Tính giá trị của biểu thức:

B = cos^2 10° + cos^2 110° + cos^2 130°

`B=cos^2 10^o +cos^2 110^o +cos^2 130^o`

`=cos^2 10^o +cos^2 [90^o -(-20^o)]+cos^2 [90^o -(-40^o)]`

`=cos^2 10^o +sin^2 (-20^o)+sin^2 (-40^o)`

`=cos^2 10^o +sin^2 20^o +sin^2 40^o` 

`=(1+cos 20^o )/2+(1-cos 40^o)/2 +(1-cos 80^o)/2`

`=(3+cos 20^o -(cos 40^o +cos 80^o))/2`

`=(3+cos 20^o -2cos 20^o cos 60^o)/2`

`=(3+cos 20^o - 2. 1/2 cos20^o)/2`

`=(3+cos 20^o -cos20^o )/2`

`=3/2`

Vậy `B=3/2`
$\bullet$ Công thức lượng giác :

`+` `sin^2 a=(1-cos 2a)/2`

`+` `cos^2 a=(1+cos 2a)/2`

`+` `cos a+cos b=2cos (a-b)/2 cos (a+b)/2` 

`B=cos^2 10^o +cos^2 110^o +cos^2 130^o`

`B=cos^2 10^o +sin^2 20^o +sin^2 40^o`

`B=(1+cos20^o)/2+(1-cos40^o)/2+(1-cos80^o)/2`

`B=3/2+1/2(cos20^o -cos40^o -cos80^o)`

`B=3/2+1/2(cos20^o -2cos60^ocos20^o)`

`B=3/2+1/2(cos20^o -cos20^o)`

`B=3/2`