cái câu 1 kia lạ thật, phần phía trc có ngoặc thì phải nhân vs hạng tử nào đó chứ nhỉ? Và mk tính ra kq là \(-\cos^22\alpha\)

\(VT=\cos^4\alpha+\sin^4\alpha-2\cos^6\alpha-2\sin^6\alpha\)

\(=\sin^4\alpha\left(1-2\sin^2\alpha\right)-\cos^4\alpha\left(2\cos^2\alpha-1\right)\)

\(=\sin^4\alpha.\cos2\alpha-\cos^4\alpha.\cos2\alpha\)

\(=\cos2\alpha\left(\sin^2\alpha.\sin^2\alpha-\cos^4\alpha\right)\)

\(=\cos2\alpha.\left[\left(1-\cos^2\alpha\right)^2-\cos^4\alpha\right]\)

\(=\cos2\alpha.\left(1-2\cos^2\alpha\right)\)

\(=-\cos^22\alpha\)

2/ \(VT=\frac{1-\cos^2\alpha+\cos^2\alpha}{1+\sin2\alpha}=\frac{1}{1+\sin2\alpha}\)

\(VP=\frac{\frac{\sin\alpha}{\cos\alpha}-1}{\frac{\sin\alpha}{\cos\alpha}+1}=\frac{\frac{\sin\alpha-\cos\alpha}{\cos\alpha}}{\frac{\sin\alpha+\cos\alpha}{\cos\alpha}}=\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}\)

hmm, câu 2 có vẻ vô lí, bn thử nhân chéo lên mà xem, nó ko ra KQ = nhau đâu

1)

\((\cos^4a+\sin ^4a)-2(\cos^6a+\sin ^6a)=(\cos ^4a+\sin ^4a)-2(\cos ^2a+\sin ^2a)(\cos ^4a-\cos ^2a\sin ^2a+\sin ^4a)\)

\(=(\cos ^4a+\sin ^4a)-2(\cos ^4a-\cos ^2a\sin ^2a+\sin ^4a)\)

\(=-(\cos ^4a-2\sin ^2a\cos ^2a+\sin ^4a)=-(\cos ^2a-\sin ^2a)^2=-\cos ^22a\)

(bạn xem lại đề. Nếu thay $(\cos ^4a+\sin ^4a)$ thành $3(\cos ^4a+\sin ^4a)$ thì kết quả thu được là $(\cos ^2a+\sin ^2a)^2=1$ như yêu cầu)

2) Sửa đề:

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

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

Bạn lưu ý viết đề bài chuẩn hơn.

3)

\(\sin ^4a+\cos ^4a-\sin ^6a-\cos ^6a=\sin ^4a+\cos ^4a-[(\sin ^2a)^3+(\cos ^2a)^3]\)

\(=\sin ^4a+\cos ^4a-(\sin ^2a+\cos ^2a)(\sin ^4a-\sin ^2a\cos ^2a+\cos ^4a)\)

\(=\sin ^4a+\cos ^4a-(\sin ^4a-\sin ^2a\cos ^2a+\cos ^4a)\)

\(=\sin ^2a\cos ^2a\) (đpcm)

4)

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

5)

\(\frac{\tan a}{1-\tan ^2a}.\frac{\cot ^2a-1}{\cot a}=\frac{\tan a}{(tan a\cot a)^2-\tan ^2a}.\frac{\cot ^2a-1}{\cot a}\)

\(=\frac{\tan a}{\tan ^2a(\cot ^2a-1)}.\frac{\cot ^2a-1}{\cot a}=\frac{1}{\tan a\cot a}=\frac{1}{1}=1\)

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Mấu chốt của các bài này là bạn sử dụng 2 công thức sau:

1. \(\sin ^2x+\cos^2x=1\)

2. \(\tan x.\cot x=1\)

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.

