利用半角公式，求下列各式的值1，sin(pai/12)-cos(pai/12). 2，tan(pai/8)+tan(3pai/8) 求下列各式的值cos7π/12.(2）cos（-61π/12...
\u6c42\u503csinpai/12cospai/12=?cos2pai/8-1/2=?sin(\u03c0/12)cos(\u03c0/12)
=(1/2)\u00b72sin(\u03c0/12)cos(\u03c0/12)
=(1/2)sin(\u03c0/6)
=1/2\u00d71/2
=1/4.
cos²(\u03c0/8)-1/2
=[1+cos(\u03c0/4)]/2-1/2
=(1/2)cos(\u03c0/4)
=1/2\u00d7\u221a2/2
=\u221a2/4\u3002
\u9605\u8bfb\u4e0b\u9762\u6750\u6599\uff1a\u6839\u636e\u4e24\u89d2\u548c\u4e0e\u5dee\u7684\u6b63\u5f26\u516c\u5f0f\uff0c\u6709
sin\uff08\u03b1+\u03b2\uff09=sin\u03b1cos\u03b2+cos\u03b1sin\u03b2------\u2460
sin\uff08\u03b1-\u03b2\uff09=sin\u03b1cos\u03b2-cos\u03b1sin\u03b2------\u2461
\u7531\u2460+\u2461\u5f97sin\uff08\u03b1+\u03b2\uff09+sin\uff08\u03b1-\u03b2\uff09=2sin\u03b1cos\u03b2------\u2462
\u4ee4\u03b1+\u03b2=A\uff0c\u03b1-\u03b2=\u03b2 \u6709\u03b1=
A+B
2
\uff0c\u03b2=
A-B
2
\u4ee3\u5165\u2462\u5f97 sinA+subB=2sin
A+B
2
cos
A-B
2
\uff0e
\uff08\u2160\uff09 \u7c7b\u6bd4\u4e0a\u8ff0\u63a8\u7406\u65b9\u6cd5\uff0c\u6839\u636e\u4e24\u89d2\u548c\u4e0e\u5dee\u7684\u4f59\u5f26\u516c\u5f0f\uff0c\u8bc1\u660e\uff1acosA-cosB=-2sin
A+B
2
sin
A-B
2
\uff1b
\uff08\u2161\uff09\u6c42\u503c\uff1asin220\u00b0+cos250\u00b0+sin20\u00b0cos50\u00b0\uff08\u63d0\u793a\uff1a\u5982\u679c\u9700\u8981\uff0c\u4e5f\u53ef\u4ee5\u76f4\u63a5\u5229\u7528\u9605\u8bfb\u6750\u6599\u53ca\uff08\u2160\uff09\u4e2d\u7684\u7ed3\u8bba\uff09
\u8003\u70b9\uff1a\u4e09\u89d2\u51fd\u6570\u7684\u79ef\u5316\u548c\u5dee\u516c\u5f0f\uff1b\u4e09\u89d2\u51fd\u6570\u7684\u548c\u5dee\u5316\u79ef\u516c\u5f0f\uff0e
\u4e13\u9898\uff1a\u4e09\u89d2\u51fd\u6570\u7684\u6c42\u503c\uff0e
\u5206\u6790\uff1a\uff08\u2160\uff09\u628a\u4e24\u89d2\u548c\u7684\u4f59\u9879\u516c\u5f0f\u51cf\u53bb\u4e24\u89d2\u5dee\u7684\u4f59\u9879\u516c\u5f0f\uff0c\u518d\u628a\u03b1+\u03b2=A\uff0c\u03b1-\u03b2=B\u4ee3\u5165\u5316\u7b80\u53ef\u5f97\u7ed3\u8bba\uff0e
\uff08\u2161\uff09\u5229\u7528\u534a\u89d2\u516c\u5f0f\u4ee5\u53ca\u79ef\u5316\u548c\u5dee\u516c\u5f0f\u5316\u7b80\u8981\u6c42\u7684\u5f0f\u5b50\uff0c\u5373\u53ef\u6c42\u5f97\u7ed3\u679c\uff0e
\u89e3\u7b54\uff1a\u89e3 \uff08\u2160\uff09\u8bc1\u660e\uff1a\u56e0\u4e3acos\uff08\u03b1+\u03b2\uff09=cos\u03b1cos\u03b2-sin\u03b1sin\u03b2\uff0c------\u2460
cos\uff08\u03b1-\u03b2\uff09=cos\u03b1cos\u03b2+sin\u03b1sin\u03b2\uff0c------\u2461\u2026\uff081\u5206\uff09
\u2460-\u2461\u5f97cos\uff08\u03b1+\u03b2\uff09-cos\uff08\u03b1-\u03b2\uff09=-2sin\u03b1sin\u03b2\uff0e------\u2462\u2026\uff082\u5206\uff09
\u4ee4\u03b1+\u03b2=A\uff0c\u03b1-\u03b2=B\uff0c\u6709 \u03b1=
A+B
2
\uff0c\u03b2=
A-B
2
\uff0c
\u4ee3\u5165\u2462\u5f97 cosA-cosB=-2sin
A+B
2
sin
A-B
2
\uff0e\u2026\uff085\u5206\uff09
\uff08\u2161\uff09sin220\u00b0+cos250\u00b0+sin20\u00b0cos50\u00b0=1+
1
2
\uff08cos100\u00b0-cos40\u00b0\uff09+
1
2
\uff08sin70\u00b0-sin30\u00b0\uff09\u2026\uff088\u5206\uff09
=1-sin70\u00b0sin30\u00b0+
1
2
sin70\u00b0-
1
2
sin30\u00b0=
3
4
\uff0e\u2026\uff0812\u5206\uff09
\u70b9\u8bc4\uff1a\u672c\u9898\u4e3b\u8981\u8003\u67e5\u4e09\u89d2\u51fd\u6570\u7684\u79ef\u5316\u548c\u5dee\u4e0e\u548c\u5dee\u5316\u79ef\u516c\u5f0f\u7684\u8bc1\u660e\uff0c\u534a\u89d2\u516c\u5f0f\u7684\u5e94\u7528\uff0c\u5c5e\u4e8e\u4e2d\u6863\u9898\uff0e
1、0<π/12<π/4 cos(π/12)>sin(π/12)
sin(π/12)-cos(π/12)<0
[sin(π/12)-cos(π/12)]²
=sin²(π/12)+cos²(π/12)-2sin(π/12)cos(π/12)
=1-sin(π/6)
=1-1/2
=1/2
sin(π/12)-cos(π/12)=-√(1/2)=-√2/2
2、tan(π/8)+tan(3π/8)
= sin(π/4)/(1+cos(π/4)) + sin(3π/4)/(1+cos(3π/4))
=(√2/2) / (1+√2/2) + (√2/2)/(1-√2/2)
=√2/(2+√2)+√2/(2-√2)
=√2(2-√2)/2+√2(2+√2)/2
=2√2
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