已知:sin^4x/a+cos^4x/b=1/(a+b) (a>0,b>0) 证明:对于任何正整数n都有sin^(2n)x/a^(n-1)+cos^(2n)x/b^(n-1)=1/(a+b)^(n-1)
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![已知:sin^4x/a+cos^4x/b=1/(a+b) (a>0,b>0) 证明:对于任何正整数n都有sin^(2n)x/a^(n-1)+cos^(2n)x/b^(n-1)=1/(a+b)^(n-1)](/uploads/image/z/5448623-23-3.jpg?t=%E5%B7%B2%E7%9F%A5%EF%BC%9Asin%5E4x%2Fa%2Bcos%5E4x%2Fb%3D1%2F%28a%2Bb%29+%28a%3E0%2Cb%3E0%29+%E8%AF%81%E6%98%8E%EF%BC%9A%E5%AF%B9%E4%BA%8E%E4%BB%BB%E4%BD%95%E6%AD%A3%E6%95%B4%E6%95%B0n%E9%83%BD%E6%9C%89sin%5E%282n%29x%2Fa%5E%28n-1%29%2Bcos%5E%282n%29x%2Fb%5E%28n-1%29%3D1%2F%28a%2Bb%29%5E%28n-1%29)
已知:sin^4x/a+cos^4x/b=1/(a+b) (a>0,b>0) 证明:对于任何正整数n都有sin^(2n)x/a^(n-1)+cos^(2n)x/b^(n-1)=1/(a+b)^(n-1)
已知:sin^4x/a+cos^4x/b=1/(a+b) (a>0,b>0)
证明:对于任何正整数n都有sin^(2n)x/a^(n-1)+cos^(2n)x/b^(n-1)=1/(a+b)^(n-1)
已知:sin^4x/a+cos^4x/b=1/(a+b) (a>0,b>0) 证明:对于任何正整数n都有sin^(2n)x/a^(n-1)+cos^(2n)x/b^(n-1)=1/(a+b)^(n-1)
证明 由 sin^4x/a +cos4^x/b =1(a+b),
得( a+b)/a sin^4x +(a+b)/b cos^4x=1,
即 b/asin^4x+a/bcos^4x+sin4x+cos^4x=1.
又 sin^4x +cos^4x =(sin² x +cos² x ) ² -2 sin² xcos² x=1- 2 sin² xcos² x,
则 b/asin^4x+a/bcos^4x- 2 sin ² xcos ² x=0 ,
即 b/asin^4x+a/bcos^4x- 2[根号下 ( b/a)]sin² x [根号下(a/b)]cos² x=0 .则{ [根号下(b/a)]sin² x- [根号下(b/a)]cos² x) }²=0 ,
{[根号下( b/a)]sin² x - [根号下(a/b)]cos² x}² =0,
所以有sin² x = a/(a+,b).cos²x = b/(a+b).
带入不等式 易得.