设f(sinx)=cos2x 1,求f(x)及f(cosx)不好意思,打错了,是f(sinx)=cos2x+1
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![设f(sinx)=cos2x 1,求f(x)及f(cosx)不好意思,打错了,是f(sinx)=cos2x+1](/uploads/image/z/8518230-54-0.jpg?t=%E8%AE%BEf%EF%BC%88sinx%EF%BC%89%3Dcos2x+1%2C%E6%B1%82f%EF%BC%88x%EF%BC%89%E5%8F%8Af%EF%BC%88cosx%EF%BC%89%E4%B8%8D%E5%A5%BD%E6%84%8F%E6%80%9D%EF%BC%8C%E6%89%93%E9%94%99%E4%BA%86%EF%BC%8C%E6%98%AFf%EF%BC%88sinx%EF%BC%89%3Dcos2x%2B1)
设f(sinx)=cos2x 1,求f(x)及f(cosx)不好意思,打错了,是f(sinx)=cos2x+1
设f(sinx)=cos2x 1,求f(x)及f(cosx)
不好意思,打错了,是f(sinx)=cos2x+1
设f(sinx)=cos2x 1,求f(x)及f(cosx)不好意思,打错了,是f(sinx)=cos2x+1
f(sinx)=cos2x +1 = 1-2(sinx)^2 +1 =2-2(sinx)^2
所以
f(x)=2-2x^2
f(cosx)=2-2(cosx)^2= - (2(cosx)^2-1) +1= 1- cos2x
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