Đáp án:

Giải thích các bước giải:

Ta có:
`sin^2 . 10^@- sin^2. 20^@+sin2^2. 30^@ -sin2^(40^@) -sin^2. 50^@-sin^2. 50^@+sin^2. 60^@-sin^2. 70^@+sin^2. 80^@`

`S = \sin^2(10^\circ) - \sin^2(20^\circ) + \sin^2(30^\circ) - \sin^2(40^\circ) - 2\sin^2(50^\circ) + \sin^2(60^\circ) - \sin^2(70^\circ) + \sin^2(80^\circ)`
`=>\sin(90^\circ - x) = \cos(x)`
`+)\sin^2(70^\circ) = \cos^2(20^\circ)`
`+) \sin^2(80^\circ) = \cos^2(10^\circ)` 
`=>S = \sin^2(10^\circ) - \sin^2(20^\circ) + \sin^2(30^\circ) - \sin^2(40^\circ) - 2\sin^2(50^\circ) + \sin^2(60^\circ) - \cos^2(20^\circ) + \cos^2(10^\circ)`
`=>S = (\sin^2(10^\circ) + \cos^2(10^\circ)) + (\sin^2(30^\circ) + \cos^2(60^\circ)) - (\sin^2(20^\circ) + \cos^2(20^\circ)) - (\sin^2(40^\circ) + \cos^2(40^\circ)) - 2\sin^2(50^\circ)`
Ta có tính chất:
`\sin^2(x) + \cos^2(x) = 1`
`+)\sin^2(10^\circ) + \cos^2(10^\circ) = 1`
`+)\sin^2(30^\circ) + \cos^2(60^\circ) = \sin^2(30^\circ) + \sin^2(30^\circ) = \frac{1}{4} + \frac{3}{4} = 1`
`+)\sin^2(20^\circ) + \cos^2(20^\circ) = 1`
`+)\sin^2(40^\circ) + \cos^2(40^\circ) = 1`
`=>S = 1 + 1 - 1 - 1 - 2\sin^2(50^\circ)` 
`=>S = 0 - 2\sin^2(50^\circ)` 
`=>S = \sin^2(10^\circ) + \sin^2(80^\circ) - \sin^2(20^\circ) - \sin^2(70^\circ) + \sin^2(30^\circ) - \sin^2(40^\circ) - 2\sin^2(50^\circ)`
Vì `\sin^2(80^\circ) = \cos^2(10^\circ)\) và \(\sin^2(70^\circ) = \cos^2(20^\circ)`nên:

Tổng của những phần bù nhau này bằng `0`
`=>sin^2 . 10^@- sin^2. 20^@+sin2^2. 30^@ -sin2^(40^@) -sin^2. 50^@-sin^2. 50^@+sin^2. 60^@-sin^2. 70^@+sin^2. 80^@=0.`