Đáp án:
) Ta có
A=3sin−4cos4sin+2cos=3tan−44tan+2=3/2−42+2=−58A=3sin−4cos4sin+2cos=3tan−44tan+2=3/2−42+2=−58
b) Ta có
B=2sin2+3cos24sin2−2cos2=2tan2+34tan2−2=2.(4/9)+34.(4/9)−2=−352B=2sin2+3cos24sin2−2cos2=2tan2+34tan2−2=2.(4/9)+34.(4/9)−2=−352.
Giải thích các bước giải: