\(D=2\left(sin^2x+cos^2x\right)\left(sin^4x+cos^4x-sin^2x.cos^2x\right)-3\left(sin^4x+cos^4x\right)\)

\(=2\left(sin^4x+cos^4x\right)-2sin^2x.cos^2x-3\left(sin^4x+cos^4x\right)\)

\(=-\left(sin^4x+2sin^2x.cos^2x+cos^4x\right)\)

\(=-\left(sin^2x+cos^2x\right)^2=-1\)

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

\(=\dfrac{1}{2}+\dfrac{1}{2}\left[cos\left(2a-\dfrac{\pi}{3}\right)-cos\left(2a-\dfrac{2\pi}{3}\right)\right]\)

\(=\dfrac{1}{2}-sin\left(2a-\dfrac{\pi}{2}\right)sin\left(\dfrac{\pi}{6}\right)\)

\(=\dfrac{1}{2}+\dfrac{1}{2}cos2a=\dfrac{1}{2}+\dfrac{1}{2}\left(2cos^2a-1\right)=cos^2a\)

\(a)sin^4x+cos^4x=1-2sin^2x\cdot cos^2x\) 

\(\Leftrightarrow sin^4x+2sin^2x\cdot cos^2x+cos^4x=1\)

\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2=1\)(luôn đúng)

a) \sin ^{4} x+\cos ^{4} x=\sin ^{4} x+\cos ^{4} x+2 \sin ^{2} x \cos ^{2} x-2 \sin ^{2} x \cos ^{2} xsin4x+cos4x=sin4x+cos4x+2sin2xcos2x−2sin2xcos2x
\begin{aligned}&=\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x \\&=1-2 \sin ^{2} x \cos ^{2} x\end{aligned}​=(sin2x+cos2x)2−2sin2xcos2x=1−2sin2xcos2x​

b) \dfrac{1+\cot x}{1-\cot x}=\dfrac{1+\dfrac{1}{\tan x}}{1-\dfrac{1}{\tan x}}=\dfrac{\dfrac{\tan x+1}{\tan x}}{\dfrac{\tan x-1}{\tan x}}=\dfrac{\tan x+1}{\tan x-1}1−cotx1+cotx​=1−tanx1​1+tanx1​​=tanxtanx−1​tanxtanx+1​​=tanx−1tanx+1​

c) \dfrac{\cos x+\sin x}{\cos ^{3} x}=\dfrac{1}{\cos ^{2} x}+\dfrac{\sin x}{\cos ^{3} x}=\tan ^{2} x+1+\tan x\left(\tan ^{2} x+1\right)cos3xcosx+sinx​=cos2x1​+cos3xsinx​=tan2x+1+tanx(tan2x+1)
=\tan ^{3} x+\tan ^{2} x+\tan x+1=tan3x+tan2x+tanx+1

a) VT=(sin²x + cos ² x)² - 2sin² x . cos² x = VP 

b) VT= \(\dfrac{1+\dfrac{1}{tanx}}{1-\dfrac{1}{tanx}}\)=VP

c) VT= \(\dfrac{1}{cos^2x}+\dfrac{sinx}{cosx}.\dfrac{1}{cos^2x}=1+tan^2x+tanx.\left(1+tan^2x\right)=VP\)

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

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

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

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

\(2\left(sin^6x+cos^6x\right)+1=2\left(sin^2x+cos^2x\right)^3-6sin^2x.cos^2x\left(sin^2x+cos^2x\right)+1\)

\(=3-6sin^2x.cos^2x\) (1)

\(3\left(sin^4x+cos^4x\right)=3\left(sin^2x+cos^2x\right)^2-6sin^2x.cos^2x\)

\(=3-6sin^2x.cos^2x\) (2)

(1);(2) \(\Rightarrow\) đpcm

a, Ta có: \(sin^2\alpha+cos^2\alpha=1\Leftrightarrow\left(\dfrac{3}{5}\right)^2+cos^2\alpha=1\Leftrightarrow cos\alpha=\pm\dfrac{4}{5}\)

