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NV
8 tháng 5 2020

\(\frac{sin^3a+cos^3a}{sina+cosa}=\frac{\left(sina+cosa\right)\left(sin^2a+cos^2a-sina.cosa\right)}{sina+cosa}\)

\(=sin^2a+cos^2a-sina.cosa\)

\(=1-sina.cosa\)

25 tháng 5 2020

mình cám ơn ạ^^

NV
25 tháng 5 2020

\(\frac{sin2a-2sina}{sin2a+2sina}=\frac{2sina.cosa-2sina}{2sina.cosa+2sina}=\frac{2sina\left(cosa-1\right)}{2sina\left(cosa+1\right)}=\frac{cosa-1}{cosa+1}\)

\(=\frac{1-2sin^2\frac{a}{2}-1}{2cos^2\frac{a}{2}-1+1}=\frac{-sin^2\frac{a}{2}}{cos^2\frac{a}{2}}=-tan^2\frac{a}{2}\)

\(\frac{sin^4x-sin^2x+cos^2x}{cos^4x-cos^2x+sin^2x}=\frac{sin^2x\left(sin^2x-1\right)+cos^2x}{cos^2x\left(cos^2x-1\right)+sin^2x}=\frac{-sin^2x.cos^2x+cos^2x}{-cos^2x.sin^2x+sin^2x}\)

\(=\frac{cos^2x\left(1-sin^2x\right)}{sin^2x\left(1-cos^2x\right)}=\frac{cos^4x}{sin^4x}=cot^4x\)

\(\frac{sin^3a-cos^3a}{sina-cosa}=\frac{\left(sina-cosa\right)\left[sin^2a+cos^2a+sina.cosa\right]}{sina-cosa}=1+sina.cosa=1+\frac{1}{2}sin2a\)

NV
28 tháng 11 2019

\(\frac{cosa}{1+sina}+\frac{sina}{cosa}=\frac{cos^2a+sina\left(1+sina\right)}{cosa\left(1+sina\right)}=\frac{1+sina}{cosa\left(1+sina\right)}=\frac{1}{cosa}\)

\(\frac{sin^2a+cos^2a+2sina.cosa}{\left(sina-cosa\right)\left(sina+cosa\right)}=\frac{\left(sina+cosa\right)^2}{\left(sina-cosa\right)\left(sina+cosa\right)}=\frac{sina+cosa}{sina-cosa}=\frac{\frac{sina}{cosa}+1}{\frac{sina}{cosa}-1}=\frac{tana+1}{tana-1}\)

\(\left(sin^2a\right)^3+\left(cos^2a\right)^3=\left(sin^2a+cos^2a\right)^3-3sin^2a.cos^2a\left(sin^2a+cos^2a\right)\)

\(=1-3sin^2a.cos^2a\)

\(sin^2a-tan^2a=tan^4a\left(\frac{sin^2a}{tan^4a}-\frac{1}{tan^2a}\right)=tan^4a\left(sin^2a.\frac{cos^2a}{sin^2a}-\frac{1}{tan^2a}\right)\)

\(=tan^4a\left(cos^2a-cot^2a\right)\) bạn ghi sai đề câu này

\(\frac{tan^3a}{sin^2a}-\frac{1}{sina.cosa}+\frac{cot^3a}{cos^2a}=tan^3a\left(1+cot^2a\right)-\frac{1}{sina.cosa}+cot^3a\left(1+tan^2a\right)\)

\(=tan^3a+tana-\frac{1}{sina.cosa}+cot^3a+cota\)

\(=tan^3a+cot^3a+\frac{sina}{cosa}+\frac{cosa}{sina}-\frac{1}{sina.cosa}\)

\(=tan^3a+cot^3a+\frac{sin^2a+cos^2a-1}{sina.cosa}=tan^3a+cot^3a\)

NV
3 tháng 6 2020

\(D=\frac{\frac{sina}{cos^3a}+\frac{5cosa}{cos^3a}}{\frac{sin^3a}{cos^3a}-\frac{2cos^3a}{cos^3a}}=\frac{tana.\frac{1}{cos^2a}+\frac{5}{cos^2a}}{tan^3a-2}=\frac{tana\left(1+tan^2a\right)+5\left(1+tan^2a\right)}{tan^3a-2}\)

Bạn thay số và bấm máy

AH
Akai Haruma
Giáo viên
28 tháng 4 2019

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.

12 tháng 5 2020

A = 2(1 - sin2α)2 - sin4α + sin2α (1-sin2α) + 3sin2α

=2 - 4sin2α + 2sin4α - sin4α + sin2α - sin4α + 3sin2α

= 2

12 tháng 5 2020

\(A=2\cos^4\alpha-\sin^4\alpha+\sin^2\alpha.\cos^2\alpha+3\sin^4\alpha+3\cos^2\alpha.\sin^2\alpha\)

\(A=2\sin^4\alpha+2\cos^4\alpha+4\sin^2\alpha.\cos^2\alpha\)

\(A=2\left[\left(\sin^2\alpha+\cos^2\alpha\right)^2-2\sin^2\alpha.\cos^2\alpha\right]+4\cos^2\alpha\sin^2\alpha=2\)

16 tháng 5 2021

\(cot\alpha=3\Leftrightarrow\dfrac{cos\alpha}{sin\alpha}=3\Leftrightarrow cos\alpha=3sin\alpha\)

Khi đó: 

\(\dfrac{3sin\alpha-2cos\alpha}{12sin^3\alpha+4cos^3\alpha}=\dfrac{3sin\alpha-6sin\alpha}{12sin^3\alpha+108sin^3\alpha}=-\dfrac{3sin\alpha}{120sin^3\alpha}=-\dfrac{1}{40sin^2\alpha}\)

NV
21 tháng 5 2020

\(\frac{cos\left(a-b\right)}{sin\left(a+b\right)}=\frac{cosa.cosb+sina.sinb}{sina.cosb+cosa.sinb}=\frac{\frac{cosa.cosb}{sina.sinb}+1}{\frac{sina.cosb}{sina.sinb}+\frac{cosa.sinb}{sina.sinb}}=\frac{cota.cotb+1}{cota+cotb}\)

Bạn ghi đề ko đúng

\(sin\left(a+b\right)sin\left(a-b\right)=\frac{1}{2}\left[cos2b-cos2a\right]\)

\(=\frac{1}{2}\left[1-2sin^2b-1+2sin^2a\right]\)

\(=sin^2a-sin^2b\)

\(=1-cos^2a-1+cos^2b=cos^2b-cos^2a\)

Câu này bạn cũng ghi đề ko đúng

\(cos\left(a+b\right)cos\left(a-b\right)=\frac{1}{2}\left[cos2a+cos2b\right]\)

\(=\frac{1}{2}\left[2cos^2a-1+1-2sin^2b\right]=cos^2a-sin^2b\)

\(=1-sin^2a-1+cos^2b=cos^2b-sin^2a\)