2018-2019学年人教B版必修4 3.3三角函数的积化和差与和差化积 学案
2018-2019学年人教B版必修4 3.3三角函数的积化和差与和差化积 学案第3页

=++-4×××=1;

②cosA+cosB+cosC-4sinsinsin=cos60°+cos90°+cos30°-4sin30°sin45°sin15°

=+0+-4×××=1;

③cosA+cosB+cosC-4sinsinsin=cos120°+cos30°+cos30°-4sin60°sin15°sin15°

=-++-4×sin215°

=-+-×(1-cos30°)=1.

(2)在(1)①中A+B+C=180°,有cosA+cosB+cosC-4sinsinsin=1;

在(1)②中A+B+C=180°,有cosA+cosB+cosC-4sinsinsin=1;

在(1)③中A+B+C=180°,有cosA+cosB+cosC-4sinsinsin=1.

猜想:当A+B+C=180°时,有cosA+cosB+cosC=1+4sinsinsin.

证明:当A+B+C=180°时,有A+B=180°-C,即=90°-,

∴cosA+cosB+cosC=2coscos+1-2sin2

=2cos(90°-)cos+1-2sin2

=2sincos-2sin2+1

=2sin(cos-sin)+1

=2sin(cos-cos)+1

=2sin(-2)sinsin(-)+1

=4sinsinsin+1.

∴cosA+cosB+cosC=1+4sinsinsin.

(3)∵10°+99.8°+70.2°=180°,

∴cos10°+cos99.8°+cos70.2°-4sin5°sin49.9°sin35.1°=1.

∴-2cos10°-2cos99.8°-2cos70.2°+8sin5°sin49.9°sin35.1°=-2.