Question

化简 sin²α+sin²β-sin²αsin²β+cos²αcos²β

急求

Answer 1

原式=sin²α+（1-sin²α）sin²β+cos²αcos²β
=sin²α+cos²αsin²β+cos²αcos²β=sin²α+cos²α（sin²β+cos²β）=sin²α+cos²α=1

Answer 2

sin^2 a+sin^2 b-sin^2asin^2b+cos^2 acos^2b
=sin^2 a+sin^2 b+(cosacosb-sinasinb)(cosacosb+sinasinb)
=sin^2a+sin^2 b+cos(a+b)cos(a-b)
=1/2 *(1-cos2a)+1/2 *(1-cos2b)+cos(a+b) *cos(a-b)
=1-1/2(cos2a+cos2b)+cos(a+b)*cos(a-b)
=1-1/2*2cos(2a+2b)/2 *cos(2a-2b)/2 +cos(a+b)*cos(a-b)
=1

Answer 3

化简 sin²α+sin²β-sin²αsin²β+cos²αcos²β
sin²α＝x
cos²α=1-x
sin²β=y
cos²β=1-y

x+y-xy+(1-x)(1-y)
=1