sin(a+b)cos(a-b)=sina*cosa+sinb*cosb
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sin(a+b)cos(a-b)=sina*cosa+sinb*cosb
先学基础知识:
积化和差公式:sin(A)*cos(B) = (1/2)*[sin(A+B) + sin(A-B)]
二倍角公式 :sin(2A) = 2*sinAcosA
再做题:
sin(a+b)cos(a-b)
= (1/2)*{sin[(a+b)+(a-b)] + sin[(a+b)-(a-b)]}
= (1/2)*[sin(2a) + sin(2b)]
= (1/2)*[2*sin(a)*cos(a) + 2*sin(b)*cos(b)]
= sin(a)*cos(a) + sin(b)*cos(b)
积化和差公式:sin(A)*cos(B) = (1/2)*[sin(A+B) + sin(A-B)]
二倍角公式 :sin(2A) = 2*sinAcosA
再做题:
sin(a+b)cos(a-b)
= (1/2)*{sin[(a+b)+(a-b)] + sin[(a+b)-(a-b)]}
= (1/2)*[sin(2a) + sin(2b)]
= (1/2)*[2*sin(a)*cos(a) + 2*sin(b)*cos(b)]
= sin(a)*cos(a) + sin(b)*cos(b)
已知cos(a-b)=3/1,求(sina+sinb)(sina+sinb)+(cosa+cosb)(cosa+cosb
证明下列恒等式:sin(a+b)*cos(a-b)=sina*cosa+sinb*cosb
证明:cos(a+b)=cosa*cosb-sina*sinb
求证:cos(a+b)=cosa*cosb-sina*sinb
求证: sina+sinb=2sin[(a+b)/2]cos[(a-b)/2] cosa+cosb=2cos[(a+b)
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cosb=cos[(a+b)-a]=cos(a+b)cosa+sin(a+b)sina
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