一道三角比求过程tana=-1/3求:2+5cos2a/3+4sin2a

2+5cos(2α)/3+4sin(2α)=2+5[(1-tan²α)/(1+tan²α)]/3+4·2tanα/(1+tan²α)=2+5[1-(-1/3)²]/[1+(-1/3)²]/3+8·(-1/3)/[1+(-1/3)²]=14/15万能公式sinα=2tan(α/2)/[1+tan²(α...