TanA=3. Tính
B=(sinA-cosA)/ (sin^3A+3.cos^3A+2sinA
Chứng minh
\(\left(1+cota\right)sin^3a+\left(1+tana\right)cos^3a=sina+cosa\)
Lời giải:
\((1+\cot a)\sin ^3a+(1+\tan a)\cos ^3a\)
\(=(1+\frac{\cos a}{\sin a})\sin ^3a+(1+\frac{\sin a}{\cos a})\cos ^3a\)
\(=(\sin a+\cos a)\sin ^2a+(\cos a+\sin a)\cos ^2a\)
\(=(\sin a+\cos a)(\sin ^2a+\cos ^2a)=(\sin a+\cos a).1=\sin a+\cos a\)
Chứng minh:
\(a,\frac{cosa}{1+sina}+tana=\frac{1}{cosa}\)
\(b,\frac{1+2sina.cosa}{sin^2a-cos^2a}=\frac{tana+1}{tana-1}\)
c,\(sin^6a+cos^6a=1-3sin^2a.cos^2a\)
d,\(sin^2a-tan^2a=tan^6a\left(cos^2a-cot^2a\right)\)
e.\(\frac{tan^3a}{sin^2a}-\frac{1}{sina.cosa}+\frac{cot^3a}{cos^2a}=tan^3a+cot^3a\)
\(\frac{cosa}{1+sina}+\frac{sina}{cosa}=\frac{cos^2a+sina\left(1+sina\right)}{cosa\left(1+sina\right)}=\frac{1+sina}{cosa\left(1+sina\right)}=\frac{1}{cosa}\)
\(\frac{sin^2a+cos^2a+2sina.cosa}{\left(sina-cosa\right)\left(sina+cosa\right)}=\frac{\left(sina+cosa\right)^2}{\left(sina-cosa\right)\left(sina+cosa\right)}=\frac{sina+cosa}{sina-cosa}=\frac{\frac{sina}{cosa}+1}{\frac{sina}{cosa}-1}=\frac{tana+1}{tana-1}\)
\(\left(sin^2a\right)^3+\left(cos^2a\right)^3=\left(sin^2a+cos^2a\right)^3-3sin^2a.cos^2a\left(sin^2a+cos^2a\right)\)
\(=1-3sin^2a.cos^2a\)
\(sin^2a-tan^2a=tan^4a\left(\frac{sin^2a}{tan^4a}-\frac{1}{tan^2a}\right)=tan^4a\left(sin^2a.\frac{cos^2a}{sin^2a}-\frac{1}{tan^2a}\right)\)
\(=tan^4a\left(cos^2a-cot^2a\right)\) bạn ghi sai đề câu này
\(\frac{tan^3a}{sin^2a}-\frac{1}{sina.cosa}+\frac{cot^3a}{cos^2a}=tan^3a\left(1+cot^2a\right)-\frac{1}{sina.cosa}+cot^3a\left(1+tan^2a\right)\)
\(=tan^3a+tana-\frac{1}{sina.cosa}+cot^3a+cota\)
\(=tan^3a+cot^3a+\frac{sina}{cosa}+\frac{cosa}{sina}-\frac{1}{sina.cosa}\)
\(=tan^3a+cot^3a+\frac{sin^2a+cos^2a-1}{sina.cosa}=tan^3a+cot^3a\)
Chứng minh các đẳng thức lượng giác sau:
a, \(\frac{sin2a-2sina}{sin2a+2sina}=-tan^2\frac{a}{2}\)
b, \(\frac{sin^4x+cos^2x-sin^2x}{cos^4x+sin^2x-cos^2x}=cot^4x\)
c, \(\frac{sin^3a-cos^3a}{sina-cosa}=1+\frac{sin2a}{2}\)
