(1) 6 sin(-90°)+3 sin 0°-8 sin 270°+12 cos 180°

2个回答

  • 解(1):6sin(-90°)+3sin0°-8sin270°+12cos180°

    =6×(-1)+3×0-8×(-1)+12×(-1)

    =-6+0+8-12

    =-10

    (2) 10cos270°+4sin0°+9tan0°+15cos360°

    =10×0+4×0+9×0+15×1

    =0+0+0+15

    =15

    (3) 2cos π/2-tan π/4+3/4 tan² π/6-sin π/6+cos² π/6+sin 3π/2

    =2×cos90°-tan45°+3/4×tan²30°-sin30°+coa²30°+sin270°

    =2×0-1+3/4×(√3/3)²-1/2+(√3/2)²+(-1)

    =0-1+3/4×1/3-1/2+3/4-1

    =-1+1/4-1/2+3/4-1

    =-3/2

    (4) sin²π/3+cos^4 3π/2-tan² π/3

    =sin²60°+cos^4 270°-tan²60°

    =(√3/2)²+0^4-(√3)²

    =3/4+0-3

    =3/4-3

    =-9/4