已知a是锐角且cos^4a-sin^4a=3/5,求(sin^2a+3sinacosa-cos^2a)/(2sin^2a+cos^2a)的值
问题描述：
已知a是锐角且cos^4a-sin^4a=3/5,求(sin^2a+3sinacosa-cos^2a)/(2sin^2a+cos^2a)的值
答
cos^4a-sin^4a=3/5
(cos²a-sin²a)( cos²a+sin²a)=3/5
cos²a-sin²a=3/4
cos²a+sin²a=1
cos²a=7/8,sin²a=1/8
tan²a=1/7
tana=√7/7
(sin^2a+3sinacosa-cos^2a)/(2sin^2a+cos^2a) 同除以cos²a
=(tan²a+3tana-1)/(2tan²a+1)
=(1/7+3√7/7-1)/(2*1/7+1)
=(3√7-6)/9
=(√7-2)/3
答
cos^4a-sin^4a=3/5
(cos²a-sin²a)( cos²a+sin²a)=3/5
因为 cos²a+sin²a=1
所以cos²a-sin²a=3/5
cos²a=4/5,sin²a=1/5
cos²a*sin²a=4/25
a是锐角cosa*sina=2/5
所以(sin²a+3sinacosa-cos²a)/(2sin²a+cos²a)
=(1/5+6/5-4/5)/(2/5+4/5)=1/2