giúp mình với ạ:((
\(\frac{sin2a-2sina}{sin2a+2sina}=\frac{2sina.cosa-2sina}{2sina.cosa+2sina}=\frac{2sina\left(cosa-1\right)}{2sina\left(cosa+1\right)}=\frac{cosa-1}{cosa+1}\)
\(=\frac{1-2sin^2\frac{a}{2}-1}{2cos^2\frac{a}{2}-1+1}=\frac{-sin^2\frac{a}{2}}{cos^2\frac{a}{2}}=-tan^2\frac{a}{2}\)
\(\frac{sin^4x-sin^2x+cos^2x}{cos^4x-cos^2x+sin^2x}=\frac{sin^2x\left(sin^2x-1\right)+cos^2x}{cos^2x\left(cos^2x-1\right)+sin^2x}=\frac{-sin^2x.cos^2x+cos^2x}{-cos^2x.sin^2x+sin^2x}\)
\(=\frac{cos^2x\left(1-sin^2x\right)}{sin^2x\left(1-cos^2x\right)}=\frac{cos^4x}{sin^4x}=cot^4x\)
\(\frac{sin^3a-cos^3a}{sina-cosa}=\frac{\left(sina-cosa\right)\left[sin^2a+cos^2a+sina.cosa\right]}{sina-cosa}=1+sina.cosa=1+\frac{1}{2}sin2a\)
Bài 1 CM các đẳng thức sau:
a, 1+ sin2a / sina + cosa - 1-tan ²a/2 / 1+ tan ²a/2 = sina
b, cota - tana = 2cot2a
c, 1+ cosa +cos2a + cos3a/ 2cos²a + cosa-1 = 2cosa
d, sin²a / sina- cosa - sina + cosa / tan²a = sina + cosa
e, sin²a - cos²(a-b ) + 2coscosb ×cos(a-b) = cos2a
f, cos²a - 2sina × ( 1-sina ) × cosa +( 1 + sina) × cosa - 2×(1+sina ) / 1- sina = cosa
Bài 2 CM các đẳng thức sau ko phụ thuộc vào x
a, A= sin⁶x + cos⁶x - 1 / sin⁴x + cos ⁴x -1
b, B = ( 2sin ⁶x - 3sin ⁴x - 4sin²x ) +( 2cos⁶x - 3 cos⁴x- 4cos⁴x
c, C= sin⁴x + 3cos⁴x -1 / sin⁶x + cos⁶x + 3cos⁴x-1
Giải giúp tớ 2 bài này vs tớ cảm ơn nhìu
cho sina+cosa=5/4
a, A=sina.cosa b, B= sina-cosa c,C=sin^3a-cos^3a
help me
\(sina+cosa=\frac{5}{4}\Rightarrow\left(sina+cosa\right)^2=\frac{25}{16}\)
\(\Rightarrow sin^2a+cos^2a+2sina.cosa=\frac{25}{16}\)
\(sina.cosa=\frac{\frac{25}{16}-1}{2}=\frac{9}{32}\)
b/ \(\left(sina-cosa\right)^2=sin^2a+cos^2a-2sinacosa\)
\(\left(sina-cosa\right)^2=1-2.\frac{9}{32}=\frac{7}{16}\)
\(\Rightarrow sina-cosa=\pm\frac{\sqrt{7}}{4}\)
c/ \(sin^3a-cos^3a=\left(sina-cosa\right)\left(sin^2a+cos^2a+sina.cosa\right)\)
\(=\left(sina-cosa\right)\left(1+\frac{9}{32}\right)=\pm\frac{41\sqrt{7}}{128}\)
Giúp mình với các bạn ơi!!!!!!!!!!!!!!
