sin(A+B)*sin(A-B)=(sinAcosB+cosAsinB)(sinAcosB-cosAsinB)=sin²Acos²B-cos²Asin²B=sin²Acos²B-(1-sin²A)sin²B=sin²Acos²B-sin²B+sin²Asin²B=sin²A(cos²B+sin²B)-sin²B=sin²A-sin²B 结果...
由正弦定理 (sina)^2=(sinb)^2+(sinc)^2 等价于a^2=b^2+c^2 可知△abc直角三角形 a=π/2 sina=2sinbcosc 1=2sinbcos(π/2-b)1=2sinbsinb sinb=1/√2 可知b=π/4 △abc等腰直角三角形
解：cos2A=cos²A-sin²A=1-2sin²A =>sin²A=(1-cos2A)/2 同理得：sin²B=(1-cos2B)/2 所以：sin²A-sin²B=(1-cos2A)/2-=(1-cos2B)/2
答案 cos2A=cos²A-sin²A=1-2sin²A=>sin²A=(1-cos2A)/2同理得:sin²B=(1-cos2B)/2所以:sin²A-sin²B=(1-cos2A)/2-=(1-cos2B)/2相关推荐 1(sinA)的平方减去(sinB)的平方运用哪个公式可以等于(1-cos2A)/2-(1-cos2B)/2 反馈...
sin(A+B)*sin(A-B)=(sinAcosB+cosAsinB)(sinAcosB-cosAsinB)=sin²Acos²B-cos²Asin²B=sin²Acos²B-(1-sin²A)sin²B=sin²Acos²B-sin²B+sin²Asin²B=sin²A(cos²B+sin²B)-sin²B=sin²A-sin²B 解析看不懂?免费查看同类题视频解析查看解答 更多答案(...
由和差化积公式sinA+sinB=2*sin[(A+B)/2]*cos[(A-B)/2]sinA-sinB=2*sin[(A-B)/2]*cos[(A+B)/2]得:（sinA）＾2-（sinB）＾2=2*sin[(A+B)/2]*cos[(A-B)/2]*2*sin[(A-B)/2]*cos[(A+B)/2]=2*sin[(A+B)/2]*cos[(A+B)/2] * 2*sin[(A-B)/2]*cos...
sin(A+B)*sin(A-B)=(sinAcosB+cosAsinB)(sinAcosB-cosAsinB)=sin²Acos²B-cos²Asin²B=sin²Acos²B-(1-sin²A)sin²B=sin²Acos²B-sin²B+sin²Asin²B=sin²A(cos²B+sin²B)-sin²B=sin²A-sin²B 2020-06-09 18:39:42 大家...
(sin(a) + sin(b))(sin(a) - sin(b)) - sin(c)^2 = sin(b)sin(c)再利用三角恒等式 sin(x)sin(y) = (1/2)(cos(x - y) - cos(x + y))，将上式中的右边部分进行展开：(sin(a) + sin(b))(sin(a) - sin(b)) - sin(c)^2 = (1/2)(cos(b - c) - cos(...
sinA的平方=sinB的平方+sinBsinC+sinC的平方,则A=? 相关知识点: 试题来源: 解析 A=60度推导不难,但在这里写起来比较麻烦.简便起见,^ 就代表平方.sin^A=sin^(B+C)=[sinBcosC+sinCcosB]^=sin^Bcos^C+sin^Ccos^B+2sinBsinCcosBcosC=右边=sin^B+sinBsinC+sin^C化简得:2sin^Bsin^C+sinBsinC(1-2...