sin  α  - sin  α   c o s 2 α  = sin  α (1 – c o s 2 α )

= sin  α [( sin 2 α  +  c o s 2 α ) –  c o s 2 α ]

= sin  α .( sin 2 α  +  c o s 2 α  –  c o s 2 α )

= sin  α . sin 2 α  =  sin 3 α

bài 1: ta có : \(cos^220+cos^240+cos^250+cos^270\)

\(=cos^220+cos^270+cos^240+cos^250\)

\(=cos^220+cos^2\left(90-20\right)+cos^240+cos^2\left(90-40\right)\)

\(=cos^220+sin^220+cos^240+sin^240=1+1=2\)

bài 2: a) ta có : \(cot^2\alpha-cos^2\alpha=cos^2\alpha\left(\dfrac{1}{sin^2\alpha}-1\right)=cos^2\alpha.\left(\dfrac{1-sin^2\alpha}{sin^2\alpha}\right)\)

\(=cos^2\alpha.\left(\dfrac{cos^2\alpha}{sin^2\alpha}\right)=cos^2\alpha.cot^2\alpha\left(đpcm\right)\)

b) ta có : \(sin^2\alpha+cos^2\alpha=1\Leftrightarrow sin^2\alpha=1-cos^2\alpha\)

\(\Leftrightarrow sin^2\alpha=\left(1-cos\alpha\right)\left(1+cos\alpha\right)\Leftrightarrow\dfrac{1+cos\alpha}{sin\alpha}=\dfrac{sin\alpha}{1-cos\alpha}\left(đpcm\right)\)

dạ e cảm ơn nh ạ!!!!

a: \(A=cos^4a+2\cdot cos^2a\cdot\sin^2a+\sin^4a\)

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

=1

=>A không phụ thuộc vào biến

b: \(B=\sin^4a+cos^2a\cdot\sin^2a+cos^2a\)

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

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

=1

=>B không phụ thuộc vào biến

c: \(C=2\left(\sin a-cosa\right)^2-\left(\sin a+cosa\right)^2+6\cdot\sin a\cdot cosa\)

\(=2\left(1-2\cdot\sin a\cdot cosa\right)-\left(1+2\cdot\sin a\cdot cosa\right)+6\cdot\sin a\cdot cosa\)

\(=2-4\cdot\sin a\cdot cosa-1-2\cdot\sin a\cdot cosa+6\cdot\sin a\cdot cosa\)

=2-1

=1

=>C không phụ thuộc vào biến

d: \(D=\left(\tan a-\cot a\right)^2-\left(\tan a+\cot a\right)^2\)

\(=\tan^2a-2\cdot\tan a\cdot\cot a+\cot^2a-\left(\tan^2a+2\cdot\tan a\cdot\cot a+\cot^2a\right)\)

\(=-4\cdot\tan a\cdot\cot a=-4\)

=>D không phụ thuộc vào biến

e: \(E=4\cdot cos^2a+\left(\sin a-cosa\right)^2+\left(\sin a+cosa\right)^2+2\left(\sin^2a-cos^2a\right)\)

\(=4\cdot cos^2a+\sin^2a+cos^2a-2\cdot\sin a\cdot cosa+\sin^2a+cos^2a+2\cdot\sin a\cdot cosa+2\left(\sin^2a-cos^2a\right)\)

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

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

=>E không phụ thuộc vào biến

f: \(F=\frac{1}{1+\sin a}+\frac{1}{1-\sin a}-2\cdot\tan^2a\)

\(=\frac{1-\sin a+1+\sin a}{\left(1+\sin a\right)\left(1-\sin a\right)}-2\cdot\tan^2a\)

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

=>F không phụ thuộc vào biến

\(A=\left(\sin\alpha+\cos\alpha+\sin\alpha-\cos\alpha\right)^2-2\left(\sin\alpha+\cos\alpha\right)\left(\sin\alpha-\cos\alpha\right)\)

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

\(B=\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)=\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha\)

\(=\left(\sin^2\alpha+\cos^2\alpha\right)^2-1=0\)

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

\(=3\left(\sin^2\alpha+\cos^2\alpha-\frac{1}{9}\right)^2-\frac{1}{9}=\frac{61}{27}\)

a: \(\dfrac{\cos\alpha}{1-\sin\alpha}=\dfrac{1+\sin\alpha}{\cos\alpha}\)

\(\Leftrightarrow\cos^2\alpha=1-\sin^2\alpha\)(đúng)

b: Ta có: \(\dfrac{\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha-\cos\alpha\right)^2}{\sin\alpha\cdot\cos\alpha}\)