Cho sina*cosa=0.22. Tính giá trị của biểu thức M=\(\sin^3a+\cos^3a-2.\sin a.\cos a\)
rút gọn A=\(\frac{sin^3a-cos^3a}{sina-cosa}+sina+cosa\)
\(A=\frac{\left(sina-cosa\right)\left(sin^2a+cos^2a+sina.cosa\right)}{sina-cosa}+sina+cosa\)
\(=1+sina.cosa+sina+cosa\)
\(=\left(sina+1\right)\left(cosa+1\right)\)
Cho sina + cosa =2. Tính sin^3a + cos^3a
ta có : \(sin^3a+cos^3a=\left(sina+cosa\right)^3-3sina.cosa\left(sina+cosa\right)\)
\(=2^3-3sina.cosa\left(2\right)=8-6sina.cosa\)
\(=11-3sin^2a-6sina.cosa-3cos^2a=11-3\left(sin+cos\right)^2=11-3.2^2=11-12=-1\)
tan =\(\sqrt{3}\).Tính A=\(\dfrac{sin^3a-cos^3a}{sina-cosa}\)
\(tana=\sqrt{3}\)
nên \(\dfrac{sina}{cosa}=\sqrt{3}\)
=>\(sina=\sqrt{3}\cdot cosa\)
=>a=60 độ
\(A=\dfrac{\left(sina-cosa\right)\left(sin^2a+cos^2a+sina\cdot cosa\right)}{sina-cosa}\)
\(=1+sina\cdot cosa=1+\dfrac{1}{2}sin2a\)
\(=1+\dfrac{1}{2}\cdot sin120=\dfrac{4+\sqrt{3}}{4}\)
chứng minh các đẳng thức sau :
a)\(\frac{cos\left(a-b\right)}{cos\left(a+b\right)}=\frac{cota.cotb+1}{cota.cotb-1}\)
b)\(2\left(sin^6a+cos^6a\right)+1=3\left(sin^4a+cos^4a\right)\)
c)\(\frac{tana-tanb}{cotb-cota}=tanatanb\)
d)\(\left(cotx+tanx\right)^2-\left(cotx-tanx\right)^2=4\)
e)\(\frac{sin^3a+cos^3a}{sina+cosa}=1-sinacosa\)
Lời giải:
a)
\(\frac{\cos (a-b)}{\cos (a+b)}=\frac{\cos a\cos b+\sin a\sin b}{\cos a\cos b-\sin a\sin b}=\frac{\frac{\cos a\cos b}{\sin a\sin b}+1}{\frac{\cos a\cos b}{\sin a\sin b}-1}=\frac{\cot a\cot b+1}{\cot a\cot b-1}\)
b)
\(2(\sin ^6a+\cos ^6a)+1=2(\sin ^2a+\cos ^2a)(\sin ^4a-\sin ^2a\cos ^2a+\cos ^4a)+1\)
\(=2(\sin ^4a-\sin ^2a\cos ^2a+\cos ^4a)+1\)
\(=3(\sin ^4a+\cos ^4a)-(\sin ^4a+\cos ^4a+2\sin ^2a\cos ^2a)+1\)
\(=3(\sin ^4a+\cos ^4a)-(\sin ^2a+\cos ^2a)^2+1\)
\(=3(\sin ^4a+\cos ^4a)-1^2+1=3(\sin ^4a+\cos ^4a)\)
c)
\(\frac{\tan a-\tan b}{cot b-\cot a}=\frac{\tan a-\tan b}{\frac{1}{\tan b}-\frac{1}{\tan a}}\) (nhớ rằng \(\tan x.\cot x=1\rightarrow \cot x=\frac{1}{\tan x}\) )
\(=\frac{\tan a-\tan b}{\frac{\tan a-\tan b}{\tan a\tan b}}=\tan a\tan b\)
d)
\((\cot x+\tan x)^2-(\cot x-\tan x)^2=(\cot ^2x+\tan ^2x+2\cot x\tan x)-(\cot ^2x-2\cot x\tan x+\tan ^2x)\)
\(=4\cot x\tan x=4.1=4\)
e)
\(\frac{\sin ^3a+\cos ^3a}{\sin a+\cos a}=\frac{(\sin a+\cos a)(\sin ^2a-\sin a\cos a+\cos ^2a)}{\sin a+\cos a}\)
\(=\sin ^2a-\sin a\cos a+\cos ^2a=(\sin ^2a+\cos ^2a)-\sin a\cos a=1-\sin a\cos a\)
Vậy ta có đpcm.