已知α,β∈(0,π/2),且sin ^ 4α/cos ^ 2β+cos ^ 4α/sin ^ 2β=1求α+β的值

显示全部楼层 · 2011-10-31 17:35:13

两边同乘以sin^2bcos^2b得sin^4asin^2b+cos^4acos^2b=sin^2bcos^2b将sin^4a=(1-cos^2a)^2代入上式化简得（1-2cos^2a+cos^4a）sin^2b+cos^4a(1-sin^2b)=sin^2bcos^2bsin^2b-2cos^2asin^2b+cos^4a=sin^2bcos^2bsin^2b(1-cos^2b)-2cos^2asin^2b+cos^4a=0sin^4b-2cos^2asin^2b+cos^4a=0(sin^2b-cos^2a)^2=0(cosa+sinb)(cosa-sinb)=0因为a,b属于（0，π/2）所以cosa+sinb不等于0所...