Ảnh 1 là bài 1,3. Ảnh 2 là bài 2 nhé bạn.

Bài 3: 

Ta có: \(A=\cos^220^0+\cos^240^0+\cos^250^0+\cos^270^0\)

\(=\left(\sin^270^0+\cos^270^0\right)+\left(\sin^250^0+\cos^250^0\right)\)

=1+1

=2

\(a,A=\left(\cos^220^0+\cos^270^0\right)+\left(\cos^240^0+\cos^250^0\right)\\ A=\left(\cos^220^0+\sin^220^0\right)+\left(\cos^240^0+\sin^240^0\right)=1+1=2\\ b,B=\left(\cos^2\alpha\right)^3+\left(\sin^2\alpha\right)^3+3\sin^2\alpha\cdot\cos^2\alpha\cdot\left(\sin^2\alpha+\cos^2\alpha\right)\\ B=\left(\sin^2\alpha+\cos^2\alpha\right)^3=1^3=1\)

\(A=\cos^210^0+\cos^220^0+\sin^220^0+\sin^210^0\\ A=1+1=2\)

cos^2(a-b)-cos^2(a+b)

=[cos(a-b)-cos(a+b)]*[cos(a-b)+cos(a+b)]

=[cosa*cosb+sina*sinb-cosa*cosb+sina*sinb]*[cosa*cosb+sina*sinb+cosa*cosb-sina*sinb]

=2*sina*sin*b*2*cosa*cosb

=sin2a*sin2b

\(A=cos^2x+\dfrac{1+cos\left(\dfrac{2\pi}{3}+2x\right)}{2}+\dfrac{1+cos\left(\dfrac{2\pi}{3}-2x\right)}{2}\\ =cos^2x+1+\dfrac{cos\left(\dfrac{2\pi}{3}+2x\right)+cos\left(\dfrac{2\pi}{3}-2x\right)}{2}\\ =cos^2x+1+cos\left(\dfrac{2\pi}{3}\right).cos2x\\ =cos^2x+1-\dfrac{1}{2}.cos2x=\dfrac{1+cos2x}{2}+1-\dfrac{cos2x}{2}=\dfrac{3}{2}.\)

a) sin²30 độ - sin²40 độ - sin²50 độ + sin² 60 độ

= cos²60^o - cos²50^o - sin²50^o + sin²60^o

= (cos²60^o + sin²60^o) - (cos²50^o + sin²50^o)

= 1 - 1 = 0

b) cos²25 độ - cos²35độ + cos²45 độ -cos² 55 độ + cos² 65 độ

= sin²65^o - sin²55^o + cos²45^o - cos²55^o + cos²65^o

= (sin²65^o + cos²65^o) - (sin²55^o + cos²55^o) + cos²45^o

=  1 - 1 +1/2

= 1/2