找不正确的一项 请简要说明理由!A cos(a+b)cos(a-b)=cos^2a-sin^2b(cosa的平方减sinb的平方)B sin(a+b)sin(a-b)=sin^2a-sin^2bC tan(a)+tanb/tan(a)-tanb=sin(a+b)/sin(a-b)C tan(a)+tanb/tan(a)-tanb改为C (tana+tanb)/(tana-t
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![找不正确的一项 请简要说明理由!A cos(a+b)cos(a-b)=cos^2a-sin^2b(cosa的平方减sinb的平方)B sin(a+b)sin(a-b)=sin^2a-sin^2bC tan(a)+tanb/tan(a)-tanb=sin(a+b)/sin(a-b)C tan(a)+tanb/tan(a)-tanb改为C (tana+tanb)/(tana-t](/uploads/image/z/918666-18-6.jpg?t=%E6%89%BE%E4%B8%8D%E6%AD%A3%E7%A1%AE%E7%9A%84%E4%B8%80%E9%A1%B9+%E8%AF%B7%E7%AE%80%E8%A6%81%E8%AF%B4%E6%98%8E%E7%90%86%E7%94%B1%21A+cos%EF%BC%88a%2Bb%29cos%28a-b%EF%BC%89%3Dcos%5E2a-sin%5E2b%28cosa%E7%9A%84%E5%B9%B3%E6%96%B9%E5%87%8Fsinb%E7%9A%84%E5%B9%B3%E6%96%B9%EF%BC%89B+sin%EF%BC%88a%2Bb%29sin%28a-b%EF%BC%89%3Dsin%5E2a-sin%5E2bC+tan%28a%29%2Btanb%2Ftan%28a%29-tanb%3Dsin%28a%2Bb%29%2Fsin%28a-b%29C+tan%28a%29%2Btanb%2Ftan%28a%29-tanb%E6%94%B9%E4%B8%BAC+%EF%BC%88tana%2Btanb%EF%BC%89%2F%EF%BC%88tana-t)
找不正确的一项 请简要说明理由!A cos(a+b)cos(a-b)=cos^2a-sin^2b(cosa的平方减sinb的平方)B sin(a+b)sin(a-b)=sin^2a-sin^2bC tan(a)+tanb/tan(a)-tanb=sin(a+b)/sin(a-b)C tan(a)+tanb/tan(a)-tanb改为C (tana+tanb)/(tana-t
找不正确的一项 请简要说明理由!
A cos(a+b)cos(a-b)=cos^2a-sin^2b(cosa的平方减sinb的平方)
B sin(a+b)sin(a-b)=sin^2a-sin^2b
C tan(a)+tanb/tan(a)-tanb=sin(a+b)/sin(a-b)
C tan(a)+tanb/tan(a)-tanb改为C (tana+tanb)/(tana-tanb)
找不正确的一项 请简要说明理由!A cos(a+b)cos(a-b)=cos^2a-sin^2b(cosa的平方减sinb的平方)B sin(a+b)sin(a-b)=sin^2a-sin^2bC tan(a)+tanb/tan(a)-tanb=sin(a+b)/sin(a-b)C tan(a)+tanb/tan(a)-tanb改为C (tana+tanb)/(tana-t
A是正确的,cos(a+b)cos(a-b)=(cos2a+cos2b)/2=cos^2a-sin^2b(cosa的平方减sinb的平方)
B是正确的,sin(a+b)sin(a-b)=(cos2b-cos2a)/2=(1-2^sin^2b-1+2sin^2a)/2=sin^2a-sin^2b
前两项用到了积化和差公式
原来的C是错误的,tan(a)+tanb/tan(a)-tanb=/sin(a+b)/sin(a-b),但你改过的答案后那本题没有错误答案
因为(tana+tanb)/(tana-tanb)=(sina/cosa+sinb/cosb)/(sina/cosa-sinb/cosb)=(sinacosb+cosasinb)/(sinacosb-cosasinb)=sin(a+b)/sin(a-b)
如果你非要选出一个答案,那只能是C,因为运算过程中用到了除法,定义域拓宽了,可以说不能完全等同
C.
因为等号左右两边的定义域不一样,
虽然对应法则是一样的.
因该是a
选A吧
A、cos(a+b)cos(a-b)=(cos2a+cos2b)/2=cos^2a-sin^2b
B、sin(a+b)sin(a-b)=(sinacosb-cosasinb)(sinacosb+cosasinb)
=(sinacosb)^2-(cosasinb)^2
=sin^2a(1-sin^2b)-(1-sin^2a)sin^2b
=sin^2a-sin^2asin^2b)-sin^2b+sin^2asin^2b
=sin^2a-sin^2b
所以选C
A 是正确的,因为:当a=0时,cos(a+b)cos(a-b)=cosbcos(-b)=cos^2b ,而
A选项中等式的右边cos^2a-sin^2b也等于 1-sin^2b,即等于cos^2b 。因为cos^2b +sin^2b=1,所以成立。