\(cos^3xsinx-sin^3xcosx=sinx.cosx\left(cos^2x-sin^2x\right)=\dfrac{1}{2}sin2x.cos2x=\dfrac{1}{4}sin4x\)

\(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=1-\dfrac{1}{2}\left(2sinx.cosx\right)^2=1-\dfrac{1}{2}sin^22x\)

\(=1-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)=\dfrac{3}{4}+\dfrac{1}{4}cos4x=\dfrac{1}{4}\left(3+cos4x\right)\)

1.

\(2cos\left(a+b\right)=cosa.cos\left(\pi+b\right)\)

\(\Leftrightarrow2cosa.cosb-2sina.sinb=-cosa.cosb\)

\(\Leftrightarrow2sina.sinb=3cosa.cosb\Rightarrow4sin^2a.sin^2b=9cos^2a.cos^2b\)

\(\Rightarrow4\left(1-cos^2a\right)\left(1-cos^2b\right)=9cos^2a.cos^2b\)

\(\Leftrightarrow4-4\left(cos^2a+cos^2b\right)=5cos^2a.cos^2b\)

\(A=\dfrac{1}{cos^2a+2\left(sin^2a+cos^2a\right)}+\dfrac{1}{cos^2b+2\left(sin^2b+cos^2b\right)}\)

\(=\dfrac{1}{2+cos^2a}+\dfrac{1}{2+cos^2b}=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+cos^2a.cos^2b}\)

\(=\dfrac{4+cos^2a+cos^2b}{4+2\left(cos^2a+cos^2b\right)+\dfrac{4}{5}-\dfrac{4}{5}\left(cos^2a+cos^2b\right)}=\dfrac{4+cos^2a+cos^2b}{\dfrac{24}{5}+\dfrac{6}{5}\left(cos^2a+cos^2b\right)}=\dfrac{5}{6}\)

2.

\(A=2cos\dfrac{2x}{3}\left(cos\dfrac{2\pi}{3}+cos\dfrac{4x}{3}\right)=2cos\dfrac{2x}{3}\left(cos\dfrac{4x}{3}-\dfrac{1}{2}\right)\)

\(=2cos\dfrac{2x}{3}.cos\dfrac{4x}{3}-cos\dfrac{2x}{3}\)

\(=cos3x+cos\dfrac{2x}{3}-cos\dfrac{2x}{3}\)

\(=cos3x\)

\(B=\dfrac{cos2b-cos2a}{cos^2a.sin^2b}-tan^2a.cot^2b=\dfrac{1-2sin^2b-\left(1-2sin^2a\right)}{cos^2a.sin^2b}-tan^2a.cot^2b\)

\(=\dfrac{2sin^2a-2sin^2b}{cos^2a.sin^2b}-tan^2a.cot^2b=2tan^2a\left(1+cot^2b\right)-2\left(1+tan^2a\right)-tan^2a.cot^2b\)

\(=2tan^2a+2tan^2a.cot^2b-2-2tan^2a-tan^2a.cot^2b\)

\(=tan^2a.cot^2b-2\)

3.

\(\dfrac{2sina}{cosa}=\dfrac{sin\left(a+b\right)}{cos\left(a+b\right)}\Leftrightarrow2sina.cos\left(a+b\right)=cosa.sin\left(a+b\right)\)

\(\Leftrightarrow sina.cos\left(a+b\right)=sin\left(a+b\right).cosa-cos\left(a+b\right)sina\)

\(\Leftrightarrow sina.cos\left(a+b\right)=sin\left(a+b-a\right)\)

\(\Leftrightarrow sina.cos\left(a+b\right)=sinb\)

b.

