1-tan^2x}{tanx}=1\)
5.
\(\dfrac{1+sin^2x}{1-sin^2x}=\dfrac{1+sin^2x}{cos^2x}=\dfrac{1}{cos^2x}+tan^2x=\dfrac{sin^2x+cos^2x}{cos^2x}+tan^2x\)
\(=tan^2x+1+tan^2x=1+2tan^2x\)
a)\(\eqalign{ & A\sin {x \over 5} = \sin {x \over 5}\cos {x \over 5}\cos {{2x} \over 5}\cos {{4x} \over 5}\cos {{8x} \over 5} \cr & = {1 \over 2}\sin {{2x} \over 5}\cos {{2x} \over 5}\cos {{4x} \over 5}\cos {{8x} \over 5} \cr & = {1 \over 4}\sin {{4x} \over 5}\cos {{4x} \over 5}\cos {{8x} \over 5} = {1 \over 8}\sin {{8x} \over 5}\cos {{8x} \over 5} \cr & = {1 \over {16}}\sin {{16x} \over 5} \cr} \)
Suy ra biểu thức rút gọn \(A =\sin{{16x} \over 5}:16\sin {x \over 5}\)
b)\(\eqalign{ & B = \sin {x \over 7} + 2\sin {{3x} \over 7} + \sin {{5x} \over 7} = 2\sin {{3x} \over 7} + (\sin {x \over 7} + \sin {{5x} \over 7}) \cr & = 2\sin {{3x} \over 7} + 2\sin {1 \over 2}({{5x} \over 7} + {x \over 7})cos{1 \over 2}({{5x} \over 7} - {x \over 7}) \cr & = 2\sin {{3x} \over 7}(1 + \cos {{2x} \over 7}) = 4\sin {{3x} \over 7}{\cos ^2}{x \over 7} \cr}\)
Giả sử các biểu thức đã cho đều xác định
a/ \(\dfrac{1+sin^2x}{1-sin^2x}=\dfrac{1+sin^2x}{cos^2x}=\dfrac{1}{cos^2x}+\dfrac{sin^2x}{cos^2x}+1+tan^2x+tan^2x=1+2tan^2x\)
b/ \(\dfrac{sinx}{1+cosx}+\dfrac{1+cosx}{sinx}=\dfrac{sin^2x+\left(1+cosx\right)^2}{\left(1+cosx\right)sinx}=\dfrac{sin^2x+cos^2x+2cosx+1}{\left(1+cosx\right)sinx}\)
\(=\dfrac{1+2cosx+1}{\left(1+cosx\right)sinx}=\dfrac{2+2cosx}{\left(1+cosx\right)sinx}=\dfrac{2\left(1+cosx\right)}{\left(1+cosx\right)sinx}=\dfrac{2}{sinx}\)
c/ \(\dfrac{1-sinx}{cosx}=\dfrac{\left(1-sinx\right)cosx}{cos^2x}=\dfrac{\left(1-sinx\right)cosx}{1-sin^2x}\)
\(\dfrac{\left(1-sinx\right)cosx}{\left(1-sinx\right)\left(1+sinx\right)}=\dfrac{cosx}{1+sinx}\)
d/ \(\left(1-cosx\right)\left(1+cot^2x\right)=\left(1-cosx\right).\dfrac{1}{sin^2x}\)
\(=\dfrac{1-cosx}{1-cos^2x}=\dfrac{1-cosx}{\left(1-cosx\right)\left(1+cosx\right)}=\dfrac{1}{1+cosx}\)
e/ \(1-\dfrac{sin^2x}{1+cotx}-\dfrac{cos^2x}{1+tanx}=1-\dfrac{sin^3x}{sinx\left(1+\dfrac{cosx}{sinx}\right)}-\dfrac{cos^3x}{cosx\left(1+\dfrac{sinx}{cosx}\right)}\)
\(=1-\left(\dfrac{sin^3x}{sinx+cosx}+\dfrac{cos^3x}{sinx+cosx}\right)=1-\left(\dfrac{sin^3x+cos^3x}{sinx+cosx}\right)\)
\(=1-\left(\dfrac{\left(sinx+cosx\right)\left(sin^2x-sinx.cosx+cos^2x\right)}{sinx+cosx}\right)\)
\(=1-\left(1-sinx.cosx\right)=sinx.cosx\)
f/ Bạn ghi đề sai à?
a)
\(\sin ^4a-\cos ^4a+1=(\sin ^2a-\cos ^2a)(\sin ^2a+\cos^2a)+1\)
\(=(\sin ^2a-\cos ^2a).1+1=\sin ^2a-\cos ^2a+\sin ^2a+\cos ^2a\)
\(=2\sin ^2a\)
b) \(\sin ^2a+2\cos ^2a-1=(\sin ^2a+\cos^2a)+\cos ^2a-1\)
\(=1+\cos ^2a-1=\cos ^2a\)
\(\Rightarrow \frac{\sin ^2a+2\cos ^2a-1}{\cot ^2a}=\frac{\cos ^2a}{\cot ^2a}=\frac{\cos ^2a}{\frac{\cos ^2a}{\sin ^2a}}=\sin ^2a\)
c)
\(\frac{1-\sin ^2a\cos ^2a}{\cos ^2a}-\cos ^2a=\frac{1}{\cos ^2a}-\sin ^2a-\cos ^2a\)
\(=\frac{1}{\cos ^2a}-(\sin ^2a+\cos ^2a)=\frac{1}{\cos ^2a}-1\)
\(=\frac{1-\cos ^2a}{\cos ^2a}=\frac{\sin ^2a}{\cos ^2a}=\tan ^2a\)
d)
\(\frac{\sin ^2a-\tan ^2a}{\cos ^2a-\cot ^2a}=\frac{\sin ^2a-\frac{\sin ^2a}{\cos ^2a}}{\cos ^2a-\frac{\cos ^2a}{\sin ^2a}}\) \(=\frac{\sin ^2a(1-\frac{1}{\cos ^2a})}{\cos ^2a(1-\frac{1}{\sin ^2a})}\)
\(=\frac{\sin ^2a.\frac{\cos ^2a-1}{\cos ^2a}}{\cos ^2a.\frac{\sin ^2a-1}{\sin ^2a}}\) \(=\frac{\sin ^2a.\frac{-\sin ^2a}{\cos ^2a}}{\cos ^2a.\frac{-\cos ^2a}{\sin ^2a}}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)
f)
\(\frac{(\sin a+\cos a)^2-1}{\cot a-\sin a\cos a}=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\frac{\cos a}{\sin a}-\sin a\cos a}\)
\(=\sin a.\frac{1+2\sin a\cos a-1}{\cos a-\cos a\sin ^2a}\)
\(=\sin a. \frac{2\sin a\cos a}{\cos a(1-\sin ^2a)}=\sin a. \frac{2\sin a\cos a}{\cos a. \cos^2 a}=\frac{2\sin ^2a}{\cos ^2a}=2\tan ^2a\)
\(A=\dfrac{2tan^2a+\dfrac{5}{cos^2a}}{4-\dfrac{3}{cos^2a}}=\dfrac{2tan^2a+5\left(1+tan^2a\right)}{4-3\left(1+tan^2a\right)}=...\) (bạn tự thay số bấm máy nhé)
\(B=\dfrac{3cot^2a-1}{cot^2a+2}=...\)