Đáp án:

 $\sin^2\alpha+\sin^2\left(\alpha-\dfrac{\pi}{3}\right)-\sin\alpha\sin\left(\alpha-\dfrac{\pi}{3}\right)=\dfrac{3}{4}$

Giải thích các bước giải:

$\sin^2\alpha+\sin^2\left(\alpha-\dfrac{\pi}{3}\right)-\sin\alpha\sin\left(\alpha-\dfrac{\pi}{3}\right)$

$=\left[\sin^2\alpha-2\sin\alpha\sin\left(\alpha-\dfrac{\pi}{3}\right)+\sin^2\left(\alpha-\dfrac{\pi}{3}\right)\right]+\sin\alpha\sin\left(\alpha-\dfrac{\pi}{3}\right)$

$=\left[\sin\alpha-\sin\left(\alpha-\dfrac{\pi}{3}\right)\right]^2+\sin\alpha\sin\left(\alpha-\dfrac{\pi}{3}\right)$

$=\left[2\cos\left(\alpha-\dfrac{\pi}{6}\right).\sin\dfrac{\pi}{6}\right]^2+\dfrac{1}{2}.\left[\cos\dfrac{\pi}{3}-\cos\left(2\alpha-\dfrac{\pi}{3}\right)\right]$

$=\left[2\cos\left(\alpha-\dfrac{\pi}{6}\right).\dfrac{1}{2}\right]^2+\dfrac{1}{2}.\dfrac{1}{2}-\dfrac{1}{2}.\cos\left(2\alpha-\dfrac{\pi}{3}\right)$

$=\cos^2\left(\alpha-\dfrac{\pi}{6}\right)+\dfrac{1}{4}-\dfrac{1}{2}.\left[2\cos^2\left(\alpha-\dfrac{\pi}{6}\right)-1\right]$

$=\cos^2\left(\alpha-\dfrac{\pi}{6}\right)+\dfrac{1}{4}-\cos^2\left(\alpha-\dfrac{\pi}{6}\right)+\dfrac{1}{2}$

$=\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{3}{4}$.

`b)` `sin^2α+sin^2(α-π/3)-sinαsin(α-π/3)`

`=sin^2α+sin(α-π/3).[sin(α-π/3)-sinα]`

`=sin^2α+sin(α-π/3).2.cos\ {α-π/3+α}/2sin\ (α-π/3-α)/2`

`=sin^2α+sin(α-π/3).2cos\ (α-π/6)sin\ ({-π}/6)`

`=sin^2α+2.{-1}/ 2 . sin(α-π/3).cos(α-π/6)`

`=sin^2α-1 . 1/ 2.[sin(α-π/3+α-π/6)+sin(α-π/3-α+π/6)]`

`=sin^2α-1/ 2 [sin(2α-π/2)+sin(-π/6)]`

`=sin^2α-1/ 2 [-sin(π/2-2α)-1/2]`

`=sin^2α+1/ 2 cos2α+1/4`

`=sin^2α+1/ 2 (1-2sin^2α)+1/ 4`

`=sin^2α+1/ 2 -sin^2α+1/ 4`

`=3/ 4` là hằng số

Vậy giá trị của biểu thức đã cho là hằng số không phụ thuộc `α`