Question

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

Answer 1

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^2b
sin^2b-2cos^2asin^2b+cos^4a=sin^2bcos^2b
sin^2b(1-cos^2b)-2cos^2asin^2b+cos^4a=0
sin^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
所以cosa=sinb
所以a+b=π/2