Question

急求答案及过程

(1) 6 sin(-90°)+3 sin 0°-8 sin 270°+12 cos 180°
（2）10cos 270°+4sin 0°+9tan 0°+15 cos 360°
(3 2 cos π/2- tan π/4+3/4 tan^2 π/6- sin π/6+ cos^2 π/6+ sin3π/2
(4) sin^2 π/3+ cos^4 3π/2- tan^2 π/3
帮忙解一下 需要过程

Answer 1

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

Answer 2

如图所示，[说明:下面的x2就是x的平方] 连接OC，设∠BOC=θ，矩形ABCD的 面积为S，则BC=30sinθ，OB=30cosθ（其中0＜θ＜π/2）；∴S=AB BC=