计算sin(-π+π/3)+tan(3π/2-π/6)+sin15,
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计算sin(-π+π/3)+tan(3π/2-π/6)+sin15,
计算sin(-π+π/3)+tan(3π/2-π/6)+sin15,
计算sin(-π+π/3)+tan(3π/2-π/6)+sin15,
楼主题目中的角度有：-π+π/3、3π/2-π/6、15.
前边的-π+π/3、3π/2-π/6,单位应该是弧度吧?
那……后边的15,单位应该是什么呀?是弧度?还是度?
sin(-π+π/3)+tan(3π/2-π/6)+sin15
=sin(-2π/3)+tan(π+π/3)+sin15
=-sin(2π/3)+tan(π/3)+sin15
=-sin(π-π/3)+tan(π/3)+sin15
=-(1/2)√3+√3+sin15
=(√3)/2+sin15
再要往下算,就必须知道15的单位了.
如果不能确定单位,就没办法算了.
1、假设是弧度：
因为：15弧度=2700/π（°）,不是特殊角,只能借助于计算器了.
有：
sin(-π+π/3)+tan(3π/2-π/6)+sin15
=(√3)/2+sin15
≈(√3)/2+0.65028784
≈1.51631324
2、如果是度：
sin15°=sin(30°/2)
=√[(1-cos30°)/2]
=√[(2-√3)/4]
=[√(2-√3)]/2
=[√(√3-1)²/2]/2
=(√3-1)/(2√2)
=(√6-√2)/4
因此：
sin(-π+π/3)+tan(3π/2-π/6)+sin15°
=(√3)/2+sin15°
=(√3)/2+(√6-√2)/4
=(2√3+√6-√2)/4
sin15°=（√6-√2）/4
sin(-π+π/3)+tan(3π/2-π/6)+sin15°
=sin(-2π/3)+tan(4π/3)+（√6-√2）/4
=-sin(2π/3)+tan(π+π/3)+（√6-√2）/4
=-sin(π-π/3)+tan(π/3)+（√6-√2）/4
=-sin(π/3)+√3+（...
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sin15°=（√6-√2）/4
sin(-π+π/3)+tan(3π/2-π/6)+sin15°
=sin(-2π/3)+tan(4π/3)+（√6-√2）/4
=-sin(2π/3)+tan(π+π/3)+（√6-√2）/4
=-sin(π-π/3)+tan(π/3)+（√6-√2）/4
=-sin(π/3)+√3+（√6-√2）/4
=-√3/2+√3+（√6-√2）/4
=√3/2+（√6-√2）/4
=2√3/4+（√6-√2）/4
=（2√3+√6-√2）/4
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