Vậy đẳng thức có thể đồng thời xảy ra.

b, Ta có: \(1+cot^2\alpha=\dfrac{1}{sin^2\alpha}\Rightarrow1+cot^2\alpha=\dfrac{1}{\left(\dfrac{1}{3}\right)^2}\Rightarrow cot\alpha=\pm2\sqrt{2}\)

Hai đẳng thức không thể đồng thời xảy ra.

c, Ta có: \(tan\alpha\cdot cot\alpha=1\Rightarrow3\cdot cot\alpha=1\Rightarrow cot\alpha=\dfrac{1}{3}\)

Đẳng thức có thể đồng thời xảy ra.

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

Câu 3:

\(A=cos\frac{\pi}{7}.cos\frac{5\pi}{7}.cos\frac{4\pi}{7}=cos\frac{\pi}{7}.cos\left(\pi-\frac{2\pi}{7}\right).cos\frac{4\pi}{7}\)

\(A=-cos\frac{\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}\)

\(\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{2}.2sin\frac{\pi}{7}.cos\frac{\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}\)

\(\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{2}.sin\frac{2\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}\)

\(\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{4}sin\frac{4\pi}{7}.cos\frac{4\pi}{7}\)

\(\Rightarrow sin\frac{\pi}{7}.A=-\frac{1}{8}sin\frac{8\pi}{7}=-\frac{1}{8}sin\left(\pi+\frac{\pi}{7}\right)=\frac{1}{8}sin\frac{\pi}{7}\)

\(\Rightarrow A=\frac{1}{8}\)

Câu 4:

Đầu tiên ta chứng minh công thức:

\(tana+tanb=\frac{sina}{cosa}+\frac{sinb}{cosb}=\frac{sina.cosb+cosa.sinb}{cosa.cosb}=\frac{sin\left(a+b\right)}{cosa.cosb}\)

Áp dụng để biến đổi tử số:

\(tan30+tan60+tan40+tan50=\frac{sin90}{cos30.cos60}+\frac{sin90}{cos40.cos50}=\frac{1}{cos30.cos60}+\frac{1}{cos40.cos50}\)

\(=\frac{2}{cos90+cos30}+\frac{2}{cos90+cos10}=\frac{2}{cos30}+\frac{2}{cos10}=2\left(\frac{cos30+cos10}{cos30.cos10}\right)\)

\(=2\left(\frac{2cos20.cos10}{cos30.cos10}\right)=\frac{4.cos20}{cos30}=\frac{8\sqrt{3}}{3}.cos20\)

\(\Rightarrow A=\frac{\frac{8\sqrt{3}}{3}cos20}{cos20}=\frac{8\sqrt{3}}{3}\)

Câu 5:

\(cos54.cos4-cos36.cos86=cos54.cos4-cos\left(90-54\right).cos\left(90-4\right)\)

\(=cos54.cos4-sin54.sin4=cos\left(54+4\right)=cos58\)

Câu 1:

\(A=\frac{1}{2sin10}-2sin70=\frac{1-4sin10.sin70}{2sin10}=\frac{1+2\left(cos80-cos60\right)}{2sin10}\)

\(=\frac{1+2cos80-1}{2sin10}=\frac{2cos80}{2sin10}=\frac{sin10}{sin10}=1\)

Câu 2:

\(cos10.cos30.cos50.cos70=cos10.cos30.\frac{1}{2}\left(cos120+cos20\right)\)

\(=\frac{1}{2}cos30\left(cos10.cos120+cos10.cos20\right)\)

\(=\frac{1}{2}cos30\left(cos10.cos120+\frac{1}{2}\left(cos30+cos10\right)\right)\)

\(=\frac{1}{2}cos30\left(cos10.cos120+\frac{1}{2}cos30+\frac{1}{2}cos10\right)\)

\(=\frac{1}{2}.\frac{\sqrt{3}}{2}\left(-\frac{1}{2}cos10+\frac{1}{2}\frac{\sqrt{3}}{2}+\frac{1}{2}cos10\right)\)

\(=\frac{3}{16}\)