\(=\dfrac{4\cdot\sin\alpha\cdot\cos\alpha}{\sin\alpha\cdot\cos\alpha}\)

=4

a: pi/2<a<pi

=>sin a>0

\(sina=\sqrt{1-\left(-\dfrac{1}{\sqrt{3}}\right)^2}=\dfrac{\sqrt{2}}{\sqrt{3}}\)

\(sin\left(a+\dfrac{pi}{6}\right)=sina\cdot cos\left(\dfrac{pi}{6}\right)+sin\left(\dfrac{pi}{6}\right)\cdot cosa\)

\(=\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{\sqrt{3}}+\dfrac{1}{2}\cdot-\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{6}-2}{2\sqrt{3}}\)

b: \(cos\left(a+\dfrac{pi}{6}\right)=cosa\cdot cos\left(\dfrac{pi}{6}\right)-sina\cdot sin\left(\dfrac{pi}{6}\right)\)

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

c: \(sin\left(a-\dfrac{pi}{3}\right)\)

\(=sina\cdot cos\left(\dfrac{pi}{3}\right)-cosa\cdot sin\left(\dfrac{pi}{3}\right)\)

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

d: \(cos\left(a-\dfrac{pi}{6}\right)\)

\(=cosa\cdot cos\left(\dfrac{pi}{6}\right)+sina\cdot sin\left(\dfrac{pi}{6}\right)\)

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

\(1+\sin^2\alpha+\cos^2\alpha=1+1=2\)

1-sin²α=cos2α

\(F=cos\left(\frac{\pi}{4}+a\right)\cdot cos\left(\frac{\pi}{4}-a\right)\)

\(=\frac12\cdot\left\lbrack cos\left(\frac{\pi}{4}+a-\frac{\pi}{4}+a\right)+cos\left(\frac{\pi}{4}+a+\frac{\pi}{4}+a\right)\right\rbrack\)

\(=\frac12\cdot\left\lbrack cos\left(2a\right)+cos\left(\frac{\pi}{2}\right)\right\rbrack=\frac12\cdot cos2a\)

\(G=\sin\left(\frac{\pi}{3}+a\right)\cdot cos\left(\frac{\pi}{3}-a\right)\)

\(=\frac12\cdot\left\lbrack\sin\left(\frac{\pi}{3}+a+\frac{\pi}{3}-a\right)+\sin\left(\frac{\pi}{3}+a-\frac{\pi}{3}+a\right)\right\rbrack\)

\(=\frac12\cdot\left\lbrack\sin\left(\frac23\pi\right)+\sin2a\right\rbrack=\frac12\cdot\left\lbrack\frac12+\sin2a\right\rbrack\)

\(H=cos\left(\frac{\pi}{2}-a\right)\cdot\sin\left(\frac{\pi}{2}+a\right)\)

\(=\frac12\cdot\left\lbrack\sin\left(\frac{\pi}{2}+a+\frac{\pi}{2}-a\right)+\sin\left(\frac{\pi}{2}+a-\frac{\pi}{2}+a\right)\right\rbrack\)

\(=\frac12\cdot\left\lbrack\sin\left(\pi\right)+\sin2a\right\rbrack=\frac12\left\lbrack2\cdot\sin a\cdot cosa\right\rbrack=\sin a\cdot cosa\)

\(I=\sin\left(\frac{\pi}{4}+a\right)-cos\left(\frac{\pi}{4}-a\right)\)

\(=\sin\left(\frac{\pi}{4}+a\right)-\sin\left(\frac{\pi}{2}-\frac{\pi}{4}+a\right)=\sin\left(\frac{\pi}{4}+a\right)-\sin\left(\frac{\pi}{4}+a\right)\)

=0

\(K=cos\left(\frac{\pi}{6}-x\right)-\sin\left(\frac{\pi}{3}+x\right)\)

\(=\sin\left(\frac{\pi}{2}-\frac{\pi}{6}+x\right)-\sin\left(\frac{\pi}{3}+x\right)=\sin\left(\frac{\pi}{3}+x\right)-\sin\left(\frac{\pi}{3}+x\right)\)

=0

Bài 1 :

\(C=cos^2a\left(cos^2a+sin^2a\right)+sin^2a=cos^2a+sin^2a=1\)

a) Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)

\(\Leftrightarrow\cos^2\alpha=1-\dfrac{9}{25}=\dfrac{16}{25}\)

Ta có: \(A=5\cdot\sin^2\alpha+6\cdot\cos^2\alpha\)

\(=5\left(\sin^2\alpha+\cos^2\alpha\right)+\cos^2\alpha\)

\(=5+\dfrac{16}{25}=\dfrac{141}{25}\)

phần b ?