\(\dfrac{2sina}{cosa}=\dfrac{sin\left(a+b\right)}{cos\left(a+b\right)}\Leftrightarrow2sina.cos\left(a+b\right)=cosa.sin\left(a+b\right)\)

\(\Leftrightarrow sin\left(2a+b\right)+sin\left(-b\right)=\dfrac{1}{2}sin\left(2a+b\right)+\dfrac{1}{2}sinb\)

\(\Leftrightarrow\dfrac{1}{2}sin\left(2a+b\right)=\dfrac{3}{2}sinb\)

\(\Leftrightarrow sin\left(2a+b\right)=3sinb\)

a) Ta có: \({\left( {\sin \alpha  + \cos \alpha } \right)^2} = {\sin ^2}\alpha  + 2\sin \alpha \cos \alpha  + {\cos ^2}\alpha  = 1 + \sin 2\alpha \;\)

b) \({\cos ^4}\alpha  - {\sin ^4}\alpha  = \left( {{{\cos }^2}\alpha  - {{\sin }^2}\alpha } \right)\left( {{{\cos }^2}\alpha  + {{\sin }^2}\alpha } \right) = \cos 2\alpha \;\)

.jkilfo,o7m5ijk

 Ta có $\sin 5\alpha -2\sin \alpha \left(\cos 4\alpha +\cos 2\alpha \right)=\sin 5\alpha -2\sin \alpha .\cos 4\alpha -2\sin \alpha .\cos 2\alpha$

$=\sin 5\alpha -\left(\sin 5\alpha -\sin 3\alpha \right)-\left(\sin 3\alpha -\sin \alpha \right)$

$=\sin \alpha .$

Vậy $\sin 5\alpha -2\sin \alpha \left(\cos 4\alpha +\cos 2\alpha \right)=\sin \alpha$

a: 

b: \(B=3-sin^290^0+2\cdot cos^260^0-3\cdot tan^245^0\)

\(=3-1+2\cdot\left(\dfrac{1}{2}\right)^2-3\cdot1^2\)

\(=2-3+2\cdot\dfrac{1}{4}=-1+\dfrac{1}{2}=-\dfrac{1}{2}\)

c: \(C=sin^245^0-2\cdot sin^250^0+3\cdot cos^245^0-2\cdot sin^240^0+4\cdot tan55\cdot tan35\)

\(=\left(\dfrac{\sqrt{2}}{2}\right)^2+3\cdot\left(\dfrac{\sqrt{2}}{2}\right)^2-2\cdot\left(sin^250^0+sin^240^0\right)+4\)

\(=\dfrac{1}{2}+3\cdot\dfrac{1}{2}-2+4\)

\(=2-2+4=4\)

a)    Ta có:

\(\begin{array}{l}{\sin ^4}\alpha  - {\cos ^4}\alpha  = 1 - 2{\cos ^2}\alpha \\ \Leftrightarrow \left( {{{\sin }^2}\alpha  + {{\cos }^2}\alpha } \right)\left( {{{\sin }^2}\alpha  - {{\cos }^2}\alpha } \right) = 1 - 2{\cos ^2}\alpha \\ \Leftrightarrow {\sin ^2}\alpha  - {\cos ^2}\alpha  - 1 + 2{\cos ^2}\alpha  = 0\\ \Leftrightarrow {\sin ^2}\alpha  + {\cos ^2}\alpha  - 1 = 0\\ \Leftrightarrow 1 - 1 = 0\\ \Leftrightarrow 0 = 0\end{array}\)

Đẳng thức luôn đúng

b)    Ta có:

\(\begin{array}{l}\tan \alpha  + \cot \alpha  = \frac{1}{{\sin \alpha .\cos \alpha }}\\ \Leftrightarrow \frac{{\sin \alpha }}{{\cos \alpha }} + \frac{{\cos \alpha }}{{\sin \alpha }} = \frac{1}{{\sin \alpha .\cos \alpha }}\\ \Leftrightarrow \frac{{{{\sin }^2}\alpha  + {{\cos }^2}\alpha }}{{\cos \alpha .\sin \alpha }} = \frac{1}{{\sin \alpha .\cos \alpha }}\\ \Leftrightarrow \frac{1}{{\sin \alpha .\cos \alpha }} = \frac{1}{{\sin \alpha .\cos \alpha }}\end{array}\)

Đẳng thức luôn đúng