化简COSa〔COSa—COSb〕加SINa〔SINa—SINb〕—1
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/07/14 07:51:23
化简COSa〔COSa—COSb〕加SINa〔SINa—SINb〕—1
![化简COSa〔COSa—COSb〕加SINa〔SINa—SINb〕—1](/uploads/image/z/16454684-20-4.jpg?t=%E5%8C%96%E7%AE%80COSa%E3%80%94COSa%E2%80%94COSb%E3%80%95%E5%8A%A0SINa%E3%80%94SINa%E2%80%94SINb%E3%80%95%E2%80%941)
COSa(COSa-COSb)+SINa(SINa-SINb)-1
=(COSa)^2-COSa*COSb+(SINa)^2-SINa*SINb-1
=(COSa)^2+(SINa)^2-COSa*COSb-SINa*SINb-1
=1-COSa*COSb-SINa*SINb-1
=-COSa*COSb-SINa*SINb
=-(COSa*COSb+SINa*SINb)
=-COS(a-b)
=(COSa)^2-COSa*COSb+(SINa)^2-SINa*SINb-1
=(COSa)^2+(SINa)^2-COSa*COSb-SINa*SINb-1
=1-COSa*COSb-SINa*SINb-1
=-COSa*COSb-SINa*SINb
=-(COSa*COSb+SINa*SINb)
=-COS(a-b)
sinA/sinB=cosA/cosB?
已知cos(a-b)=3/1,求(sina+sinb)(sina+sinb)+(cosa+cosb)(cosa+cosb
三角恒等公式cosa cosb sina cosb - cosa cosb cosa sinb - sina sinb
sinA+sinB+sinC>=cosA+cosB+cosC
sina cosa=sinb cosb是什么三角形
数学三角函数:sinA+sinB+sinC>1+cosA+cosB+cosC
已知锐角三角形ABC,证明sinA+sinB+sinB>cosA+cosB+cosC
化简:2(sinA)^2(sinB)^2+2(cosA)^2(cosB)^2-cos2A(cosB)^2
锐角三角形ABC中,证明sinA+sinB+sinC>cosA+cosB+cosC
证明在锐角三角形ABC中sinA+sinB>cosA+cosB
已知向量a=(cosa,sina),b=(cosb,sinb)
求证:在锐角三角形中,sinA+sinB+sinC>cosA+cosB+cosC