怎么证明两角和的余弦公式Cos(x+y)=CosxCosy-SinxSiny
\u600e\u4e48\u8bc1\u660e\u4e24\u89d2\u548c\u7684\u4f59\u5f26\u516c\u5f0f Cos(x+y)=CosxCosy-SinxSiny\u89e3\u7b54\uff1a
\u600e\u4e48\u8bc1\u660e\u4e24\u89d2\u548c\u7684\u4f59\u5f26\u516c\u5f0f Cos(x+y)=CosxCosy-SinxSiny
\u90a3\u4e2a\u7b54\u6848\u8c01\u5199\u7684\uff1f\u600e\u4e48\u7528\u540e\u9762\u7684\u516c\u5f0f\uff0c\u8bc1\u524d\u9762\u7684\u7ed3\u8bba\u4e86\u3002
\u8fd9\u4e2a\u8bc1\u660e\u65b9\u6cd5\u5e94\u8be5\u662f\u89e3\u6790\u6cd5
\u5982\u4e0a\u56fe\u6240\u793a\uff0c\u8fd9\u662f\u4e00\u4e2a\u534a\u5f84\u4e3a 1 \u7684\u5355\u4f4d\u5706\u3002
AB^2=OA^2 + OB^2 - 2*OA * OB * cos(\u03b1-\u03b2) = 1 + 1 - 2*1*1*cos(\u03b1-\u03b2) = 2 - 2 *cos(\u03b1-\u03b2)
\u6839\u636e\u52fe\u80a1\u5b9a\u7406\uff0c
AB^2 = DE^2 + CF^2
= (OD - OE)^2 + (OF + OC)^2
= (OA * sin\u03b1 - OB * sin\u03b2) ^2 + (OB * cos\u03b2 - OA * cos\u03b1)^2
= (sin\u03b1 - sin\u03b2) ^2 + (cos\u03b2 - cos\u03b1)^2
= (sin\u03b1)^2 + (sin\u03b2)^2 - 2*sin\u03b1*sin\u03b2 + (cos\u03b1)^2 + (cos\u03b2)^2 - 2*cos\u03b1*cos\u03b2
= (sin\u03b1)^2 + (cos\u03b1)^2 + (sin\u03b2)^2 + (cos\u03b2)^2 - 2*sin\u03b1*sin\u03b2 - 2*cos\u03b1*cos\u03b2
= 1 + 1 - 2*sin\u03b1*sin\u03b2 - 2*cos\u03b1*cos\u03b2
\u7ed3\u5408\u8fd9\u4e24\u4e2a\u516c\u5f0f\uff0c\u6211\u4eec\u53ef\u4ee5\u5f97\u5230\uff1a
2 - 2 *cos(\u03b1-\u03b2) = 2 - 2*sin\u03b1*sin\u03b2 - 2*cos\u03b1*cos\u03b2
\u2234 cos(\u03b1-\u03b2) = cos\u03b1*cos\u03b2 + sin\u03b1*sin\u03b2
\u4e0a\u9762\u662f\u4e24\u89d2\u5dee\u7684\u4f59\u5f26\u516c\u5f0f\u3002\u4e24\u89d2\u548c\u7684\u4f59\u5f26\u516c\u5f0f\u5982\u4e0b\uff1a
cos(\u03b1+\u03b2) = cos(\u03b1 - (-\u03b2))
= cos\u03b1*cos(-\u03b2) + sin\u03b1*sin(-\u03b2)
\u2235 cos(-\u03b2) = cos\u03b2, sin(-\u03b2) = - sin\u03b2
\u2234 cos(\u03b1+\u03b2) = cos\u03b1*cos\u03b2 - sin\u03b1*sin\u03b2
右边=2*sin[(A+B)/2]*cos[(A-B)/2]
=2*[sin(A/2)*cos(B/2)+cos(A/2)sin(B/2)]*[cos(A/2)cos(B/2)+sin(A/2)sin(B/2)]
=2*sin(A/2)*cos(A/2)*cos(B/2)*cos(B/2)+2*cos(A/2)*cos(A/2)*sin(B/2)*cos(B/2)+2*sin(A/2)*sin(A/2)*cos(B/2)*sin(B/2)+2*sin(A/2)*cos(A/2)*sin(B/2)*sin(B/2)
=sinA*[cos(B/2)*cos(B/2)+sin(B/2)*sin(B/2)]+sin(B/2)*[cos(B/2)*cos(B/2)+sin(B/2)*sin(B/2)]
=sinA+sinB=左边
证毕
其中用到公式:
sinA=2*sin(A/2)*cos(A/2),sinB=2*cos(B/2)*sin(B/2)
cos(B/2)*cos(B/2)+sin(B/2)*sin(B/2)=1
其他的公式依此类推,自己推推看吧!
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