(1)设tanα=−12,求[1sin2α−sinαcosα−2cos2α的值;
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  • (1)∵1=sin2α+cos2α,tanα=−

    1

    2.

    ∴原式=

    sin2α+cos2α

    sin2α−sinαcosα−2cos2α=

    sin2α

    cos2α+

    cos 2α

    cos2α

    sin2α

    cos2α−

    sin αcosα

    cos2α−

    2cos 2α

    cos2α=

    tan2α+1

    tan2α−tanα−2=

    1

    4+1

    1

    4+

    1

    2−2=−1;

    (2)∵由-180°<α<-90°,得-105°<α+75°<-15°,

    ∴sin(75°+α)=−

    1−cos2(75°+α)=−

    2

    2

    3,

    ∵cos(15°-α)=cos[90°-(75°+α)]=sin(75°+α)

    ∴cos(15°-α)=−

    2

    2

    3.